Jquery checking success of ajax post

Question

how do i define the success and failure function of an ajax $.post?

Answer 1

If you need a failure function, you can't use the $.get or $.post functions; you will need to call the $.ajax function directly. You pass an options object that can have "success" and "error" callbacks.

Instead of this:

$.post("/post/url.php", parameters, successFunction);

you would use this:

$.ajax({
    url: "/post/url.php",
    type: "POST",
    data: parameters,
    success: successFunction,
    error: errorFunction
});

There are lots of other options available too. The documentation lists all the options available.

Answer 2

using jQuery 1.8 and above, should use the following:

var request = $.ajax({
    type: 'POST',
    url: 'mmm.php',
    data: { abc: "abcdefghijklmnopqrstuvwxyz" } })
    .done(function(data) { alert("success"+data.slice(0, 100)); })
    .fail(function() { alert("error"); })
    .always(function() { alert("complete"); });

check out the docs as @hitautodestruct stated.

Answer 3

This style is also possible:

$.get("mypage.html")
    .done(function(result){
        alert("done. read "+result.length+" characters.");
    })
    .fail(function(jqXHR, textStatus, errorThrown){
        alert("fail. status: "+textStatus);
    })

Answer 4

The documentation is here: http://docs.jquery.com/Ajax/jQuery.ajax

But, to summarize, the ajax call takes a bunch of options. the ones you are looking for are error and success.

You would call it like this:

$.ajax({
  url: 'mypage.html',
  success: function(){
    alert('success');
  },
  error: function(){
    alert('failure');
  }
});

I have shown the success and error function taking no arguments, but they can receive arguments.

The error function can take three arguments: XMLHttpRequest, textStatus, and errorThrown.

The success function can take two arguments: data and textStatus. The page you requested will be in the data argument.

Answer 5

I was wondering, why they didnt provide in jquery itself, so i made a few changes in jquery file ,,, here are the changed code block:

original Code block:

    post: function( url, data, callback, type ) {
    // shift arguments if data argument was omited
    if ( jQuery.isFunction( data ) ) {
        type = type || callback;
        callback = data;
        data = {};
    }

    return jQuery.ajax({
        type: "POST",
        url: url,
        data: data,
        success: callback,
        dataType: type
    });  

Changed Code block:

        post: function (url, data, callback, failcallback, type) {
        if (type === undefined || type === null) {
            if (!jQuery.isFunction(failcallback)) { 
            type=failcallback
        }
        else if (!jQuery.isFunction(callback)) {
            type = callback
        }
        }
        if (jQuery.isFunction(data) && jQuery.isFunction(callback)) {
            failcallback = callback;

        }
        // shift arguments if data argument was omited
        if (jQuery.isFunction(data)) {
            type = type || callback;
            callback = data;
            data = {};

        }


        return jQuery.ajax({
            type: "POST",
            url: url,
            data: data,
            success: callback,
            error:failcallback,
            dataType: type
        });
    },

This should help the one trying to catch error on $.Post in jquery.

Updated: Or there is another way to do this is :

   $.post(url,{},function(res){
        //To do write if call is successful
   }).error(function(p1,p2,p3){
             //To do Write if call is failed
          });