From: Tommy Pham >> From: Tim Legg >> >> 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... > I think the problem is that he didn't check that the key he used actually existed before using the value it pointed to. So he got an empty string for $row['Number']; because the key should have been 'Part_Number'. I don't know that even E_STRICT would catch that one. Bob McConnell -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php