Ajax调用靠山的方法带参数的方法

Ajax调用后台的方法带参数的方法

    <script src="jquery.js"></script>    <script type="text/javascript">        var a = "aaaa";        $(function () {                      $.ajax({                type: "POST",                  contentType: "application/json",                 url: "Default.aspx/Time",                 data: '{"abc":"' + a + '"}',                dataType: 'json',                success: function (result) {                        alert(result.d);                }            });        });    </script>

     [WebMethod]     public static string Time(string abc)    {                return abc;           }

        var a = "aaaa";        var b = "bbbb";        $(function () {                      $.ajax({                type: "POST",                  contentType: "application/json",                 url: "Default.aspx/Time",                 data: "{a1:'" + a+ "',a2:'" + b + "'}",                              dataType:'json',                success: function (result) {                        alert(result.d);                }            });        });

   [WebMethod]     public static string Time(string a1,string a2)    {                return a1+a2;           }