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JS : Return result from nested ajax success function

How do I return the result of the nested ajax call as the result of the parent function?

//Declaring the function
var myFunction = function(myData){  
  $.ajax({
  type:'post',
  url:"/ajaxPage.php",
  data:{data:myData},
  success:function(r){
  return r;
  });
}

//Calling the function
var theResult = myFunction(myData);

I want the variable 'theResult' to hold the contents of the ajax call.

like image 732
wilsonpage Avatar asked Dec 13 '22 14:12

wilsonpage


2 Answers

Since the ajax is asynchronous you cannot return it in the parent function. What you can do, is to provide a callback function, and you call it as well with the result.

 //Declaring the function

var myFunction = function(myData, callback){  
  $.ajax({
  type:'post',
  url:"/ajaxPage.php",
  data:{data:myData},
  success:function(r){
    callback(r);
  });
}


//Calling the function
var theResult = myFunction(myData, function(res) {
    // deal with it..
});
like image 65
José Leal Avatar answered Dec 27 '22 09:12

José Leal


If you're using jQuery, you should definitely be using $.Deferred() and Promises/A. I've made this more verbose that you really need to demonstrate some of the functionality. $.ajax itself is already returning a $.Deferred().promise() so you can actually just cut out the extra step if you only need to make the one XHR request (see bottom example).

//Declaring the function
var myFunction = function(myData){
  var deferredResponse = new $.Deferred();
  $.ajax({
    type:  'post',
    url:   '/ajaxPage.php',
    data: {'data': myData}
  }).done(deferredResponse.resolve).fail(deferredResponse.reject);
  return deferredResponse.promise();
}

//Calling the function
$.when(myFunction(myData)).done(function(response){
  var theResult = response;
});

The verbose syntax comes in handy when you have multiple nested XHR requests and want to return the response of the inner-most call: deferredResponse.resolve(innerResponse). Here's the simply version.

//Declaring the function
var myFunction = function(myData){
  return $.ajax({
    type:  'post',
    url:   '/ajaxPage.php',
    data: {'data': myData}
  });
}

//Calling the function
myFunction(myData).done(function(response){
  var theResult = response;
});
like image 33
idbehold Avatar answered Dec 27 '22 07:12

idbehold