There is not a way to do this that I can think of. Whenever I deal with AJAX, I simply turn debugging off. If I need to debug it, I just keep debug on and read the debug message in the response (even though it will break the json).
On Jul 22, 1:51 pm, Dennis <[email protected]> wrote: > On Jul 22, 9:03 pm, cricket <[email protected]> wrote: > > > $this->log($this->I_DONT_KNOW_THIS_VAR); > > Sorry but I don't understand how this should help me? (Or do I miss > something here?) > I don't want to log something. I want to prevent Cake from just > printing out its debug messages, but instead add them to my json > output. > > But after spending half of the day with debugging cake I think thats > just not possible (without modifying cake's own classes > > thanks anyway! Check out the new CakePHP Questions site http://cakeqs.org and help others with their CakePHP related questions. You received this message because you are subscribed to the Google Groups "CakePHP" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/cake-php?hl=en
