How to set a function's return value to the json response of its ajax call?

Question

I'm struggling in setting a function's return value to the json response of its ajax call.

function callMe() {    
    $.ajax({
        url: "someJson.php",
        type: "GET",
        dataType: 'json',              
        success: function(response){
            return response;
    }
}

 var jsonResponse = callMe();

How get I get jsonReponse to the hold the json response?

see http://stackoverflow.com/questions/1572610/ajax-jquery-synchronous-callback-success — bnieland, Mar 31 '12 at 02:55

Josh Mein · Accepted Answer · 2012-03-31T03:03:05.670

1

This is not possible by default because ajax is asynchronous; therefore, once you send the ajax request, the function continues and does not wait until the ajax call returns before ending the function. It is best to do whatever you need to inside the success of the ajax call.

Edit:

You can make an ajax call synchronous, however, it is not recommended because it can lock up the browser.

Note that synchronous requests may temporarily lock the browser, disabling any actions while the request is active.

jQuery Ajax Info

edited Mar 31 '12 at 03:03

answered Mar 31 '12 at 02:45

Josh Mein

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So, this _is_ possible just not recommended. – iambriansreed Mar 31 '12 at 03:23
It is not possible by default. How would it be good to tell him how to do something that is bad practice and is frowned upon? – Josh Mein Mar 31 '12 at 03:25
Depends what the definition of is **is**. – iambriansreed Mar 31 '12 at 03:27
What is that supposed to mean? – Josh Mein Mar 31 '12 at 03:31

wedgwood · Answer 2 · 2012-03-31T03:43:51.807

Because the default behave of jquery ajax is async mode, you Can't directly get the response. At Least, You have two choice:

pass the func as parameter into "callme":

function callMe(callback) {
   $.ajax({
      url: "someJson.php",
      type: "GET",
      dataType: 'json',              
      success: function(response){
          callback(response);
      }
   });
}

function dealResponse(response) {
  // do something with response
}

callMe(dealResponse);

use sync mode ajax, and use your code directly get response and deal with it. but if you go this way, you script will block and may temporarily lock the browser when ajax request start, until the request end.

function callMe() {  
  var ret;

  $.ajax({
    url: "someJson.php",
    type: "GET",
    async
    dataType: 'json',              
    success: function(response){
      ret = response;
    }
  });

  retun ret;
}

var jsonResponse = callMe();

Hope these could help you :)

Tamik Soziev · Answer 3 · 2012-03-31T02:50:47.613

-1

Do it like this:

function callMe() {  
    var result = false;  
    $.ajax({
        url: "someJson.php",
        type: "GET",
        async: false,
        dataType: 'json',              
        success: function(response){
            result = response;
            return response;
    } 
    return result;
}

var jsonResponse = callMe();

edited Mar 31 '12 at 02:50

answered Mar 31 '12 at 02:45

Tamik Soziev

14,307
5
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because the return will be called at the same time as the return. It's not being fired as a callback – nickw444 Mar 31 '12 at 02:55
@antisanity see my answer...its not asynchronous. – Tamik Soziev Mar 31 '12 at 02:55
1

Setting async to false is usually a bad move, too. Not saying there's never a time to do it, but if you can at ALL just do what needs doing in the success function, fewer kittens will need to die. – Greg Pettit Mar 31 '12 at 03:05
i've answered his question...he asked how can he do it, i explain how, why downvote? – Tamik Soziev Mar 31 '12 at 17:21

How to set a function's return value to the json response of its ajax call?

3 Answers3