jQuery return ajax result into outside variable

Question

I have some problem using ajax.

How can I assign all result from ajax into outside variable ?

I google it up and found this code..

var return_first = (function () {     var tmp = null;     $.ajax({         'async': false,         'type': "POST",         'global': false,         'dataType': 'html',         'url': "ajax.php?first",         'data': { 'request': "", 'target': arrange_url, 'method': method_target },         'success': function (data) {             tmp = data;         }     });     return tmp; }); 

but not work for me..

Can anybody tell what is wrong about that code ?

Answer 1

You are missing a comma after

'data': { 'request': "", 'target': 'arrange_url', 'method': 'method_target' } 

Also, if you want return_first to hold the result of your anonymous function, you need to make a function call:

var return_first = function () {     var tmp = null;     $.ajax({         'async': false,         'type': "POST",         'global': false,         'dataType': 'html',         'url': "ajax.php?first",         'data': { 'request': "", 'target': 'arrange_url', 'method': 'method_target' },         'success': function (data) {             tmp = data;         }     });     return tmp; }(); 

Note () at the end.

Answer 2

This is all you need to do:

var myVariable;  $.ajax({     'async': false,     'type': "POST",     'global': false,     'dataType': 'html',     'url': "ajax.php?first",     'data': { 'request': "", 'target': 'arrange_url', 'method': 'method_target' },     'success': function (data) {         myVariable = data;     } }); 

NOTE: Use of "async" has been depreciated. See https://xhr.spec.whatwg.org/.