I need a SQL...

[Bjørn T Johansen]

I need to write a SQL that calculates the interval between a start time
and a stop time. This is the easy part. The problem is that I only have
the time part, i.e. no date, so how can I be sure to also calculate the
interval if the start time is before midnight and the stop time is after
midnight?
Regards,

BTJ

--
-----------------------------------------------------------------------------------------------
Bjørn T Johansen (BSc,MNIF)
Executive Manager
bt*@havleik.no Havleik Consulting
Phone : [phone] Conradisvei 4
Fax : [phone] N-1338 Sandvika
Cellular : [phone] http://www.havleik.no
-----------------------------------------------------------------------------------------------
"The stickers on the side of the box said "Supported Platforms: Windows
98, Windows NT 4.0,
Windows 2000 or better", so clearly Linux was a supported platform."
-----------------------------------------------------------------------------------------------

[Mattias Kregert]

Solution:

SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff FROM mytable;

/Mattias

----- Original Message -----
From: Bjørn T Johansen
To: Andrew L. Gould
Cc: PostgreSQL general list
Sent: Thursday, September 11, 2003 2:12 PM
Subject: Re: [GENERAL] I need a SQL...
On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:
On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:

I need to write a SQL that calculates the interval between a start time
and a stop time. This is the easy part. The problem is that I only have
the time part, i.e. no date, so how can I be sure to also calculate the
interval if the start time is before midnight and the stop time is after
midnight?
Regards,

BTJ

If the activity or period you are measuring can equal or exceed 12 hours, you
won't be able to calculate it reliably without a start date and a stop date..
If the periods are always less than 12 hours (and you assume all the data is
good), then stop times that are less than start times would indicate an
intervening midnight.

The dates do not have to be in the same fields as the times, since you can add
date and time data to create a timestamp for datetime calculations:

(stop_date + stop_time) - (start_date + start_time)

Best of luck,

Andrew Gould

Yes, the period can exceed 12 hours, so what you are saying is that this isnot possible to solve
without the date part? I can write this logic in my business logic but I was hoping to
solve this in my database layer...
BTJ

[Andrew L. Gould]

On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:

I need to write a SQL that calculates the interval between a start time
and a stop time. This is the easy part. The problem is that I only have
the time part, i.e. no date, so how can I be sure to also calculate the
interval if the start time is before midnight and the stop time is after
midnight?
Regards,

BTJ

If the activity or period you are measuring can equal or exceed 12 hours, you
won't be able to calculate it reliably without a start date and a stop date.
If the periods are always less than 12 hours (and you assume all the data is
good), then stop times that are less than start times would indicate an
intervening midnight.

The dates do not have to be in the same fields as the times, since you can add
date and time data to create a timestamp for datetime calculations:

(stop_date + stop_time) - (start_date + start_time)

Best of luck,

Andrew Gould

---------------------------(end of broadcast)---------------------------
TIP 8: explain analyze is your friend

[Bjørn T Johansen]

Well, it's close... :)

But it looks like the case doesn't work..
If I run your sql, the timediff is negative.

But if I run this:
SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
the timediff has correct value..

Do you see any error in the case, cause I don't?
BTJ

On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:

Solution:

SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
FROM mytable;

/Mattias

----- Original Message -----
From: Bjørn T Johansen
To: Andrew L. Gould
Cc: PostgreSQL general list
Sent: Thursday, September 11, 2003 2:12 PM
Subject: Re: [GENERAL] I need a SQL...

On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:

> On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:

> > I need to write a SQL that calculates the interval between a start time
> > and a stop time. This is the easy part. The problem is that I only have
> > the time part, i.e. no date, so how can I be sure to also calculate the
> > interval if the start time is before midnight and the stop time is after
> > midnight?
> >
> >
> > Regards,
> >
> > BTJ

>
> If the activity or period you are measuring can equal or exceed 12 hours, you
> won't be able to calculate it reliably without a start date and a stop date.
> If the periods are always less than 12 hours (and you assume all the data is
> good), then stop times that are less than start times would indicate an
> intervening midnight.
>
> The dates do not have to be in the same fields as the times, since you can add
> date and time data to create a timestamp for datetime calculations:
>
> (stop_date + stop_time) - (start_date + start_time)
>
> Best of luck,
>
> Andrew Gould

Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
without the date part? I can write this logic in my business logic but I was hoping to
solve this in my database layer...
BTJ

---------------------------(end of broadcast)---------------------------
TIP 4: Don't 'kill -9' the postmaster

[Bjørn T Johansen]

On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:

On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:

I need to write a SQL that calculates the interval between a start time
and a stop time. This is the easy part. The problem is that I only have
the time part, i.e. no date, so how can I be sure to also calculate the
interval if the start time is before midnight and the stop time is after
midnight?
Regards,

BTJ

If the activity or period you are measuring can equal or exceed 12 hours, you
won't be able to calculate it reliably without a start date and a stop date.
If the periods are always less than 12 hours (and you assume all the data is
good), then stop times that are less than start times would indicate an
intervening midnight.

The dates do not have to be in the same fields as the times, since you can add
date and time data to create a timestamp for datetime calculations:

(stop_date + stop_time) - (start_date + start_time)

Best of luck,

Andrew Gould

Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
without the date part? I can write this logic in my business logic but I was hoping to
solve this in my database layer...
BTJ

[Mattias Kregert]

When i run it, it works as intended (on pg 7.3.3). Which version do you use?

Are you absolutely sure you copied it exactly? You typed in '>=' and not '=', right?

/Mattias

----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>
To: "Mattias Kregert" <ma*****@kregert.se>
Cc: <pg***********@postgresql.org>
Sent: Thursday, September 11, 2003 3:02 PM
Subject: Re: [GENERAL] I need a SQL...

Well, it's close... :)

But it looks like the case doesn't work..
If I run your sql, the timediff is negative.

But if I run this:
SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
the timediff has correct value..

Do you see any error in the case, cause I don't?
BTJ

On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:

Solution:

SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
FROM mytable;

/Mattias

----- Original Message -----
From: Bjørn T Johansen
To: Andrew L. Gould
Cc: PostgreSQL general list
Sent: Thursday, September 11, 2003 2:12 PM
Subject: Re: [GENERAL] I need a SQL...

On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:

> On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:
> > I need to write a SQL that calculates the interval between a start time
> > and a stop time. This is the easy part. The problem is that I only have
> > the time part, i.e. no date, so how can I be sure to also calculate the
> > interval if the start time is before midnight and the stop time is after
> > midnight?
> >
> >
> > Regards,
> >
> > BTJ
>
> If the activity or period you are measuring can equal or exceed 12 hours, you
> won't be able to calculate it reliably without a start date and a stop date.
> If the periods are always less than 12 hours (and you assume all the data is
> good), then stop times that are less than start times would indicate an
> intervening midnight.
>
> The dates do not have to be in the same fields as the times, since you can add
> date and time data to create a timestamp for datetime calculations:
>
> (stop_date + stop_time) - (start_date + start_time)
>
> Best of luck,
>
> Andrew Gould

Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
without the date part? I can write this logic in my business logic but I was hoping to
solve this in my database layer...
BTJ

---------------------------(end of broadcast)---------------------------
TIP 4: Don't 'kill -9' the postmaster

---------------------------(end of broadcast)---------------------------
TIP 6: Have you searched our list archives?

http://archives.postgresql.org

[Mattias Kregert]

dsfdfd
----- Original Message -----
From: Bjørn T Johansen
To: Andrew L. Gould
Cc: PostgreSQL general list
Sent: Thursday, September 11, 2003 2:12 PM
Subject: Re: [GENERAL] I need a SQL...
On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:

Yes, the period can exceed 12 hours, so what you are saying is that this isnot possible to solve
without the date part? I can write this logic in my business logic but I was hoping to
solve this in my database layer...
BTJdsfdf

[Bjørn T Johansen]

Yes, I am sure, I just use copy-and-paste and I have double checked....
I am running on 7.3.4 but that shouldn't make any difference?

