How do I catch an Ajax query post error?

Question

I would like to catch the error and show the appropriate message if the Ajax request fails.

My code is like the following, but I could not manage to catch the failing Ajax request.

function getAjaxData(id)
{
     $.post("status.ajax.php", {deviceId : id}, function(data){

        var tab1;

        if (data.length>0) {
            tab1 = data;
        }
        else {
            tab1 = "Error in Ajax";
        }

        return tab1;
    });
}

I found out that, "Error in Ajax" is never executed when the Ajax request failed.

How do I handle the Ajax error and show the appropriate message if it fails?

Answer 1

Since jQuery 1.5 you can use the deferred objects mechanism:

$.post('some.php', {name: 'John'})
    .done(function(msg){  })
    .fail(function(xhr, status, error) {
        // error handling
    });

Another way is using .ajax:

$.ajax({
  type: "POST",
  url: "some.php",
  data: "name=John&location=Boston",
  success: function(msg){
        alert( "Data Saved: " + msg );
  },
  error: function(XMLHttpRequest, textStatus, errorThrown) {
     alert("some error");
  }
});

Answer 2

jQuery 1.5 added deferred objects that handle this nicely. Simply call $.post and attach any handlers you'd like after the call. Deferred objects even allow you to attach multiple success and error handlers.

Example:

$.post('status.ajax.php', {deviceId: id})
    .done( function(msg) { ... } )
    .fail( function(xhr, textStatus, errorThrown) {
        alert(xhr.responseText);
    });

Prior to jQuery 1.8, the function done was called success and fail was called error.

Answer 3

$.ajax({
  type: 'POST',
  url: 'status.ajax.php',
  data: {
     deviceId: id
  },
  success: function(data){
     // your code from above
  },
  error: function(xhr, textStatus, error){
      console.log(xhr.statusText);
      console.log(textStatus);
      console.log(error);
  }
});

Answer 4

$.post('someUri', { }, 
  function(data){ doSomeStuff })
 .fail(function(error) { alert(error.responseJSON) });

Answer 5

A simple way is to implement ajaxError:

Whenever an Ajax request completes with an error, jQuery triggers the ajaxError event. Any and all handlers that have been registered with the .ajaxError() method are executed at this time.

For example:

$('.log').ajaxError(function() {
  $(this).text('Triggered ajaxError handler.');
});

I would suggest reading the ajaxError documentation. It does more than the simple use-case demonstrated above - mainly its callback accepts a number of parameters:

$('.log').ajaxError(function(e, xhr, settings, exception) {
  if (settings.url == 'ajax/missing.html') {
    $(this).text('Triggered ajaxError handler.');
  }
});

Answer 6

You have to log the responseText:

$.ajax({
    type: 'POST',
    url: 'status.ajax.php',
    data: {
    deviceId: id
  }
})
.done(
 function (data) {
  //your code
 }
)
.fail(function (data) {
      console.log( "Ajax failed: " + data['responseText'] );
})

Answer 7

In case you want to utilize .then() which has a subtle difference in comparison with .done() :

return $.post(url, payload)
.then(
    function (result, textStatus, jqXHR) {
        return result;
    },
    function (jqXHR, textStatus, errorThrown) {
        return console.error(errorThrown);
    });

Answer 8

you attach the .onerror handler to the ajax object, why people insist on posting JQuery for responses when vanila works cross platform...

quickie example:

ajax = new XMLHttpRequest();
ajax.open( "POST", "/url/to/handler.php", true );
ajax.onerror = function(){
    alert("Oops! Something went wrong...");
}
ajax.send(someWebFormToken );