How to return data in jQuery Ajax?

Question

let's say i have a simple function:

    function test(url){
var AjaxCall = $.ajax({
        type: GET,
        url: url,
        data: {},
        success: function(data,status,xhr){
             return xhr
        },
        error: function(xhr,e){
             return xhr
        }
});


    if(xhr.status === 200)console.log('success');


}

So what i want is to treat the errors inside the function not after calling it and adding the callback in the call to treat those error for example, is this possible?

Answer 1

If you are using jQuery lower than 1.8, yes you can by setting async = false

  function test(url){
var AjaxCall = $.ajax({
        type: GET,
        url: url,
        async : false
        data: {},
        success: function(data,status,xhr){
             return xhr
        },
        error: function(xhr,e){
             return xhr
        }
});
}

Please note that this is deprecated, and for good reason

As of jQuery 1.8, the use of async: false with jqXHR ($.Deferred) is deprecated; you must use the success/error/complete callback options instead of the corresponding methods of the jqXHR object such as jqXHR.done() or the deprecated jqXHR.success().

http://api.jquery.com/jquery.ajax/

function test(url){
    var AjaxCall = $.ajax({
            type: GET,
            url: url,
            data: {},
            complete: function(xhr, status){
                 if(xhr.status === 200)console.log('success');
            }
    });
}

And no, you cannot return a value from your function with this method. That's not how asynchronous works. If you want to return, you have to set async = false and that kind of defeats the purpose of using ajax.