BTJ

On Thu, 2003-09-11 at 14:56, Mattias Kregert wrote:

When i run it, it works as intended (on pg 7.3.3). Which version do you use?

Are you absolutely sure you copied it exactly? You typed in '>=' and not '=', right?

/Mattias

----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>
To: "Mattias Kregert" <ma*****@kregert.se>
Cc: <pg***********@postgresql.org>
Sent: Thursday, September 11, 2003 3:02 PM
Subject: Re: [GENERAL] I need a SQL...

Well, it's close... :)

But it looks like the case doesn't work..
If I run your sql, the timediff is negative.

But if I run this:
SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
the timediff has correct value..

Do you see any error in the case, cause I don't?
BTJ

On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:

Solution:

SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
FROM mytable;

/Mattias

----- Original Message -----
From: Bjørn T Johansen
To: Andrew L. Gould
Cc: PostgreSQL general list
Sent: Thursday, September 11, 2003 2:12 PM
Subject: Re: [GENERAL] I need a SQL...

On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:
> On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:
> > I need to write a SQL that calculates the interval between a start time
> > and a stop time. This is the easy part. The problem is that I only have
> > the time part, i.e. no date, so how can I be sure to also calculate the
> > interval if the start time is before midnight and the stop time is after
> > midnight?
> >
> >
> > Regards,
> >
> > BTJ
>
> If the activity or period you are measuring can equal or exceed 12 hours, you
> won't be able to calculate it reliably without a start date and a stop date.
> If the periods are always less than 12 hours (and you assume all the data is
> good), then stop times that are less than start times would indicate an
> intervening midnight.
>
> The dates do not have to be in the same fields as the times, since you can add
> date and time data to create a timestamp for datetime calculations:
>
> (stop_date + stop_time) - (start_date + start_time)
>
> Best of luck,
>
> Andrew Gould

Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
without the date part? I can write this logic in my business logic but I was hoping to
solve this in my database layer...
BTJ

---------------------------(end of broadcast)---------------------------
TIP 4: Don't 'kill -9' the postmaster

--
-----------------------------------------------------------------------------------------------
Bjørn T Johansen (BSc,MNIF)
Executive Manager
bt*@havleik.no Havleik Consulting
Phone : [phone] Conradisvei 4
Fax : [phone] N-1338 Sandvika
Cellular : [phone] http://www.havleik.no
-----------------------------------------------------------------------------------------------
"The stickers on the side of the box said "Supported Platforms: Windows
98, Windows NT 4.0,
Windows 2000 or better", so clearly Linux was a supported platform."
-----------------------------------------------------------------------------------------------

---------------------------(end of broadcast)---------------------------
TIP 7: don't forget to increase your free space map settings

[Mattias Kregert]

Solution:

SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff FROM mytable;

/Mattias

----- Original Message -----
From: Bjørn T Johansen
To: Andrew L. Gould
Cc: PostgreSQL general list
Sent: Thursday, September 11, 2003 2:12 PM
Subject: Re: [GENERAL] I need a SQL...
On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:
On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:

I need to write a SQL that calculates the interval between a start time
and a stop time. This is the easy part. The problem is that I only have
the time part, i.e. no date, so how can I be sure to also calculate the
interval if the start time is before midnight and the stop time is after
midnight?
Regards,

BTJ

If the activity or period you are measuring can equal or exceed 12 hours, you
won't be able to calculate it reliably without a start date and a stop date..
If the periods are always less than 12 hours (and you assume all the data is
good), then stop times that are less than start times would indicate an
intervening midnight.

The dates do not have to be in the same fields as the times, since you can add
date and time data to create a timestamp for datetime calculations:

(stop_date + stop_time) - (start_date + start_time)

Best of luck,

Andrew Gould

Yes, the period can exceed 12 hours, so what you are saying is that this isnot possible to solve
without the date part? I can write this logic in my business logic but I was hoping to
solve this in my database layer...
BTJ

[Bjørn T Johansen]

Well, it's close... :)

But it looks like the case doesn't work..
If I run your sql, the timediff is negative.

But if I run this:
SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
the timediff has correct value..

Do you see any error in the case, cause I don't?
BTJ

On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:

Solution:

SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
FROM mytable;

/Mattias

----- Original Message -----
From: Bjørn T Johansen
To: Andrew L. Gould
Cc: PostgreSQL general list
Sent: Thursday, September 11, 2003 2:12 PM
Subject: Re: [GENERAL] I need a SQL...

On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:

> On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:

> > I need to write a SQL that calculates the interval between a start time
> > and a stop time. This is the easy part. The problem is that I only have
> > the time part, i.e. no date, so how can I be sure to also calculate the
> > interval if the start time is before midnight and the stop time is after
> > midnight?
> >
> >
> > Regards,
> >
> > BTJ

>
> If the activity or period you are measuring can equal or exceed 12 hours, you
> won't be able to calculate it reliably without a start date and a stop date.
> If the periods are always less than 12 hours (and you assume all the data is
> good), then stop times that are less than start times would indicate an
> intervening midnight.
>
> The dates do not have to be in the same fields as the times, since you can add
> date and time data to create a timestamp for datetime calculations:
>
> (stop_date + stop_time) - (start_date + start_time)
>
> Best of luck,
>
> Andrew Gould

Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
without the date part? I can write this logic in my business logic but I was hoping to
solve this in my database layer...
BTJ

---------------------------(end of broadcast)---------------------------
TIP 4: Don't 'kill -9' the postmaster

[Mattias Kregert]

When i run it, it works as intended (on pg 7.3.3). Which version do you use?

Are you absolutely sure you copied it exactly? You typed in '>=' and not '=', right?

/Mattias

----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>
To: "Mattias Kregert" <ma*****@kregert.se>
Cc: <pg***********@postgresql.org>
Sent: Thursday, September 11, 2003 3:02 PM
Subject: Re: [GENERAL] I need a SQL...

Well, it's close... :)

But it looks like the case doesn't work..
If I run your sql, the timediff is negative.

But if I run this:
SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
the timediff has correct value..

Do you see any error in the case, cause I don't?
BTJ

On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:

Solution:

SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
FROM mytable;

/Mattias

----- Original Message -----
From: Bjørn T Johansen
To: Andrew L. Gould
Cc: PostgreSQL general list
Sent: Thursday, September 11, 2003 2:12 PM
Subject: Re: [GENERAL] I need a SQL...

On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:

> On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:
> > I need to write a SQL that calculates the interval between a start time
> > and a stop time. This is the easy part. The problem is that I only have
> > the time part, i.e. no date, so how can I be sure to also calculate the
> > interval if the start time is before midnight and the stop time is after
> > midnight?
> >
> >
> > Regards,
> >
> > BTJ
>
> If the activity or period you are measuring can equal or exceed 12 hours, you
> won't be able to calculate it reliably without a start date and a stop date.
> If the periods are always less than 12 hours (and you assume all the data is
> good), then stop times that are less than start times would indicate an
> intervening midnight.
>
> The dates do not have to be in the same fields as the times, since you can add
> date and time data to create a timestamp for datetime calculations:
>
> (stop_date + stop_time) - (start_date + start_time)
>
> Best of luck,
>
> Andrew Gould

Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
without the date part? I can write this logic in my business logic but I was hoping to
solve this in my database layer...
BTJ

---------------------------(end of broadcast)---------------------------
TIP 4: Don't 'kill -9' the postmaster

---------------------------(end of broadcast)---------------------------
TIP 6: Have you searched our list archives?

http://archives.postgresql.org

[Bjørn T Johansen]

Yes, I am sure, I just use copy-and-paste and I have double checked....
I am running on 7.3.4 but that shouldn't make any difference?

