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

When you receive an AJAX error like that, you can replace your current document contents with the debug/error document returned from the server. For example, you can do something like the following:

$.ajax({
    url: 'failing_controller/',
    type: 'POST'
})
    .fail(function (jqXHR, textStatus, errorThrown) {
        document.open();
        document.write(jqXHR.responseText);
        document.close();
    })
    .success(function (data, textStatus, jqXHR) {
        // ... handle data ...
    });

This can be really helpful during development and debugging, because it lets you inspect the server state in real-time. You will probably want to replace the failure handler with something more proper in a production environment.