Use variable outside a $.get function

Question

I'm having problem using variables outside a $.get function, the variable isn't found and the script stops.

What can I do?

$.get("../ajax.php", function (data) {
    var json = JSON.parse(data);
    test = json['jquery']['test'];
});

alert(test)

UPDATE: (This solved my problem) I get the json with $.get. And pass it to my function and the json will be able to get later in the script.

$.get("../ajax.php", function (data) {
function(data) {                    
    loadApp(data);
});

function loadApp(data) {    
    json = JSON.parse(data);                
}

Please search before asking a question. This has been asked **so** many times previously. — Rory McCrossan, Nov 06 '13 at 09:53
I don't think it's a bad question I only find "hard ways" of doing it. Don't see the real logic in how to use them easy outside the function. — Kim, Nov 06 '13 at 09:56
Read this to clear everything up... http://stackoverflow.com/questions/14220321/how-to-return-the-response-from-an-ajax-call — Reinstate Monica Cellio, Nov 06 '13 at 09:57
@Kim there is no way to use it outside the function. Any code reliant on the result of the AJAX request must be placed in the callback. — Rory McCrossan, Nov 06 '13 at 09:58

Eric · Answer 1 · 2013-11-06T09:59:59.347

Since it's a asynchronous call, all processing will have to be made in the callback function. Put your alert(test) inside the callback function instead.

EDIT:

$.get("../ajax.php", function (data) {
    var json = JSON.parse(data);
    test = json['jquery']['test'];

    alert(test); // Needs to be processed in this scope
});

This occurs since the JavaScript processor reads line by line, detects the ajax-call and sends it, it then jumps to the alert, which will try to alert a non-existing variable. Even if you declare it gobally it wont work; since the AJAX-return is asynchronous, and will therefore be executed in parallel to the other code.

Use variable outside a $.get function

1 Answers1