BTJ

On Thu, 2003-09-11 at 14:56, Mattias Kregert wrote:

When i run it, it works as intended (on pg 7.3.3). Which version do you use?

Are you absolutely sure you copied it exactly? You typed in '>=' and not '=', right?

/Mattias

----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>
To: "Mattias Kregert" <ma*****@kregert.se>
Cc: <pg***********@postgresql.org>
Sent: Thursday, September 11, 2003 3:02 PM
Subject: Re: [GENERAL] I need a SQL...

Well, it's close... :)

But it looks like the case doesn't work..
If I run your sql, the timediff is negative.

But if I run this:
SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
the timediff has correct value..

Do you see any error in the case, cause I don't?
BTJ

On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:

Solution:

SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
FROM mytable;

/Mattias

----- Original Message -----
From: Bjørn T Johansen
To: Andrew L. Gould
Cc: PostgreSQL general list
Sent: Thursday, September 11, 2003 2:12 PM
Subject: Re: [GENERAL] I need a SQL...

On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:
> On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:
> > I need to write a SQL that calculates the interval between a start time
> > and a stop time. This is the easy part. The problem is that I only have
> > the time part, i.e. no date, so how can I be sure to also calculate the
> > interval if the start time is before midnight and the stop time is after
> > midnight?
> >
> >
> > Regards,
> >
> > BTJ
>
> If the activity or period you are measuring can equal or exceed 12 hours, you
> won't be able to calculate it reliably without a start date and a stop date.
> If the periods are always less than 12 hours (and you assume all the data is
> good), then stop times that are less than start times would indicate an
> intervening midnight.
>
> The dates do not have to be in the same fields as the times, since you can add
> date and time data to create a timestamp for datetime calculations:
>
> (stop_date + stop_time) - (start_date + start_time)
>
> Best of luck,
>
> Andrew Gould

Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
without the date part? I can write this logic in my business logic but I was hoping to
solve this in my database layer...
BTJ

---------------------------(end of broadcast)---------------------------
TIP 4: Don't 'kill -9' the postmaster

--
-----------------------------------------------------------------------------------------------
Bjørn T Johansen (BSc,MNIF)
Executive Manager
bt*@havleik.no Havleik Consulting
Phone : [phone] Conradisvei 4
Fax : [phone] N-1338 Sandvika
Cellular : [phone] http://www.havleik.no
-----------------------------------------------------------------------------------------------
"The stickers on the side of the box said "Supported Platforms: Windows
98, Windows NT 4.0,
Windows 2000 or better", so clearly Linux was a supported platform."
-----------------------------------------------------------------------------------------------

---------------------------(end of broadcast)---------------------------
TIP 7: don't forget to increase your free space map settings

[Mattias Kregert]

Very strange indeed!

This is my output.
------------------------------------

Welcome to psql 7.3.3, the PostgreSQL interactive terminal.

Type: \copyright for distribution terms
\h for help with SQL commands
\? for help on internal slash commands
\g or terminate with semicolon to execute query
\q to quit

test=# create table mytable (starttime time, stoptime time);
CREATE TABLE
test=# insert into mytable values ('10:45', '22:30');
INSERT 103688 1
test=# insert into mytable values ('19:45', '04:30');
INSERT 103689 1
test=# insert into mytable values ('00:00', '00:00');
INSERT 103690 1
test=# insert into mytable values ('23:59', '00:01');
INSERT 103691 1
test=# insert into mytable values ('00:01', '23:59');
INSERT 103692 1
test=# select starttime,stoptime,(case when stoptime-starttime >= 0 then stoptime-starttime else stoptime-starttime+'24 hours' end) as timediff from mytable;
starttime | stoptime | timediff
-----------+----------+----------
10:45:00 | 22:30:00 | 11:45
19:45:00 | 04:30:00 | 08:45
00:00:00 | 00:00:00 | 00:00
23:59:00 | 00:01:00 | 00:02
00:01:00 | 23:59:00 | 23:58
(5 rows)

test=#
------------------------------------

As you see, it all works as it should. Can you do exactly the same and send me the complete output? If you get a different result, then it's time to send in a bug report...

/Mattias
----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>

Yes, I am sure, I just use copy-and-paste and I have double checked....
I am running on 7.3.4 but that shouldn't make any difference?

BTJ

On Thu, 2003-09-11 at 14:56, Mattias Kregert wrote:

When i run it, it works as intended (on pg 7.3.3). Which version do you use?

Are you absolutely sure you copied it exactly? You typed in '>=' and not '=', right?

/Mattias

----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>
To: "Mattias Kregert" <ma*****@kregert.se>
Cc: <pg***********@postgresql.org>
Sent: Thursday, September 11, 2003 3:02 PM
Subject: Re: [GENERAL] I need a SQL...

Well, it's close... :)

But it looks like the case doesn't work..
If I run your sql, the timediff is negative.

But if I run this:
SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
the timediff has correct value..

Do you see any error in the case, cause I don't?
BTJ

On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:
> Solution:
>
> SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
> stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
> FROM mytable;
>
> /Mattias
>
> ----- Original Message -----
> From: Bjørn T Johansen
> To: Andrew L. Gould
> Cc: PostgreSQL general list
> Sent: Thursday, September 11, 2003 2:12 PM
> Subject: Re: [GENERAL] I need a SQL...
>
> On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:
> > On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:
> > > I need to write a SQL that calculates the interval between a start time
> > > and a stop time. This is the easy part. The problem is that I only have
> > > the time part, i.e. no date, so how can I be sure to also calculate the
> > > interval if the start time is before midnight and the stop time is after
> > > midnight?
> > >
> > >
> > > Regards,
> > >
> > > BTJ
> >
> > If the activity or period you are measuring can equal or exceed 12 hours, you
> > won't be able to calculate it reliably without a start date and a stop date.
> > If the periods are always less than 12 hours (and you assume all the data is
> > good), then stop times that are less than start times would indicate an
> > intervening midnight.
> >
> > The dates do not have to be in the same fields as the times, since you can add
> > date and time data to create a timestamp for datetime calculations:
> >
> > (stop_date + stop_time) - (start_date + start_time)
> >
> > Best of luck,
> >
> > Andrew Gould
>
>
>
> Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
> without the date part? I can write this logic in my business logic but I was hoping to
> solve this in my database layer...
>
>
> BTJ
---------------------------(end of broadcast)---------------------------
TIP 4: Don't 'kill -9' the postmaster

--
-----------------------------------------------------------------------------------------------
Bjørn T Johansen (BSc,MNIF)
Executive Manager
bt*@havleik.no Havleik Consulting
Phone : [phone] Conradisvei 4
Fax : [phone] N-1338 Sandvika
Cellular : [phone] http://www.havleik.no
-----------------------------------------------------------------------------------------------
"The stickers on the side of the box said "Supported Platforms: Windows
98, Windows NT 4.0,
Windows 2000 or better", so clearly Linux was a supported platform."
-----------------------------------------------------------------------------------------------

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[Nigel J. Andrews]

On 11 Sep 2003, Bjørn T Johansen wrote:

Yes, I am sure, I just use copy-and-paste and I have double checked....
I am running on 7.3.4 but that shouldn't make any difference?

