Getting Django debug information when using Ajax call?

Question

Is there a good way to get the debug information that Django provides when using jQuery to make Ajax calls? Right now when I make a call, I just see a http/200 server error in the python runserver window, but because the call is made through javascript, I don't get a debug page with all the information.

Answer 1

You can inspect the contents of the response returned to your jQuery ajax call. Using a tool like Firebug can make this pretty easy.

Django will still return the debug page, it's just that it is responding to the ajax call rather than a regular browser request.

It's often a good technique to get your stuff working with regular requests, and then "ajaxify" them only once you are sure the server side code is working.

Answer 2

You can use this snippet: http://www.djangosnippets.org/snippets/650/ to get plaintext tracebacks for viewing in firebug, instead of HTML.

Answer 3

Probably the simplest way is to turn off debug, and rely on email debug messages Django sends. However, this isn't often practical, but it works.