jquery ajax call success, how do i change a global variable in the wrapper javascript function?

Question

function ajax_test(str1){ 
  var url = \"None\" 
  jq.ajax({
    type:\'post\', 
    cache: false, 
    url: \'http://....\' + str1, 
    success: function(data, sta         


        

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
});