Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
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.
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.