解决ajax异步传输数据，return返回为undefined的问题

1. function GetUserInfo(tp) {
2. var username;
3. $.ajax({
4. type: "POST",
5. cache: false,
6. data: "type=exlogin&tp=" + tp,
7. url: "Handle/OpeartionHandler.ashx",
8. success: function(userinfo) {
9. username = userinfo;
10. }, error: function(data) {
11. username = "";
12. }
13. });
14. return username;
15. }

function GetUserInfo(tp) {
    var username;
    $.ajax({
        type: "POST",
        cache: false,
        data: "type=exlogin&tp=" + tp,
        url: "Handle/OpeartionHandler.ashx",
        success: function(userinfo) {
            username = userinfo;
        }, error: function(data) {
            username = "";
        }
    });
    return username;
}

对于此方法调用之后会一直返回undefined，原因是Jquery的ajax是异步的，所以大多时候没执行完AJAX就return htmlcontent了，所以会一直返回undefined,

解决方法：添加async: false,即修改此方法为同步

1. function GetUserInfo(tp) {
2. var username;
3. $.ajax({
4. type: "POST",
5. cache: false,
6. data: "type=exlogin&tp=" + tp,
7. async: false,
8. url: "Handle/OpeartionHandler.ashx",
9. success: function(userinfo) {
10. username = userinfo;
11. }, error: function(data) {
12. username = "";
13. }
14. });
15. return username;
16. }