BTJ

On Thu, 2003-09-11 at 14:56, Mattias Kregert wrote:

When i run it, it works as intended (on pg 7.3.3). Which version do you use?

Are you absolutely sure you copied it exactly? You typed in '>=' and not '=', right?

/Mattias

----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>

Well, it's close... :)

But it looks like the case doesn't work..
If I run your sql, the timediff is negative.

But if I run this:
SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
the timediff has correct value..

Do you see any error in the case, cause I don't?
BTJ

On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:
> Solution:
>
> SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
> stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
> FROM mytable;

...

Seems to work for me.

test=# select case when '22:03:21'::time - '10:34:01'::time >= 0 then '22:03:21'::time - '10:34:01'::time else '22:03:21'::time - '10:34:01'::time + '24 hours' end;
case
----------
11:29:20
(1 row)

test=# select version();
version
---------------------------------------------------------------
PostgreSQL 7.3.4 on i686-pc-linux-gnu, compiled by GCC 2.95.4
(1 row)
There must be something different you are doing.

However, is that really the correct result? What about intervals that cross
daylight saving changes? 24 hours won't cut it in that case and you can only
tell by storing the date not just the time of day.
--
Nigel J. Andrews
---------------------------(end of broadcast)---------------------------
TIP 3: if posting/reading through Usenet, please send an appropriate
subscribe-nomail command to ma*******@postgresql.org so that your
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[Mattias Kregert]

Very strange indeed!

This is my output.
------------------------------------

Welcome to psql 7.3.3, the PostgreSQL interactive terminal.

Type: \copyright for distribution terms
\h for help with SQL commands
\? for help on internal slash commands
\g or terminate with semicolon to execute query
\q to quit

test=# create table mytable (starttime time, stoptime time);
CREATE TABLE
test=# insert into mytable values ('10:45', '22:30');
INSERT 103688 1
test=# insert into mytable values ('19:45', '04:30');
INSERT 103689 1
test=# insert into mytable values ('00:00', '00:00');
INSERT 103690 1
test=# insert into mytable values ('23:59', '00:01');
INSERT 103691 1
test=# insert into mytable values ('00:01', '23:59');
INSERT 103692 1
test=# select starttime,stoptime,(case when stoptime-starttime >= 0 then stoptime-starttime else stoptime-starttime+'24 hours' end) as timediff from mytable;
starttime | stoptime | timediff
-----------+----------+----------
10:45:00 | 22:30:00 | 11:45
19:45:00 | 04:30:00 | 08:45
00:00:00 | 00:00:00 | 00:00
23:59:00 | 00:01:00 | 00:02
00:01:00 | 23:59:00 | 23:58
(5 rows)

test=#
------------------------------------

As you see, it all works as it should. Can you do exactly the same and send me the complete output? If you get a different result, then it's time to send in a bug report...

/Mattias
----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>

Yes, I am sure, I just use copy-and-paste and I have double checked....
I am running on 7.3.4 but that shouldn't make any difference?

BTJ

On Thu, 2003-09-11 at 14:56, Mattias Kregert wrote:

When i run it, it works as intended (on pg 7.3.3). Which version do you use?

Are you absolutely sure you copied it exactly? You typed in '>=' and not '=', right?

/Mattias

----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>
To: "Mattias Kregert" <ma*****@kregert.se>
Cc: <pg***********@postgresql.org>
Sent: Thursday, September 11, 2003 3:02 PM
Subject: Re: [GENERAL] I need a SQL...

Well, it's close... :)

But it looks like the case doesn't work..
If I run your sql, the timediff is negative.

But if I run this:
SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
the timediff has correct value..

Do you see any error in the case, cause I don't?
BTJ

On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:
> Solution:
>
> SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
> stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
> FROM mytable;
>
> /Mattias
>
> ----- Original Message -----
> From: Bjørn T Johansen
> To: Andrew L. Gould
> Cc: PostgreSQL general list
> Sent: Thursday, September 11, 2003 2:12 PM
> Subject: Re: [GENERAL] I need a SQL...
>
> On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:
> > On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:
> > > I need to write a SQL that calculates the interval between a start time
> > > and a stop time. This is the easy part. The problem is that I only have
> > > the time part, i.e. no date, so how can I be sure to also calculate the
> > > interval if the start time is before midnight and the stop time is after
> > > midnight?
> > >
> > >
> > > Regards,
> > >
> > > BTJ
> >
> > If the activity or period you are measuring can equal or exceed 12 hours, you
> > won't be able to calculate it reliably without a start date and a stop date.
> > If the periods are always less than 12 hours (and you assume all the data is
> > good), then stop times that are less than start times would indicate an
> > intervening midnight.
> >
> > The dates do not have to be in the same fields as the times, since you can add
> > date and time data to create a timestamp for datetime calculations:
> >
> > (stop_date + stop_time) - (start_date + start_time)
> >
> > Best of luck,
> >
> > Andrew Gould
>
>
>
> Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
> without the date part? I can write this logic in my business logic but I was hoping to
> solve this in my database layer...
>
>
> BTJ
---------------------------(end of broadcast)---------------------------
TIP 4: Don't 'kill -9' the postmaster

--
-----------------------------------------------------------------------------------------------
Bjørn T Johansen (BSc,MNIF)
Executive Manager
bt*@havleik.no Havleik Consulting
Phone : [phone] Conradisvei 4
Fax : [phone] N-1338 Sandvika
Cellular : [phone] http://www.havleik.no
-----------------------------------------------------------------------------------------------
"The stickers on the side of the box said "Supported Platforms: Windows
98, Windows NT 4.0,
Windows 2000 or better", so clearly Linux was a supported platform."
-----------------------------------------------------------------------------------------------

---------------------------(end of broadcast)---------------------------
TIP 7: don't forget to increase your free space map settings

---------------------------(end of broadcast)---------------------------
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[Nigel J. Andrews]

On 11 Sep 2003, Bjørn T Johansen wrote:

Yes, I am sure, I just use copy-and-paste and I have double checked....
I am running on 7.3.4 but that shouldn't make any difference?

BTJ

On Thu, 2003-09-11 at 14:56, Mattias Kregert wrote:

When i run it, it works as intended (on pg 7.3.3). Which version do you use?

Are you absolutely sure you copied it exactly? You typed in '>=' and not '=', right?

/Mattias

----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>

Well, it's close... :)

But it looks like the case doesn't work..
If I run your sql, the timediff is negative.

But if I run this:
SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
the timediff has correct value..

Do you see any error in the case, cause I don't?
BTJ

On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:
> Solution:
>
> SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
> stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
> FROM mytable;

...

Seems to work for me.

test=# select case when '22:03:21'::time - '10:34:01'::time >= 0 then '22:03:21'::time - '10:34:01'::time else '22:03:21'::time - '10:34:01'::time + '24 hours' end;
case
----------
11:29:20
(1 row)

test=# select version();
version
---------------------------------------------------------------
PostgreSQL 7.3.4 on i686-pc-linux-gnu, compiled by GCC 2.95.4
(1 row)
There must be something different you are doing.

