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View Full Version : Php error in feedback.php - 1.5rc4


Oceanwatcher
18th of August 2005 (Thu), 04:39
I think it is better to start a separate thread for this error. Probably makes it easier to find for others while searching.

After fixing another php error, it seems like the main problem left is feedback.php . I am getting these errors:

[Tue Aug 16 19:53:31 2005] [error] PHP Notice: Undefined variable: valid in /xxxx/xxxxx/xxxxxx/feedback.php on line 807
[Tue Aug 16 19:53:31 2005] [error] PHP Notice: Undefined variable: select_output in /xxxx/xxxxx/xxxxxx/feedback.php on line 750
[Tue Aug 16 19:53:31 2005] [error] PHP Notice: Undefined variable: search_row in /xxxx/xxxxx/xxxxxx/feedback.php on line 748
[Tue Aug 16 19:53:31 2005] [error] PHP Notice: Undefined variable: select_output in /xxxx/xxxxx/xxxxxx/feedback.php on line 671
[Tue Aug 16 19:53:31 2005] [error] PHP Notice: Undefined variable: search_row in /xxxx/xxxxx/xxxxxx/feedback.php on line 669

Does anyone know what this is and what can be done? Maybe this is something Pekka has to deal with?

Pekka
18th of August 2005 (Thu), 05:50
Notices are not harmful, they just point that variables have not been defined. You can add

error_reporting(0);

on top of feedback.php to hide those notices. Usually on production servers minor error reporting is disabled globally in php.ini.

Oceanwatcher
18th of August 2005 (Thu), 17:16
So these variables do not need to be defined? It is not pointing to errors that need to be corrected?

Pekka
19th of August 2005 (Fri), 03:43
There are many variables that do not need to be always defined. Above errors contain some variables which are not needed at all, that happens sometimes :)