[thelist] PHP help with a loop and array

[Jeremy Ashcraft]

try changing 'echo $picresult[$i]' to 'echo $picresult[0]'

Since you are only returning one thing from your query, only the first element of
the array is going to contain any data.  After $i>0, you are trying to display
blank array fields.

If you are trying to put all results into the $picresult array, try
$picresult[$i] = mysql_fetch_array($picsql);
or
array_push($picresult, mysql_fetch_array($picsql));

HTH

jeremy

CDitty wrote:

> Hello all.
>
> I seem to be having a little problem with a loop statement for a php
> script.  No matter what I try, it only returns data on the first loop.  The
> rest of the loops are ignored.
>
> for($i=0; $i < 6; $i++ )
> {
>         $picsql = mysql_query("select picture from scratch_odds where gameid =
> '$gameid' and id = $myotherpics[$i]");
>         echo "select picture from scratch_odds where gameid = '$gameid' and id =
> $myotherpics[$i]<br>";
>         $picresult = mysql_fetch_array($picsql);
>         echo $picresult[$i] . " - Result<br>";
>         echo $i . " - Counter<br>";
> }
>
> This is a copy of what is echoed out.  As you can see below, all the
> variables are set.
>
> select picture from scratch_odds where gameid = '1' and id = 6
> http://www.redhotsweeps.com/scratchnwin/games/15/331_11_2.gif - Result
> 0 - Counter
> select picture from scratch_odds where gameid = '1' and id = 1
> - Result
> 1 - Counter
> select picture from scratch_odds where gameid = '1' and id = 5
> - Result
> 2 - Counter
> select picture from scratch_odds where gameid = '1' and id = 2
> - Result
> 3 - Counter
> select picture from scratch_odds where gameid = '1' and id = 3
> - Result
> 4 - Counter
> select picture from scratch_odds where gameid = '1' and id = 4
> - Result
> 5 - Counter
>
> Any ideas on what is wrong?
>
> Thanks
>
> Chris
>
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