JS Ajax function not setting variable value

Question

I have a simple ajax function

  function get_country(request_term) {
    var country_list = '';
    $.ajax({
      url   : "activity/get_country", 
      type  : "POST",
      cache : false,
      data  : {
        request_term : request_term
      },
      success : function(data) {
        if(data != '') {
          country_list = $.parseJSON(data);
          alert(country_list);               ----> Got value here
        }
      }
    });
    alert(country_list);                     ----> Not getting value here
    return country_list;
  }

the problem is, i'm getting the data in the success function, but unable to return it from the main function.

possible duplicate of How to return AJAX response Text?

Quentin
– Quentin

2013-04-12 06:46:34 +00:00
Commented Apr 12, 2013 at 6:46 — Quentin
– Quentin, Commented Apr 12, 2013 at 6:46

Explosion Pills · Accepted Answer · 2013-04-12 06:45:06Z

1

Because ajax is asynchronous, you can't know when the success function will complete (or if it will ever complete). Therefore any code that requires the result of the ajax call must depend on the ajax callback too.

jQuery makes it very easy to bind additional callbacks.

return $.ajax({ /* rest of your code */

get_country(request_term).done(function (data) {
    //are you sure JSON isn't returned already?
    country_list = $.parseJSON(data);
});

answered Apr 12, 2013 at 6:45

Explosion Pills

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Comments

user1969752 · Accepted Answer · 2013-04-12 06:51:25Z

1

You can do it by making async as false. But it is not a recommented way.

This code will return country_list

function get_country(request_term) {
var country_list = '';
$.ajax({
  url   : "activity/get_country", 
  type  : "POST",
  cache : false,
  async : false,
  data  : {
    request_term : request_term
  },
  success : function(data) {
    if(data != '') {
      country_list = $.parseJSON(data);
      alert(country_list);               
    }
  }
});
alert(country_list);                     
return country_list;

}

answered Apr 12, 2013 at 6:51

user1969752

1 Comment

user1969752 Over a year ago

@DBK please reffer this link stackoverflow.com/questions/6517403/…

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