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jquery - Resource interpreted as Document but transferred with MIME type application/json warning in Chrome Developer Tools

I have the following snippet, which uses the jQuery Form plugin to post a form to the server (in ajax).

  var options = {
    dataType: "json",
    success: function(data) { 
      alert("success");
    } 
  }; 

  $form.ajaxSubmit(options);

The form:

<form enctype="multipart/form-data" id="name_change_form" method="post" action="/my_account/"> 
<div style='display:none'><input type='hidden' name='csrfmiddlewaretoken' value='6c9b552aaba88b8442077e2957e69303' /></div> 
  <table> 
    <tr> 
      <td> 
        <label for="id_first_name">First name</label>:
      </td> 
      <td> 
        <input name="first_name" value="Patrick" maxlength="30" type="text" id="id_first_name" size="30" /> 
      </td> 
    </tr> 
    <tr> 
      <td> 
        <label for="id_last_name">Last name</label>:
      </td> 
      <td> 
        <input name="last_name" value="Sung" maxlength="30" type="text" id="id_last_name" size="30" /> 
      </td> 
    </tr> 
  </table> 
  <input type="hidden" name="form_id" value="name_change_form" /> 
</form> 

The ajax implementation is working fine. But I am getting a warning

Resource interpreted as Document but transferred with MIME type application/json

in Chrome Developer Tools. I want to find out why the warning, or even better, a way to resolve it.

I changed to use $.post instead and magically the error was gone since then. I have no idea why $.post works but not $form.ajaxSubmit. If someone can offer their explanation that would be great. At the very least, this problem is resolved. Below is the new code.

var url = $form.attr("action");
$.post(
  url, 
  $form.serialize(), 
  function(data) {
    alert("success");
  },
  "json"
); 
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1 Reply

0 votes
by (71.8m points)

I was facing the same error. The solution that worked for me is:

From the server end, while returning JSON response, change the content-type: text/html

Now the browsers (Chrome, Firefox and IE8) do not give an error.


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