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date/time calculations: calculating exact difference

P: n/a
Hi all,

I have been struggling with a problem all day, I have been unable to come
up with a working solution.

I want to write a function which takes 2 unix timestamps and calculates
the difference. I want it to return the difference in years, months, days,
hours, minutes and seconds (a complete summary). Keeping into account of
course that these are 2 real dates, I dont want it to work with 30.475 as
an average number of days in a month, I want it to work with the supplied
months/years.

The output ought to be: there is 0 years, 7 months, 12 days, 5 hours,
56 minutes and 51 seconds difference.
(I dont want the difference as in 1237438424 seconds but a complete
summary).

Any of my attempts have failed. It is easy to calculate the numbers of
seconds, minutes, hours and days (total only). But I cant get passed
calculating the number of months and years. Furthermore, any of the
libraries from the internet which I have tested seem to be flawed as well,
their results are not precise.

Can this not be done? Anyone?

Kind regards,

Hans Gruber
Jul 17 '05 #1
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4 Replies


P: n/a
Hans Gruber wrote:
Hi all,

I have been struggling with a problem all day, I have been unable to come
up with a working solution.

I want to write a function which takes 2 unix timestamps and calculates
the difference. I want it to return the difference in years, months, days,
hours, minutes and seconds (a complete summary). Keeping into account of
course that these are 2 real dates, I dont want it to work with 30.475 as
an average number of days in a month, I want it to work with the supplied
months/years.

The output ought to be: there is 0 years, 7 months, 12 days, 5 hours,
56 minutes and 51 seconds difference.
(I dont want the difference as in 1237438424 seconds but a complete
summary).


Look into the date functions. There are functions in there to tell you
how many days are in the current month. Even if you wanted to do this
by hand you could create an array with each month and how many days it
has and check for a leap year.

Then use the date function to work with these.... It will work, but to
get it in your exact format might be a bit of work.

Mike
Jul 17 '05 #2

P: n/a
On Sun, 15 May 2005 20:58:54 -0500, Mike Willbanks wrote:
Hans Gruber wrote:
Hi all,

I have been struggling with a problem all day, I have been unable to come
up with a working solution.

I want to write a function which takes 2 unix timestamps and calculates
the difference. I want it to return the difference in years, months, days,
hours, minutes and seconds (a complete summary). Keeping into account of
course that these are 2 real dates, I dont want it to work with 30.475 as
an average number of days in a month, I want it to work with the supplied
months/years.

The output ought to be: there is 0 years, 7 months, 12 days, 5 hours,
56 minutes and 51 seconds difference.
(I dont want the difference as in 1237438424 seconds but a complete
summary).


Look into the date functions. There are functions in there to tell you
how many days are in the current month. Even if you wanted to do this
by hand you could create an array with each month and how many days it
has and check for a leap year.

Then use the date function to work with these.... It will work, but to
get it in your exact format might be a bit of work.

Mike


Thanks Mike, I was actually thinking up something like that right before I
went to bed last night. Another day of hard work coming up ;)

Hans
Jul 17 '05 #3

P: n/a
I noticed that Message-ID: <pa****************************@nakatomi.de>
from Hans Gruber contained the following:
Thanks Mike, I was actually thinking up something like that right before I
went to bed last night. Another day of hard work coming up ;)


It's easy as you say, to convert seconds into minutes hours days and
weeks. But if you said months what would you mean? Often the secret to
coding something is the correct identification of the problem.

Let's say I want to work out how many months it is till my next birthday
(March 5th). What I'd probably do is ignore the month length difference
and count up. It's May 16th today and so it's 9 months till February
16th. Then it's 13 days till my birthday.

If that's the way you want to work it out then you have your logic. I'm
not going to code it but I think I'd be looking carefully at mktime() if
I was.

--
Geoff Berrow (put thecat out to email)
It's only Usenet, no one dies.
My opinions, not the committee's, mine.
Simple RFDs http://www.ckdog.co.uk/rfdmaker/
Jul 17 '05 #4

P: n/a
On Mon, 16 May 2005 07:54:48 +0100, Geoff Berrow
<bl******@ckdog.co.uk> wrote:
I noticed that Message-ID: <pa****************************@nakatomi.de>
from Hans Gruber contained the following:
Thanks Mike, I was actually thinking up something like that right before I
went to bed last night. Another day of hard work coming up ;)
It's easy as you say, to convert seconds into minutes hours days and
weeks. But if you said months what would you mean? Often the secret to
coding something is the correct identification of the problem.

Let's say I want to work out how many months it is till my next birthday
(March 5th). What I'd probably do is ignore the month length difference
and count up. It's May 16th today and so it's 9 months till February
16th. Then it's 13 days till my birthday.


If your birthday was 30 June and today was 31 May would that be 1
month or just 30 days until your birthday?

Although if you decided the answer was "1 month", be aware you'd get
the same answer on 30 May.

There isn't really a "correct" answer, but the OP should consider
these edge cases.

However, it depends what data represents and how the user will use it.
If long differences are shown to a lower resolution, this anomaly may
not matter. If the time difference is several months, the user may not
be interested in the value down to the minute and second.

For example:

DIFFERENCE :- WHAT TO SHOW 60 days :- months, days.

< 60 days :- days only.
< 7 days :- days, hours.
< 1 day :- hours, minutes.
< 6 hours :- hours, minutes, seconds.

--
David ( @priz.co.uk )
Jul 17 '05 #5

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