However, is that really the correct result? What about intervals that cross
daylight saving changes? 24 hours won't cut it in that case and you can only
tell by storing the date not just the time of day.
--
Nigel J. Andrews
---------------------------(end of broadcast)---------------------------
TIP 3: if posting/reading through Usenet, please send an appropriate
subscribe-nomail command to ma*******@postgresql.org so that your
message can get through to the mailing list cleanly

[Bjørn T Johansen]

Well, here is my output..:

DT=# create table mytable (starttime time, stoptime time);
CREATE TABLE
DT=# insert into mytable values ('10:45', '22:30');
INSERT 20746 1
DT=# insert into mytable values ('19:45', '04:30');
INSERT 20747 1
DT=# insert into mytable values ('00:00', '00:00');
INSERT 20748 1
DT=# insert into mytable values ('23:59', '00:01');
INSERT 20749 1
DT=# insert into mytable values ('00:01', '23:59');
INSERT 20750 1
DT=# select starttime,stoptime,(case when stoptime-starttime >= 0 then
stoptime-starttime else stoptime-starttime+'24 hours' end) as timediff
from mytable;
starttime | stoptime | timediff
-----------+----------+----------
10:45:00 | 22:30:00 | 11:45
19:45:00 | 04:30:00 | -15:15
00:00:00 | 00:00:00 | 00:00
23:59:00 | 00:01:00 | -23:58
00:01:00 | 23:59:00 | 23:58
(5 rows)

DT=#

Strange....

On Thu, 2003-09-11 at 15:25, Mattias Kregert wrote:

Very strange indeed!

This is my output.
------------------------------------

Welcome to psql 7.3.3, the PostgreSQL interactive terminal.

Type: \copyright for distribution terms
\h for help with SQL commands
\? for help on internal slash commands
\g or terminate with semicolon to execute query
\q to quit

test=# create table mytable (starttime time, stoptime time);
CREATE TABLE
test=# insert into mytable values ('10:45', '22:30');
INSERT 103688 1
test=# insert into mytable values ('19:45', '04:30');
INSERT 103689 1
test=# insert into mytable values ('00:00', '00:00');
INSERT 103690 1
test=# insert into mytable values ('23:59', '00:01');
INSERT 103691 1
test=# insert into mytable values ('00:01', '23:59');
INSERT 103692 1
test=# select starttime,stoptime,(case when stoptime-starttime >= 0 then stoptime-starttime else stoptime-starttime+'24 hours' end) as timediff from mytable;
starttime | stoptime | timediff
-----------+----------+----------
10:45:00 | 22:30:00 | 11:45
19:45:00 | 04:30:00 | 08:45
00:00:00 | 00:00:00 | 00:00
23:59:00 | 00:01:00 | 00:02
00:01:00 | 23:59:00 | 23:58
(5 rows)

test=#
------------------------------------

As you see, it all works as it should. Can you do exactly the same and send me the complete output? If you get a different result, then it's time to send in a bug report...

/Mattias
----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>

Yes, I am sure, I just use copy-and-paste and I have double checked....
I am running on 7.3.4 but that shouldn't make any difference?

BTJ

On Thu, 2003-09-11 at 14:56, Mattias Kregert wrote:

When i run it, it works as intended (on pg 7.3.3). Which version do you use?

Are you absolutely sure you copied it exactly? You typed in '>=' and not '=', right?

/Mattias

----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>
To: "Mattias Kregert" <ma*****@kregert.se>
Cc: <pg***********@postgresql.org>
Sent: Thursday, September 11, 2003 3:02 PM
Subject: Re: [GENERAL] I need a SQL...
> Well, it's close... :)
>
> But it looks like the case doesn't work..
> If I run your sql, the timediff is negative.
>
> But if I run this:
> SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
> the timediff has correct value..
>
> Do you see any error in the case, cause I don't?
>
>
> BTJ
>
> On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:
> > Solution:
> >
> > SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
> > stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
> > FROM mytable;
> >
> > /Mattias
> >
> > ----- Original Message -----
> > From: Bjørn T Johansen
> > To: Andrew L. Gould
> > Cc: PostgreSQL general list
> > Sent: Thursday, September 11, 2003 2:12 PM
> > Subject: Re: [GENERAL] I need a SQL...
> >
> > On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:
> > > On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:
> > > > I need to write a SQL that calculates the interval between a start time
> > > > and a stop time. This is the easy part. The problem is that I only have
> > > > the time part, i.e. no date, so how can I be sure to also calculate the
> > > > interval if the start time is before midnight and the stop time is after
> > > > midnight?
> > > >
> > > >
> > > > Regards,
> > > >
> > > > BTJ
> > >
> > > If the activity or period you are measuring can equal or exceed 12 hours, you
> > > won't be able to calculate it reliably without a start date and a stop date.
> > > If the periods are always less than 12 hours (and you assume all the data is
> > > good), then stop times that are less than start times would indicate an
> > > intervening midnight.
> > >
> > > The dates do not have to be in the same fields as the times, since you can add
> > > date and time data to create a timestamp for datetime calculations:
> > >
> > > (stop_date + stop_time) - (start_date + start_time)
> > >
> > > Best of luck,
> > >
> > > Andrew Gould
> >
> >
> >
> > Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
> > without the date part? I can write this logic in my business logic but I was hoping to
> > solve this in my database layer...
> >
> >
> > BTJ
>
>
> ---------------------------(end of broadcast)---------------------------
> TIP 4: Don't 'kill -9' the postmaster
>

--
-----------------------------------------------------------------------------------------------
Bjørn T Johansen (BSc,MNIF)
Executive Manager
bt*@havleik.no Havleik Consulting
Phone : [phone] Conradisvei 4
Fax : [phone] N-1338 Sandvika
Cellular : [phone] http://www.havleik.no
-----------------------------------------------------------------------------------------------
"The stickers on the side of the box said "Supported Platforms: Windows
98, Windows NT 4.0,
Windows 2000 or better", so clearly Linux was a supported platform."
-----------------------------------------------------------------------------------------------

---------------------------(end of broadcast)---------------------------
TIP 7: don't forget to increase your free space map settings

---------------------------(end of broadcast)---------------------------
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(send "unregister YourEmailAddressHere" to ma*******@postgresql.org)

[Mattias Kregert]

This is too weird... and you are sure you haven't modified anything in pg_operator, used CREATE OPERATOR or something like that??

I think it's time for you to send in a bug report......

/Mattias

----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>

Well, here is my output..:

DT=# create table mytable (starttime time, stoptime time);
CREATE TABLE
DT=# insert into mytable values ('10:45', '22:30');
INSERT 20746 1
DT=# insert into mytable values ('19:45', '04:30');
INSERT 20747 1
DT=# insert into mytable values ('00:00', '00:00');
INSERT 20748 1
DT=# insert into mytable values ('23:59', '00:01');
INSERT 20749 1
DT=# insert into mytable values ('00:01', '23:59');
INSERT 20750 1
DT=# select starttime,stoptime,(case when stoptime-starttime >= 0 then
stoptime-starttime else stoptime-starttime+'24 hours' end) as timediff
from mytable;
starttime | stoptime | timediff
-----------+----------+----------
10:45:00 | 22:30:00 | 11:45
19:45:00 | 04:30:00 | -15:15
00:00:00 | 00:00:00 | 00:00
23:59:00 | 00:01:00 | -23:58
00:01:00 | 23:59:00 | 23:58
(5 rows)

DT=#

Strange....

On Thu, 2003-09-11 at 15:25, Mattias Kregert wrote:

Very strange indeed!

This is my output.
------------------------------------

Welcome to psql 7.3.3, the PostgreSQL interactive terminal.

