return AJAX callback return [duplicate]

You can’t return the result of your ajax request since the request is asynchronous (and synchronous ajax requests are a terrible idea).

Your best bet will be to pass your own callback into f1

var f1 = function(arg, callback) {
    $.ajax({
        success: function(data) {
            callback(data);
        }
    });
}

Then you’d call f1 like this:

f1(arg, function(data) { 
           var a = f2(data);
           alert(a);
        }
 );