05-08-2009, 01:48 PM
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#11 (permalink)
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The Gregarious
Join Date: Feb 2009
Location: New York
Posts: 645
Thanks: 64
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Quote:
Originally Posted by Kay1021
Thanks
I did exactly what you were saying...i didn't change anything else...except now i get an error
at this line
PHP Code:
if(!in_array($nt[cat_name],$first_list)) {
it says
I've double checked all my { and everything....
I'm losing my mind...lol |
Ohh...hmm, I wonder if it's because of the array operator..
Let's try this instead...
PHP Code:
while($nt=mysql_fetch_array($result))
{
$kitty = $nt['cat_name'];
if(!in_array($kitty,$first_list)) {
echo "<option value=''>$kitty</option>";
}
}
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