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Posted by Dejan on 04/02/06 15:37
ppl, tnx very much on your anwsers.
I have dealed with this problem and solved it.
I have added this error checking, buth the real error was tiping mistake.
I have named mysql table "gred", and neaded "grad". and then mysql sead
"mysql_num_rows(): supplied argument is not a valid MySQL result
resource "
tnx again
Dejan
<webramz@gmail.com> wrote in message
news:1144140487.542634.17920@t31g2000cwb.googlegroups.com...
> Write your code like this:
>
> $query = "....";
> $result = mysql_query($query);
> if($result) {
> $num_rows = mysql_num_rows($result); // notice the Result Resource
> here //
> } else {
> echo mysql_error();
> // die();
> }
>
>
> Best Regards,
>
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