CODEWITHSUNDEEP
javascriptjqueryajaxrequest
zmajeric
zmajeric

Reputation: 53

jQuery AJAX request events - done,fail,success

I have code like that

var ajaxrequest = $.ajax({
            type: "POST",
            dataType: "json",
            url: "xy.php", 
            data: {
                action : "read"
            }
            }).fail(function(){
                //something to do when ajaxreq fails
            }).done(function(data){

               //something to do when ajaxreq is done
            });

It is working no problem. My question is why this doesnt work:

var ajaxrequest = $.ajax({
            type: "POST",
            dataType: "json",
            url: "n3_vaje_api.php", //Relative or absolute path to response.php file
            data: {
                action : "read",
            },
            fail:function(){
                //something to do when ajaxreq fails
            },
            done:function(data){
              //something to do when ajaxreq is done
            }
        });

Fail and done are just examples, complete also doesnt work if used inside. But using it outside like:

ajaxrequest.complete(f(){});

is working just fine... I know instead of done I should use success, but thats not my point here. Whats the deal here?

Upvotes: 5

Views: 11764

Answers (3)

hydRAnger
hydRAnger

Reputation: 3383

Deprecation Notice: The jqXHR.success(), jqXHR.error(), and jqXHR.complete() callbacks are removed as of jQuery 3.0. You can use jqXHR.done(), jqXHR.fail(), and jqXHR.always() instead.

Upvotes: 0

Dev C
Dev C

Reputation: 1

You could simply use $.post instead of $.ajax and json as the fourth argument.

$.post("n3_vaje_api.php", {action : "read"}, function(response) {
    // Do something with the request
}, 'json')
.done(function() {
    alert( "second success" );
  })
  .fail(function() {
    alert( "error" );
  })
  .always(function() {
    alert( "finished" );
});

Upvotes: -1

Pranay Rana
Pranay Rana

Reputation: 176896

you need to use success and error is the method you need to use if you want to use your second option

this is example of ajax request without promise, where you are getting success and error function as parameter 

 $.ajax({url:"demo_test.txt"
      ,error : function (xhr,status,error)
        { //alert error}
      ,success:function(result){
      $("#div1").html(result);
    }});

In the first opetion you are using promise object return by ajax requst that is the reason you are getting done and fail method.

this is example of promise object , in below example request is promise object 

var request = $.ajax({
  url: "script.php",
  type: "POST",
  data: { id : menuId },
  dataType: "html"
});

request.done(function( msg ) {
  $( "#log" ).html( msg );
});

request.fail(function( jqXHR, textStatus ) {
  alert( "Request failed: " + textStatus );
});

Upvotes: 7

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