You're doubling up on your mysql_query's .. the first one runs, and I
don't know the ins and outs of PHP, but the second mysql_query causes
the resource to not be available... had a similar problem recently.
Take the mysql_query out of the $cartquery assignment.
-Jason
Mike At Spy wrote:
> Anyone ever have the experience of working on a site, working with MySQL and
> then having something almost inexplicable happen?
>
> I was working on a script, and had this weird event. A very simple script
> with this:
>
> mysql_connect("localhost","user","pass");
> mysql_select_db("dbname")
> or die("Query Connection Database failed");
> $old=date("z")-1;
> mysql_query("DELETE FROM Users WHERE Date < $old");
>
> $cartquery = mysql_query("SELECT CartItemsID,Date FROM CartItems")
> or die("Query failed");
> $cartnow = mysql_query($cartquery);
> $cr=0;
> $cartme = mysql_num_rows($cartnow); <<<<this is the line that is failing!
> while ($cr < $cartme){
> $cartrow=mysql_fetch_row($cartnow);
> $CII=$cartrow[0];
> $CDa=$cartrow[1];
> $pieces=explode(":",$CDa);
> $DCHK=$pieces[1];
> if ($DCHK < $old) {
> mysql_query("DELETE FROM CartItems WHERE CartItemsID = '$CII'");
> }
> }
>
> Started spitting this error at me:
>
> Warning: Supplied argument is not a valid MySQL result resource in (exact
> location snipped)
>
> This is the line containing mysql_num_rows that is getting this error.
>
> Someone please tell me I'm missing something obvious! Yes, the connects to
> MySQL database are fine. I can run the query right in phpMyAdmin with no
> issues.
>
> If mysql_num_rows returns 0, or empty, it shouldn't be complaining, should
> it? It would just skip the loop, correct?
>
> Thanks,
>
> -Mike
>
>
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
