How do I change a global variable in the wrapper JavaScript function after an ajax call? [duplicate]

Question

function ajax_test(str1){    var url = "None"    jq.ajax({     type:'post',      cache: false,      url: 'http://....' + str1,      success: function(data, status, xhr){        url=data;      },      error: function (xhr, status, e) {       },      async: true,      dataType: 'json'    });    return url  }  

How can I set the global variable url to be the returned success ajax data?

Answer 1

In Javascript, it is impossible for a function to return an asynchronous result. The function will usually return before the AJAX request is even made.

You can always force your request to be syncronous with async: false, but that's usually not a good idea because it will cause the browser to lock up while it waits for the results.

The standard way to get around this is by using a callback function.

function ajax_test(str1, callback){      jq.ajax({       //... your options      success: function(data, status, xhr){          callback(data);      }    });   }   

and then you can call it like this:

ajax_test("str", function(url) {   //do something with url });