Type: \copyright for distribution terms
\h for help with SQL commands
\? for help on internal slash commands
\g or terminate with semicolon to execute query
\q to quit

test=# create table mytable (starttime time, stoptime time);
CREATE TABLE
test=# insert into mytable values ('10:45', '22:30');
INSERT 103688 1
test=# insert into mytable values ('19:45', '04:30');
INSERT 103689 1
test=# insert into mytable values ('00:00', '00:00');
INSERT 103690 1
test=# insert into mytable values ('23:59', '00:01');
INSERT 103691 1
test=# insert into mytable values ('00:01', '23:59');
INSERT 103692 1
test=# select starttime,stoptime,(case when stoptime-starttime >= 0then stoptime-starttime else stoptime-starttime+'24 hours' end) as timediff from mytable;
starttime | stoptime | timediff
-----------+----------+----------
10:45:00 | 22:30:00 | 11:45
19:45:00 | 04:30:00 | 08:45
00:00:00 | 00:00:00 | 00:00
23:59:00 | 00:01:00 | 00:02
00:01:00 | 23:59:00 | 23:58
(5 rows)

test=#
------------------------------------

As you see, it all works as it should. Can you do exactly the same and send me the complete output? If you get a different result, then it's time to send in a bug report...

/Mattias


----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>

Yes, I am sure, I just use copy-and-paste and I have double checked.....
I am running on 7.3.4 but that shouldn't make any difference?

BTJ

On Thu, 2003-09-11 at 14:56, Mattias Kregert wrote:
> When i run it, it works as intended (on pg 7.3.3). Which version doyou use?
>
> Are you absolutely sure you copied it exactly? You typed in '>=' and not '=', right?
>
> /Mattias
>
> ----- Original Message -----
> From: "Bjørn T Johansen" <bt*@havleik.no>
> To: "Mattias Kregert" <ma*****@kregert.se>
> Cc: <pg***********@postgresql.org>
> Sent: Thursday, September 11, 2003 3:02 PM
> Subject: Re: [GENERAL] I need a SQL...
>
>
> > Well, it's close... :)
> >
> > But it looks like the case doesn't work..
> > If I run your sql, the timediff is negative.
> >
> > But if I run this:
> > SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
> > the timediff has correct value..
> >
> > Do you see any error in the case, cause I don't?
> >
> >
> > BTJ
> >
> > On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:
> > > Solution:
> > >
> > > SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
> > > stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
> > > FROM mytable;
> > >
> > > /Mattias
> > >
> > > ----- Original Message -----
> > > From: Bjørn T Johansen
> > > To: Andrew L. Gould
> > > Cc: PostgreSQL general list
> > > Sent: Thursday, September 11, 2003 2:12 PM
> > > Subject: Re: [GENERAL] I need a SQL...
> > >
> > > On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:
> > > > On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:
> > > > > I need to write a SQL that calculates the interval between a start time
> > > > > and a stop time. This is the easy part. The problemis that I only have
> > > > > the time part, i.e. no date, so how can I be sure to also calculate the
> > > > > interval if the start time is before midnight and the stop time is after
> > > > > midnight?
> > > > >
> > > > >
> > > > > Regards,
> > > > >
> > > > > BTJ
> > > >
> > > > If the activity or period you are measuring can equalor exceed 12 hours, you
> > > > won't be able to calculate it reliably without a start date and a stop date.
> > > > If the periods are always less than 12 hours (and youassume all the data is
> > > > good), then stop times that are less than start timeswould indicate an
> > > > intervening midnight.
> > > >
> > > > The dates do not have to be in the same fields as thetimes, since you can add
> > > > date and time data to create a timestamp for datetimecalculations:
> > > >
> > > > (stop_date + stop_time) - (start_date + start_time)
> > > >
> > > > Best of luck,
> > > >
> > > > Andrew Gould
> > >
> > >
> > >
> > > Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
> > > without the date part? I can write this logic in my business logic but I was hoping to
> > > solve this in my database layer...
> > >
> > >
> > > BTJ
> >
> >
> > ---------------------------(end of broadcast)---------------------------
> > TIP 4: Don't 'kill -9' the postmaster
> >
--
-----------------------------------------------------------------------------------------------
Bjørn T Johansen (BSc,MNIF)
Executive Manager
bt*@havleik.no Havleik Consulting
Phone : [phone] Conradisvei 4
Fax : [phone] N-1338 Sandvika
Cellular : [phone] http://www.havleik.no
-----------------------------------------------------------------------------------------------
"The stickers on the side of the box said "Supported Platforms: Windows
98, Windows NT 4.0,
Windows 2000 or better", so clearly Linux was a supported platform."
-----------------------------------------------------------------------------------------------



---------------------------(end of broadcast)---------------------------
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---------------------------(end of broadcast)---------------------------
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http://archives.postgresql.org

[Tom Lane]

=?ISO-8859-1?Q?Bj=F8rn?= T Johansen <bt*@havleik.no> writes:

Yes, I am sure, I just use copy-and-paste and I have double checked....
I am running on 7.3.4 but that shouldn't make any difference?

It'll probably work better if you quote the zero. Unquoted, you get
some weird textual comparison. Compare:

regression=# explain select * from time_tbl where f1-f1 >= 0;
QUERY PLAN
--------------------------------------------------------
Seq Scan on time_tbl (cost=0.00..1.14 rows=3 width=8)
Filter: (((f1 - f1))::text >= '0'::text)
(2 rows)

regression=# explain select * from time_tbl where f1-f1 >= '0';
QUERY PLAN
--------------------------------------------------------
Seq Scan on time_tbl (cost=0.00..1.12 rows=3 width=8)
Filter: ((f1 - f1) >= '00:00'::interval)
(2 rows)

In "C" locale, the textual comparison accidentally manages to give
the desired answers, but in other locales it would not.

(Just another example of why implicit coercions to text are evil.)

regards, tom lane

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[Bjørn T Johansen]

Well, here is my output..:

DT=# create table mytable (starttime time, stoptime time);
CREATE TABLE
DT=# insert into mytable values ('10:45', '22:30');
INSERT 20746 1
DT=# insert into mytable values ('19:45', '04:30');
INSERT 20747 1
DT=# insert into mytable values ('00:00', '00:00');
INSERT 20748 1
DT=# insert into mytable values ('23:59', '00:01');
INSERT 20749 1
DT=# insert into mytable values ('00:01', '23:59');
INSERT 20750 1
DT=# select starttime,stoptime,(case when stoptime-starttime >= 0 then
stoptime-starttime else stoptime-starttime+'24 hours' end) as timediff
from mytable;
starttime | stoptime | timediff
-----------+----------+----------
10:45:00 | 22:30:00 | 11:45
19:45:00 | 04:30:00 | -15:15
00:00:00 | 00:00:00 | 00:00
23:59:00 | 00:01:00 | -23:58
00:01:00 | 23:59:00 | 23:58
(5 rows)

DT=#

Strange....

On Thu, 2003-09-11 at 15:25, Mattias Kregert wrote:

Very strange indeed!

This is my output.
------------------------------------

Welcome to psql 7.3.3, the PostgreSQL interactive terminal.

Type: \copyright for distribution terms
\h for help with SQL commands
\? for help on internal slash commands
\g or terminate with semicolon to execute query
\q to quit

test=# create table mytable (starttime time, stoptime time);
CREATE TABLE
test=# insert into mytable values ('10:45', '22:30');
INSERT 103688 1
test=# insert into mytable values ('19:45', '04:30');
INSERT 103689 1
test=# insert into mytable values ('00:00', '00:00');
INSERT 103690 1
test=# insert into mytable values ('23:59', '00:01');
INSERT 103691 1
test=# insert into mytable values ('00:01', '23:59');
INSERT 103692 1
test=# select starttime,stoptime,(case when stoptime-starttime >= 0 then stoptime-starttime else stoptime-starttime+'24 hours' end) as timediff from mytable;
starttime | stoptime | timediff
-----------+----------+----------
10:45:00 | 22:30:00 | 11:45
19:45:00 | 04:30:00 | 08:45
00:00:00 | 00:00:00 | 00:00
23:59:00 | 00:01:00 | 00:02
00:01:00 | 23:59:00 | 23:58
(5 rows)

test=#
------------------------------------

As you see, it all works as it should. Can you do exactly the same and send me the complete output? If you get a different result, then it's time to send in a bug report...

