I think he meant that he is using 'Number' instead of 'Part_Number'
when accessing the array and not in the SQL, his SQL was correct, this
was wrong:
> echo $row['Number'];
E_STRICT catches that kind of error.
Jonathan
On Wed, Sep 23, 2009 at 3:28 PM, Tommy Pham <tommy...@yahoo.com> wrote:
>
>
> ----- Original Message ----
>> From: Tim Legg <kc0...@yahoo.com>
>> To: php-general@lists.php.net
>> 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
>>
>>
>>
>>
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