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:
$pic=mysql_query("Select Rune, username FROM runeRunner where
(User_ID=3)0,$connection);
$xx=mysql_fetch_array($pic);
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,
Pablo
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