How to get result of form submission with jQuery Form Plugin?

Question

How can I get status code (200, 400, 404 etc.) at showResponse below:

$("#myform").validate({
  submitHandler: function(form) {
    var options = { 
       success: showResponse
    };
    $(form).ajaxSubmit(options);

    function showResponse(responseText, statusText, xhr, $form)  {
      if (responseText == 'ok') { // status code to be used instead
        ...
      } else {
        ...
      }
    }

    return false;
  }
});

Answer 1

This is from the jQuery Ajax docs:

statusCode(added 1.5)Map Default: {}

A map of numeric HTTP codes and functions to be called when the response has the corresponding code. For example, the following will alert when the response status is a 404:

$.ajax({
  statusCode: {
    404: function() {
      alert('page not found');
    }
  }
});

If the request is successful, the status code functions take the same parameters as the success callback; if it results in an error, they take the same parameters as the error callback.

According to the jQuery form docs you can pass any of the standard $.ajax options to ajaxForm.