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Query looking back

Diamondback
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Posts: n/a
#1: Jul 23 '05
I have two tables, WIDGETS and VERSIONS. The WIDGETS table has
descriptive information about the widgets while the VERSIONS table
contains IDs relating to different iterations of those widgets over the
years. The idea is that from any widget in the database you can look
forward and backward to see different versions of the widget through
the years. Here are the tables:

WIDGETS
widget_id
name
description
year

VERSIONS
v_id
new_id
old_id


Now my query gets me somewhere in the vicinity, but not where I want to
be.

select w.widget, w.widget_id, v.new_id, v.old_id
from widgets w, versions v
where w.widget_id = 712
and w.widget_id = v.new_id
order by w.widget;

.... and I get ...

WIDGET WIDGET_ID NEW_ID OLD_ID

Widget 1 712 712 535
Widget 1 712 712 287
Widget 1 712 712 1
Widget 1 712 712 205
Widget 1 712 712 458

What I would like is a list that shows the widget names related to
OLD_ID, rather than NEW_ID. But I'm not sure how to get there. Help
is appreciated.

Bart Van der Donck
Guest
 
Posts: n/a
#2: Jul 23 '05

re: Query looking back


Diamondback wrote:
[color=blue]
> I have two tables, WIDGETS and VERSIONS. The WIDGETS table has
> descriptive information about the widgets while the VERSIONS table
> contains IDs relating to different iterations of those widgets over the
> years. The idea is that from any widget in the database you can look
> forward and backward to see different versions of the widget through
> the years.
>
> Now my query gets me somewhere in the vicinity, but not where I want to
> be.
>
> select w.widget, w.widget_id, v.new_id, v.old_id
> from widgets w, versions v
> where w.widget_id = 712
> and w.widget_id = v.new_id
> order by w.widget;
>
> ... and I get ...
>
> WIDGET WIDGET_ID NEW_ID OLD_ID
>
> Widget 1 712 712 535
> Widget 1 712 712 287
> Widget 1 712 712 1
> Widget 1 712 712 205
> Widget 1 712 712 458
>
> What I would like is a list that shows the widget names related to
> OLD_ID, rather than NEW_ID. But I'm not sure how to get there. Help
> is appreciated.[/color]

Maybe there is a plain sql solution out there, but it will definitely
not be an easy one. Maybe it's possible with a join/subquery
combination or stuff in that direction.

For code transparency and general ease of use I'ld suggest an API
approach. You could execute 2 separate queries where the first one ties
each [v]![old_id] to its corresponding [w]![name]. The second query
could then use these stored values from the first query.

I set up the following test environment:

SELECT * FROM w;

---------|--------|-----------|----
widget_id|name |description|year
---------|--------|-----------|----
40|Widget 0| |
400|Widget 1| |
500|Widget 2| |
600|Widget 3| |
700|Widget 4| |
712|Widget 5| |
---------|--------|-----------|----

SELECT * FROM v;

----|------|------
v_id|new_id|old_id
----|------|------
1|712 |400
2|712 |500
3|712 |600
4|712 |700
5|713 |50
6|713 |80
7|714 |40
----|------|------

Now the API could first perform the following query:

SELECT v.old_id, w.name
FROM v
LEFT JOIN w
ON v.old_id = w.widget_id

And then tie those results to, let's say, a hash:

$widgetname{"$var_old_id"} = $var_name;

(supposed that $var_old_id is the first field and $var_name the second
one from your query result set)

Then, doing a query like:

SELECT v.old_ID AS name, w.widget_id, v.new_id, v.old_id
FROM w, v
WHERE w.widget_id =712
AND w.widget_id = v.new_id
ORDER BY w.name

will give you:

----|---------|------|------
name|widget_id|new_id|old_id
----|---------|------|------
400 |712 |712 |400
500 |712 |712 |500
600 |712 |712 |600
700 |712 |712 |700
----|---------|------|------

Finally, before showing results to screen, replace the first column by
its descriptive name that was stored in your hash.

I realise this all sounds rather cryptic. Here is a small perl program
(tested):

#!/usr/bin/perl
use DBI;
# uncomment the following 2 lines if CGI
# print "Content-Type: text/html\n\n<html>\n<pre>\n";
# use CGI::Carp qw(fatalsToBrowser);
$db = DBI->connect("DBI:mysql:NAME:HOST","USR", "PW");
$query = $db->prepare("SELECT
v.old_id, w.name
FROM v
LEFT JOIN w
ON v.old_id = w.widget_id
");
$query->execute;
$numrows = $query->rows;
while (@array = $query->fetchrow_array)
{
my ($var_old_id, $var_name)= @array;
$widgetname{"$var_old_id"} = $var_name;
}
$query->finish;
$db->disconnect;
print "w.name|w.widget_id|v.new_id|v.old_id|\n";
$db = DBI->connect("DBI:mysql:NAME:HOST","USR","PW");
$query = $db->prepare("SELECT
v.old_ID, w.widget_id,
v.new_id, v.old_id
FROM w, v
WHERE w.widget_id =712
AND w.widget_id = v.new_id
ORDER BY v.old_ID");
$query->execute;
$numrows = $query->rows;
while (@array = $query->fetchrow_array)
{
print "$widgetname{$array[0]}|";
print $array[1]."|";
print $array[2]."|";
print $array[3]."|\n";
}
$query->finish;
$db->disconnect;
# uncomment this line if CGI
print "</pre>\n</html>";

Output result (when ran as non-CGI):

w.name|w.widget_id|v.new_id|v.old_id|
Widget 1|712|712|400|
Widget 2|712|712|500|
Widget 3|712|712|600|
Widget 4|712|712|700|

The same script can be made in your favourite api language if you're
not a perl man.

Hope this helps.

--
Bart

Diamondback
Guest
 
Posts: n/a
#3: Jul 23 '05

re: Query looking back


You're right, it does sound cryptic. But I'm giving it a try ... I'll
let you know how it turns out.

Felix Geerinckx
Guest
 
Posts: n/a
#4: Jul 23 '05

re: Query looking back


On 26/05/2005, Diamondback wrote:
[color=blue]
> select w.widget, w.widget_id, v.new_id, v.old_id
> from widgets w, versions v
> where w.widget_id = 712
> and w.widget_id = v.new_id
> order by w.widget;
> ...
> What I would like is a list that shows the widget names related to
> OLD_ID, rather than NEW_ID. But I'm not sure how to get there. Help
> is appreciated.[/color]

Something like this?

select w.name, ow.name as oldname, w.widget_id, v.new_id, v.old_id
from widgets w, versions v, widgets ow
where w.widget_id = 712
and w.widget_id = v.new_id
and ow.widget_id = v.old_id
order by w.name;

--
felix
Diamondback
Guest
 
Posts: n/a
#5: Jul 23 '05

re: Query looking back


That did it exactly. Thank you.

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