----- Original Message ----
> From: Tim Legg <[email protected]>
> To: [email protected]
> Sent: Wednesday, September 23, 2009 11:11:46 AM
> Subject: [PHP] Stricter Error Checking?
>
> 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 $row['Number'];
>
> After retyping the code 3 or 4 times over the course of the morning, I
> finally
> found where the problem was. The problem is that the database field is
> called
> 'Part_Number', not 'Number'. The field 'Number' does not exist in the
> database. I am very surprised that I didn't even get a warning that there
> might
> be a problem with the statement. All I saw is that nothing was being
> returned
> via the echo command.
if(!$result)
{
die('Error: ' . mysql_error());
}
This didn't work when you used 'Number' instead of 'Part_Number'? Strange...
>
> There really must be a stricter error checking that is turned on somewhere,
> isn't there?
>
>
> Thanks for your help.
>
> Tim
>
>
>
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
