function with ajax request

Question

can you help me with this problem? In success:.... I get a result that I need, but ajaxFunction at the end returns ' ' value.

ajaxFunction = function(lastNumber){
        var json_result = '';
        $.ajax({
            type: "POST",
            contentType: "application/json",
            dataType: "json",
            data: '{"first" : "'+lastNumber+'"}',
            url: "http://localhost:8080/some_url",
            success: function (result) {
                console.log("in success:  " + result.result);
                json_result =  result.result;
            }
        });
        console.log("before return   " + json_result);
        return json_result;
    };

Possible duplicate of [How do I return the response from an asynchronous call?](http://stackoverflow.com/questions/14220321/how-do-i-return-the-response-from-an-asynchronous-call) — dotnetom, Mar 10 '16 at 04:56

score 1 · Answer 1 · edited May 23 '17 at 11:59

The problem is that your result is returned, and then the ajax called is executed async.

Here's an example for an alternative, by using Deferred.

return value from nested function in jquery

function test(){
   var def = new $.Deferred();   
    $.ajax({
        url: 'test',
        success: function(data) {
            // Do some minor logic to determine if server response was success.
            console.log('success in callback');
            if(data == true){
              def.resolve(true);  
            }
            else{
              def.reject(); 
              // or possibly def.resolve(true) depending what you're looking for.
            }

        },
        error:function(){
          console.log('error in callback');
          def.reject();
        }
    });

  return def; 
}

EDIT: Example for your case:

var ajaxFunction = function(lastNumber){
  var def = new $.Deferred();         
        $.ajax({
            type: "POST",
            contentType: "application/json",
            dataType: "json",
            data: '{"first" : "'+lastNumber+'"}',
            url: "http://example.com",
            success: function (result) {
                console.log("in success:  " + result.result);
                def.resolve(result.result);
            },
          error: function(){
            def.reject();
            }
        });
        return def;
    };

var testMethod = function(){
  var result = ajaxFunction('some number');
  result.done(function(result){
    alert(result);
    console.log('done in deferred');
    // Do something when everything went well.

  }).fail(function(){
    alert('error');
    console.log('error in deferred');
    // Do something when something went bad.
  });

}

testMethod();

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Hi, in this case it returns an [object Object] object. How can I get this object's value? I need a value and not an object. — Maria, Mar 10 '16 at 07:22

Abhishek Dhanraj Shahdeo · Answer 2 · 2016-03-10T06:07:26.093

0

You can put this and try

    ajaxFunction = function(lastNumber){
    var json_result = '';
    $.ajax({
        async: false,
        type: "POST",
        contentType: "application/json",
        dataType: "json",
        data: '{"first" : "'+lastNumber+'"}',
        url: "http://localhost:8080/some_url",
        success: function (result) {
            console.log("in success:  " + result.result);
            json_result =  result.result;
        }
    });
    console.log("before return   " + json_result);
    return json_result;
};

It will make the call synchronous as you are setting async to false, which is by default true.

If not working with this one, you can make use of either call back or promise(works and returns like synchronous calls).

edited Mar 10 '16 at 06:07

answered Mar 10 '16 at 06:00

Abhishek Dhanraj Shahdeo

1,356
2
14
35

I've tryed like this, but async: false is depricated – Maria Mar 10 '16 at 06:04
1

Using synchronous AJAX calls is almost always a bad idea, as it freezes the UI until response from server is received – dotnetom Mar 10 '16 at 06:04

score 0 · Answer 3 · answered Mar 10 '16 at 07:37

0

always use $.Deferred() object and return it in your Ajax call.

use .resolve() to your deferred variable in your .success() callback of Ajax call.

Call your method and check with .done method of your deferred function.

answered Mar 10 '16 at 07:37

Nitin Vijay

1,414
12
28

function with ajax request

3 Answers3