----- Original Message ---- > From: Tim Legg <kc0ofc@xxxxxxxxx> > To: php-general@xxxxxxxxxxxxx > Sent: Wednesday, September 23, 2009 11:11:46 AM > Subject: 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