CODEWITHSUNDEEP
javascriptjquerypromisedeferred
Natalie
Natalie

Reputation: 297

deferred- code working with fail/done but not working with then

I have a function called func1 that's calling var dfd = $.ajax, and returning dfd.promise() , now in my main function, I have the code like this: 

function addOperation() {
   var addPromise = func1();
   addPromise.then(
     function(sender, args, msg) {
       alert("success");
     },

     function() {
       alert("fail");
     });
 }

The above code is not working, what's working for me is:

function addOperation() {
  var addPromise = func1();
  addPromise.fail(
      function(sender, args, msg) {
        alert("fail");
      })
    .done(
      function() {
        alert("success");
      });
}

From my understanding is that then(functionSuccess, functionFail) is the same as calling .done and .fail on the original promise, isn't that right? I am lost with this.

Upvotes: 0

Views: 52

Answers (2)

Jamiec
Jamiec

Reputation: 136074

you should not be returning dfd.promise() from your function, the return value from $.ajax is the promise.

Just return dfd directly (or the result of $.ajax)

The docs say it best:

The jqXHR objects returned by $.ajax() as of jQuery 1.5 implement the Promise interface, giving them all the properties, methods, and behavior of a Promise (see Deferred object for more information). These methods take one or more function arguments that are called when the $.ajax() request terminates

Upvotes: 1

Drumbeg
Drumbeg

Reputation: 1934

The jqXHR object returned by $.ajax implements the Promise interface.

The jqXHR objects returned by $.ajax() as of jQuery 1.5 implement the Promise interface, giving them all the properties, methods, and behavior of a Promise (see Deferred object for more information).

Therefore, func1 should just return the value returned by $.ajax.

Upvotes: 0

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