How to use jquery ajax error parameter

Question

I'm looking for a simple example and/or explanation of how to use the error parameter for .ajax.

This question (jQuery ajax error function) points to this jQuery documentation (http://api.jquery.com/jQuery.ajax/) which I do not understand.

I have the following code that does not work and I can't figure out why. I'm hoping the error parameter will help:

jquery:

 <script> 
    // wait for the DOM to be loaded 
    $(document).ready(function() { 
        // bind 'myForm' and provide a simple callback function 
        $("#myForm").submit(function(){
                var user_input = $("#signup_id_email").val();
                $.ajax
                ({
                      type: "POST",
                      url: "ajax_test.php",
                      dataType: 'json',
                      data: {email: user_input},
                      **error: ""**
                })
                .done(function(r) 
                {
                    $("#answer_id").append(r.email);
                });
            });
    }); 
</script>

PHP (ajax_text.php)

<?php
echo json_encode($_POST);
?>

Hi, do you even have a POST request going out to the server? — Vlad Lyga, Apr 22 '13 at 21:19
If you're familiar with Chrome Dev Tools, check under the Network tab to make sure you received a response from your request — adamb, Apr 22 '13 at 21:24

score 17 · Accepted Answer · answered Apr 22 '13 at 21:33

The error "property" of the $.ajax parameter object is used to supply a callback function known as a closure to the $.ajax method. In other words you need to provide an anonymous function to handle an error if one occurs during the ajax request. Here is a basic example.

    $.ajax({
       type: "POST",
       url: "ajax_test.php",
       dataType: 'application/json',
       data: {email: user_input},
       success: function(result) {
                 // You can use success instead of .done
       },
       error: function(requestObject, error, errorThrown) {
            alert(error);
            alert(errorThrown);
       }
     });

Keep in mind that the error callback function will only be invoked if the request actually errors out. If your code simply does not return anything but the request still returns a status of 200 you will have to handle that exception in the success callback function.

Hope this helps.

EDIT: note that I removed the use of chaining events and now all callback functions are handled inside the original parameters object passed into $.ajax.

score 3 · Answer 2 · answered Apr 22 '13 at 21:34

I doubt you're even sending a request. Your whitespace is off.

Javascript does not require semicolons at the end of statements. Because of this, it sometimes infers them where you don't want them.

Try putting the '({' on the same line as the '$.ajax'. You didn't see the error because there isn't one. The $.ajax is a valid statement, even though it does nothing. Similarly, the object that you create inside of the parentheses is a valid object, even though it does nothing as well.

$.ajax({
    type: "POST",
    ...
}).done(function(r) {
  ....
});

To answer your original question, error takes a function as a parameter. This function takes 3 parameters in this order:

the XHR (the request)
the status of the request
the error that was thrown (can be undefined)

An example looks like this:

$.ajax({
    ...
    error: function(xhr, status, error) {
        alert("oh no!");
    }
})

How to use jquery ajax error parameter

2 Answers2