"Brad Lipovsky" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> I get the following error:
>
>  $Query = "SELECT * from $TableName WHERE title LIKE "%$title%")";

If you tried 'echo $Query' right here, you wouldn't see anything.

Try 'echo "305" % "300";'

You begin to see the problem, right?  You used double quotes within your
string without escaping them (ie \") - so PHP treated it as the end of the
string.  You would have gotten an error message, except that % is PHP's
modular division operator - so it tried to convert the strings to numbers
and do modular division on them.

How 'bout
$query = "SELECT * FROM $table WHERE title LIKE '%$title%' ";




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