how about something like:
$count = 0;
$list = array();
$sql[]="select * from storyboards, ... ";
$sql[]="select * from storyboard2, ... ";
$sql[]="select * from storyboard3, ... ";
$sql[]="select * from storyboard4, ... ";
foreach ($sql as $thissql)
{ $result=MySQL_query($thissql,$db);
while($row = MySQL_fetch_array($result))
$list[$row["field_to_sort_by"] . "***" . ++$count] = $row;
}
ksort($list)
I haven't tried this precisely but the principle should work, and I have
tried this method of sorting arrays by an element of the member arrays.
Tim Ward
Senior Systems Engineer
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> -----Original Message-----
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
> Sent: 24 April 2001 21:42
> To: [EMAIL PROTECTED]
> Subject: getting rows from separate tables stacked in a single array ?
> (newbie, mysql)
>
>
> Hi.
>
> This will probably sound simple to all of you but I am
> hitting my head on
> a brick wall so far :-).
>
> I need to generate a list (sorted by descending time) of
> different objects
> referenced in multiple tables. All those objects share a key
> related to a
> project.
>
>
> This works splendidly from a single table :
>
>
>
>
> $sql="select * from storyboards, where spot_id = \"$spot_id\" order by
> date_posted desc";
> $result=MySQL_query($sql,$db);
> while($row=MySQL_fetch_array($result))
> {
> $qt_title = $row["title"];
> $qt_duration = $row["duration"];
> $qt_date_posted = $row["date_posted"]; //(timestamp)
> $qt_description = $row["description"];
> $qt_id = $row["quicktime_id"];
>
>
>
> }
>
> Is there a mysql query that would allow me to stack complete rows in
> multiple tables at once (It seems more elegant because I can sort them
> while extracting them) or is there a way in PHP to
> concatenate results in
> a single array and then sort them by time... ?
>
> I tried to use a join query, wich produced an invalid query error. (I
> think it tries to produce a table with merged data, not just
> a stack of
> rows).
>
>
> $sql4="select * from quicktimes, other_images, storyboards, where
> quicktimes.spot_id, other_images.spot_id, storyboards.spot_id =
> \"$spot_id\" order by date_posted desc";
>
> thank you for your help !
> nicolas
>
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