返回ajax调用结果的JavaScript函数

Question

Help needed. Im writing a function that returns result of ajax call but i did not get any results, i guess it's a scope issue, but is there any way to do it? Here is my code:

function Favorites() {
    var links;
    $.ajax({
        type: "GET",
        url: "/Services/Favorite.svc/Favorites",
        data: "{}",
        contentType: "application/json; charset=utf-8",
        dataType: "json",
        cache: false,
        success: function(msg) {
            links = (typeof msg.d) == 'string' ? eval('(' + msg.d + ')') : msg.d;
        }
    });
    return links;
};

Answer 1

Ajax is asynchronous, ie when return links is executed, the callback function in success might not even have been called.

Extend your function to accept a callback:

function Favorites(callback) {
    var links;
    $.ajax({
        type: "GET",
        url: "/Services/Favorite.svc/Favorites",
        data: "{}",
        contentType: "application/json; charset=utf-8",
        dataType: "json",
        cache: false,
        success: callback
    });
};

and call it with:

var callback = function(msg) {
      links = (typeof msg.d) == 'string' ? eval('(' + msg.d + ')') : msg.d;
      // do other stuff with links here
}

Favorites(callback);

Answer 2

Your problem is that the HTTP request you're making is asnychronous and your Favorites function returns before the Ajax request has come back. You will need to change your function so that it accepts a callback to be executed once the response has come back:

function Favorites(callback) {
    $.ajax({
        type: "GET",
        url: "/Services/Favorite.svc/Favorites",
        data: "{}",
        contentType: "application/json; charset=utf-8",
        dataType: "json",
        cache: false,
        success: function(msg) {
            var links = (typeof msg.d == 'string') ? eval('(' + msg.d + ')') : msg.d;
            callback(links);
        }
    });
};

Favorites( function(links) { alert(links); } );

Aside: convention is that only functions intended to be used as constructors should start with a capital letter, so your function would be better named as favorites .