On Friday 23 January 2004 09:38, Chris W. Parker wrote:
> <?php
> function query($sql, $current_line)
> {
> $this->Result = mysql_query($sql) or die($this->stop($current_line));
>
> if(!$this->Result)
> {
> echo mysql_error();
> }
The if-clause will never be evaluated because if there had been an error your
program would have dieded on the previous line.
> How is it possible that I'm getting the following errors?
>
> Warning: mysql_num_fields(): supplied argument is not a valid MySQL
> result resource in /home/cparker/www/schedulevark/lib/classes/db.php on
> line 127
>
> Warning: mysql_num_rows(): supplied argument is not a valid MySQL result
> resource in /home/cparker/www/schedulevark/lib/classes/db.php on line
> 130
> success!
>
> I can't figure it out! It sure looks like a valid MySQL result to me! Oh
I'm sure if php tells you it's invalid you can bet your *** it's invalid!
Check for errors and report with mysql_error() after _each_ and _every_ call
to the mysql_* functions.
--
Jason Wong -> Gremlins Associates -> www.gremlins.biz
Open Source Software Systems Integrators
* Web Design & Hosting * Internet & Intranet Applications Development *
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