/Mattias
----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>

Yes, I am sure, I just use copy-and-paste and I have double checked....
I am running on 7.3.4 but that shouldn't make any difference?

BTJ

On Thu, 2003-09-11 at 14:56, Mattias Kregert wrote:

When i run it, it works as intended (on pg 7.3.3). Which version do you use?

Are you absolutely sure you copied it exactly? You typed in '>=' and not '=', right?

/Mattias

----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>
To: "Mattias Kregert" <ma*****@kregert.se>
Cc: <pg***********@postgresql.org>
Sent: Thursday, September 11, 2003 3:02 PM
Subject: Re: [GENERAL] I need a SQL...
> Well, it's close... :)
>
> But it looks like the case doesn't work..
> If I run your sql, the timediff is negative.
>
> But if I run this:
> SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
> the timediff has correct value..
>
> Do you see any error in the case, cause I don't?
>
>
> BTJ
>
> On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:
> > Solution:
> >
> > SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
> > stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
> > FROM mytable;
> >
> > /Mattias
> >
> > ----- Original Message -----
> > From: Bjørn T Johansen
> > To: Andrew L. Gould
> > Cc: PostgreSQL general list
> > Sent: Thursday, September 11, 2003 2:12 PM
> > Subject: Re: [GENERAL] I need a SQL...
> >
> > On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:
> > > On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:
> > > > I need to write a SQL that calculates the interval between a start time
> > > > and a stop time. This is the easy part. The problem is that I only have
> > > > the time part, i.e. no date, so how can I be sure to also calculate the
> > > > interval if the start time is before midnight and the stop time is after
> > > > midnight?
> > > >
> > > >
> > > > Regards,
> > > >
> > > > BTJ
> > >
> > > If the activity or period you are measuring can equal or exceed 12 hours, you
> > > won't be able to calculate it reliably without a start date and a stop date.
> > > If the periods are always less than 12 hours (and you assume all the data is
> > > good), then stop times that are less than start times would indicate an
> > > intervening midnight.
> > >
> > > The dates do not have to be in the same fields as the times, since you can add
> > > date and time data to create a timestamp for datetime calculations:
> > >
> > > (stop_date + stop_time) - (start_date + start_time)
> > >
> > > Best of luck,
> > >
> > > Andrew Gould
> >
> >
> >
> > Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
> > without the date part? I can write this logic in my business logic but I was hoping to
> > solve this in my database layer...
> >
> >
> > BTJ
>
>
> ---------------------------(end of broadcast)---------------------------
> TIP 4: Don't 'kill -9' the postmaster
>

--
-----------------------------------------------------------------------------------------------
Bjørn T Johansen (BSc,MNIF)
Executive Manager
bt*@havleik.no Havleik Consulting
Phone : [phone] Conradisvei 4
Fax : [phone] N-1338 Sandvika
Cellular : [phone] http://www.havleik.no
-----------------------------------------------------------------------------------------------
"The stickers on the side of the box said "Supported Platforms: Windows
98, Windows NT 4.0,
Windows 2000 or better", so clearly Linux was a supported platform."
-----------------------------------------------------------------------------------------------

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TIP 2: you can get off all lists at once with the unregister command
(send "unregister YourEmailAddressHere" to ma*******@postgresql.org)

[Mattias Kregert]

This is too weird... and you are sure you haven't modified anything in pg_operator, used CREATE OPERATOR or something like that??

I think it's time for you to send in a bug report......

/Mattias

----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>

Well, here is my output..:

DT=# create table mytable (starttime time, stoptime time);
CREATE TABLE
DT=# insert into mytable values ('10:45', '22:30');
INSERT 20746 1
DT=# insert into mytable values ('19:45', '04:30');
INSERT 20747 1
DT=# insert into mytable values ('00:00', '00:00');
INSERT 20748 1
DT=# insert into mytable values ('23:59', '00:01');
INSERT 20749 1
DT=# insert into mytable values ('00:01', '23:59');
INSERT 20750 1
DT=# select starttime,stoptime,(case when stoptime-starttime >= 0 then
stoptime-starttime else stoptime-starttime+'24 hours' end) as timediff
from mytable;
starttime | stoptime | timediff
-----------+----------+----------
10:45:00 | 22:30:00 | 11:45
19:45:00 | 04:30:00 | -15:15
00:00:00 | 00:00:00 | 00:00
23:59:00 | 00:01:00 | -23:58
00:01:00 | 23:59:00 | 23:58
(5 rows)

DT=#

Strange....

On Thu, 2003-09-11 at 15:25, Mattias Kregert wrote:

Very strange indeed!

This is my output.
------------------------------------

Welcome to psql 7.3.3, the PostgreSQL interactive terminal.

Type: \copyright for distribution terms
\h for help with SQL commands
\? for help on internal slash commands
\g or terminate with semicolon to execute query
\q to quit

test=# create table mytable (starttime time, stoptime time);
CREATE TABLE
test=# insert into mytable values ('10:45', '22:30');
INSERT 103688 1
test=# insert into mytable values ('19:45', '04:30');
INSERT 103689 1
test=# insert into mytable values ('00:00', '00:00');
INSERT 103690 1
test=# insert into mytable values ('23:59', '00:01');
INSERT 103691 1
test=# insert into mytable values ('00:01', '23:59');
INSERT 103692 1
test=# select starttime,stoptime,(case when stoptime-starttime >= 0then stoptime-starttime else stoptime-starttime+'24 hours' end) as timediff from mytable;
starttime | stoptime | timediff
-----------+----------+----------
10:45:00 | 22:30:00 | 11:45
19:45:00 | 04:30:00 | 08:45
00:00:00 | 00:00:00 | 00:00
23:59:00 | 00:01:00 | 00:02
00:01:00 | 23:59:00 | 23:58
(5 rows)

test=#
------------------------------------

As you see, it all works as it should. Can you do exactly the same and send me the complete output? If you get a different result, then it's time to send in a bug report...

/Mattias


----- Original Message -----
From: "Bjørn T Johansen" <bt*@havleik.no>

Yes, I am sure, I just use copy-and-paste and I have double checked.....
I am running on 7.3.4 but that shouldn't make any difference?

