在Ajax success回调函数中通过超链接访问Controller

var pageData={page:1,pageSize:2 };
        
    $.ajax({
         async: true,
         type: "post",
         url: "user/showAllUser.do",
         traditional: true,
          data:pageData,
         dataType: "json",
         cache: true,
         success: function(data) {

           alert(data)

           window.location.href = "/manageSystem/user/mytest.do" //经测试这里的url是要写上项目名的  

         //$("#mybody").html("点击<a href='/manageSystem/user/mytest.do'>此处</a>跳转。");
     

         }
     });