Tim Legg wrote:
> Hello,
>
> I just spent way, way to much time trying to debug code due to a misnamed element. Here is a simplified example of the problem I dealt with.
>
>
> $test = "SELECT * FROM `Materials` WHERE `Part_Number` = '125664'";
> $result = mysql_query($test,$handle);
> if(!$result)
> {
> die('Error: ' . mysql_error());
> }
> $row = mysql_fetch_array($result);
> echo $row['Number'];
>
> After retyping the code 3 or 4 times over the course of the morning, I finally found where the problem was. The problem is that the database field is called 'Part_Number', not 'Number'. The field 'Number' does not exist in the database. I am very surprised that I didn't even get a warning that there might be a problem with the statement. All I saw is that nothing was being returned via the echo command.
>
> There really must be a stricter error checking that is turned on somewhere, isn't there?
>
>
> Thanks for your help.
>
> Tim
>
>
>
When developing you should always use something like this:
ini_set('display_errors', 'On');
error_reporting(E_ALL);
--
Thanks!
-Shawn
http://www.spidean.com
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