BTJ

On Thu, 2003-09-11 at 14:56, Mattias Kregert wrote:
> When i run it, it works as intended (on pg 7.3.3). Which version doyou use?
>
> Are you absolutely sure you copied it exactly? You typed in '>=' and not '=', right?
>
> /Mattias
>
> ----- Original Message -----
> From: "Bjørn T Johansen" <bt*@havleik.no>
> To: "Mattias Kregert" <ma*****@kregert.se>
> Cc: <pg***********@postgresql.org>
> Sent: Thursday, September 11, 2003 3:02 PM
> Subject: Re: [GENERAL] I need a SQL...
>
>
> > Well, it's close... :)
> >
> > But it looks like the case doesn't work..
> > If I run your sql, the timediff is negative.
> >
> > But if I run this:
> > SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
> > the timediff has correct value..
> >
> > Do you see any error in the case, cause I don't?
> >
> >
> > BTJ
> >
> > On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:
> > > Solution:
> > >
> > > SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
> > > stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
> > > FROM mytable;
> > >
> > > /Mattias
> > >
> > > ----- Original Message -----
> > > From: Bjørn T Johansen
> > > To: Andrew L. Gould
> > > Cc: PostgreSQL general list
> > > Sent: Thursday, September 11, 2003 2:12 PM
> > > Subject: Re: [GENERAL] I need a SQL...
> > >
> > > On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:
> > > > On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:
> > > > > I need to write a SQL that calculates the interval between a start time
> > > > > and a stop time. This is the easy part. The problemis that I only have
> > > > > the time part, i.e. no date, so how can I be sure to also calculate the
> > > > > interval if the start time is before midnight and the stop time is after
> > > > > midnight?
> > > > >
> > > > >
> > > > > Regards,
> > > > >
> > > > > BTJ
> > > >
> > > > If the activity or period you are measuring can equalor exceed 12 hours, you
> > > > won't be able to calculate it reliably without a start date and a stop date.
> > > > If the periods are always less than 12 hours (and youassume all the data is
> > > > good), then stop times that are less than start timeswould indicate an
> > > > intervening midnight.
> > > >
> > > > The dates do not have to be in the same fields as thetimes, since you can add
> > > > date and time data to create a timestamp for datetimecalculations:
> > > >
> > > > (stop_date + stop_time) - (start_date + start_time)
> > > >
> > > > Best of luck,
> > > >
> > > > Andrew Gould
> > >
> > >
> > >
> > > Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
> > > without the date part? I can write this logic in my business logic but I was hoping to
> > > solve this in my database layer...
> > >
> > >
> > > BTJ
> >
> >
> > ---------------------------(end of broadcast)---------------------------
> > TIP 4: Don't 'kill -9' the postmaster
> >
--
-----------------------------------------------------------------------------------------------
Bjørn T Johansen (BSc,MNIF)
Executive Manager
bt*@havleik.no Havleik Consulting
Phone : [phone] Conradisvei 4
Fax : [phone] N-1338 Sandvika
Cellular : [phone] http://www.havleik.no
-----------------------------------------------------------------------------------------------
"The stickers on the side of the box said "Supported Platforms: Windows
98, Windows NT 4.0,
Windows 2000 or better", so clearly Linux was a supported platform."
-----------------------------------------------------------------------------------------------



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TIP 7: don't forget to increase your free space map settings

---------------------------(end of broadcast)---------------------------
TIP 6: Have you searched our list archives?

http://archives.postgresql.org

[Bjørn T Johansen]

You are absoletely correct, quoting the zero did the trick...
And my database has been intialized with Norwegian local, so I am
guessing that that's reason....

Thanks to all fo you... :)
BTJ
On Thu, 2003-09-11 at 16:26, Tom Lane wrote:

=?ISO-8859-1?Q?Bj=F8rn?= T Johansen <bt*@havleik.no> writes:

Yes, I am sure, I just use copy-and-paste and I have double checked....
I am running on 7.3.4 but that shouldn't make any difference?

It'll probably work better if you quote the zero. Unquoted, you get
some weird textual comparison. Compare:

regression=# explain select * from time_tbl where f1-f1 >= 0;
QUERY PLAN
--------------------------------------------------------
Seq Scan on time_tbl (cost=0.00..1.14 rows=3 width=8)
Filter: (((f1 - f1))::text >= '0'::text)
(2 rows)

regression=# explain select * from time_tbl where f1-f1 >= '0';
QUERY PLAN
--------------------------------------------------------
Seq Scan on time_tbl (cost=0.00..1.12 rows=3 width=8)
Filter: ((f1 - f1) >= '00:00'::interval)
(2 rows)

In "C" locale, the textual comparison accidentally manages to give
the desired answers, but in other locales it would not.

(Just another example of why implicit coercions to text are evil.)

regards, tom lane

---------------------------(end of broadcast)---------------------------
TIP 6: Have you searched our list archives?

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[Tom Lane]

=?ISO-8859-1?Q?Bj=F8rn?= T Johansen <bt*@havleik.no> writes:

Yes, I am sure, I just use copy-and-paste and I have double checked....
I am running on 7.3.4 but that shouldn't make any difference?

It'll probably work better if you quote the zero. Unquoted, you get
some weird textual comparison. Compare:

regression=# explain select * from time_tbl where f1-f1 >= 0;
QUERY PLAN
--------------------------------------------------------
Seq Scan on time_tbl (cost=0.00..1.14 rows=3 width=8)
Filter: (((f1 - f1))::text >= '0'::text)
(2 rows)

regression=# explain select * from time_tbl where f1-f1 >= '0';
QUERY PLAN
--------------------------------------------------------
Seq Scan on time_tbl (cost=0.00..1.12 rows=3 width=8)
Filter: ((f1 - f1) >= '00:00'::interval)
(2 rows)

In "C" locale, the textual comparison accidentally manages to give
the desired answers, but in other locales it would not.

(Just another example of why implicit coercions to text are evil.)

regards, tom lane

---------------------------(end of broadcast)---------------------------
TIP 1: subscribe and unsubscribe commands go to ma*******@postgresql.org

[Bjørn T Johansen]

You are absoletely correct, quoting the zero did the trick...
And my database has been intialized with Norwegian local, so I am
guessing that that's reason....

Thanks to all fo you... :)
BTJ
On Thu, 2003-09-11 at 16:26, Tom Lane wrote:

=?ISO-8859-1?Q?Bj=F8rn?= T Johansen <bt*@havleik.no> writes:

Yes, I am sure, I just use copy-and-paste and I have double checked....
I am running on 7.3.4 but that shouldn't make any difference?

It'll probably work better if you quote the zero. Unquoted, you get
some weird textual comparison. Compare:

regression=# explain select * from time_tbl where f1-f1 >= 0;
QUERY PLAN
--------------------------------------------------------
Seq Scan on time_tbl (cost=0.00..1.14 rows=3 width=8)
Filter: (((f1 - f1))::text >= '0'::text)
(2 rows)

regression=# explain select * from time_tbl where f1-f1 >= '0';
QUERY PLAN
--------------------------------------------------------
Seq Scan on time_tbl (cost=0.00..1.12 rows=3 width=8)
Filter: ((f1 - f1) >= '00:00'::interval)
(2 rows)

In "C" locale, the textual comparison accidentally manages to give
the desired answers, but in other locales it would not.

(Just another example of why implicit coercions to text are evil.)

regards, tom lane

---------------------------(end of broadcast)---------------------------
TIP 6: Have you searched our list archives?

http://archives.postgresql.org

[Adam Kavan]

Yes, the period can exceed 12 hours, so what you are saying is that this
is not possible to solve without the date part? I can write this logic in
my business logic but I was hoping to solve this in my database layer... BTJ

How about SELECT CASE WHEN "StartTime" > "EndTime" THEN
'23:59:99.99999999999'::time - "StartTime" + "EndTime" ELSE
"EndTime"-"StartTime" END FROM "TimeTable";

--- Adam Kavan
--- ak****@cox.net

---------------------------(end of broadcast)---------------------------
TIP 9: the planner will ignore your desire to choose an index scan if your
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[Adam Kavan]

Yes, the period can exceed 12 hours, so what you are saying is that this
is not possible to solve without the date part? I can write this logic in
my business logic but I was hoping to solve this in my database layer... BTJ

How about SELECT CASE WHEN "StartTime" > "EndTime" THEN
'23:59:99.99999999999'::time - "StartTime" + "EndTime" ELSE
"EndTime"-"StartTime" END FROM "TimeTable";

--- Adam Kavan
--- ak****@cox.net

---------------------------(end of broadcast)---------------------------
TIP 9: the planner will ignore your desire to choose an index scan if your
joining column's datatypes do not match