Jquery和JS用外部变量获取Ajax返回的参数值的方法实例(超简单)

复制代码 代码如下:

 var csj_data;

      $.ajax
       ({
           cache: false,
           async: false,
           type: 'post',
           data: { aaa: "1" },
           url: "../handle/Ladder_Fee_Code.ashx?ajaxaction=Select_FangAn",
           success: function (data) {

 
               csj_data = data;
               alert(data);
           }
       });

alert(csj_data);