joyce — Wed, 28 Nov 2001 02:21:49 +0000

oops..i forgot to declare the variable inside my function. that is why not working.

thanks..mark... )

Mark Hensler — Tue, 27 Nov 2001 07:48:54 +0000

The only thing that I see, is that the link in page A is missing the value for the category. I'm guessing that this is a typo?

<?php
$sql = \"SELECT * FROM cat_review WHERE cat_ID = '$cat_ID'\";
$result = mysql_query($sql);
if (!$result) {
    echo \"\n<!--QUERY ERROR:\" . mysql_error() . \"-->\n\";
    echo \"<!--QUERY:$sql-->\n\";
}
$record = mysql_fetch_array($result);
?>

Does this print anything useful?