Return in each getJSON - Jquery

Question

I would like to know the following how do I give Return in a function created by me, in a situation below. Because it always returns Undefined the object, but when I give an alert it returns the normal object.

  function jjson(url){
            var result;
            $.getJSON( url, function(data) {
                alert(data)
                result = data;
            });

            return result;
        }

ATT

Answer 1

This function will always give Return Undefined because AJAX is asynchronous.

You have to wear a callback to do what you need to do when the server response comes in. You can use it like this:

function jjson(url, callback){
        $.getJSON( url, callback);
 }

and then call the function by passing to the callback what you want to do:

jjson('google.com', function(result){
    alert(result);
});

In practice this can be simplified if you put what you want to do directly into getJSON...

$.getJSON(url, function(data) {
    gerarGrafico(data); //por exemplo
});

'Cause then that code will only run when the server data has arrived.