Hello water_foul,

     what about this?

    If you do for the first time: $pic=mysql_fetch_array($pic) it will
    work, but the SECOND TIME that you do:
    $pic=mysql_fetch_array($pic).... it will fail, and thats because $pic
    is no longer a resource identificator... and you will get:

> > Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
> > result
    try this:
    $pic=mysql_query("Select Rune, username FROM runeRunner where

    do you follow me?...


> >> > why doesn't this work:
> >> > $pic=mysql_query('SELECT     Rune, username FROM         RuneRunner
> >> > RuneRunner_1 WHERE     (User_ID = 3)',$connection);
> >> > $pic=mysql_fetch_array($pic);    

P.S. if you need to trace, you can do  var_dump($pic);

Best regards,

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