.net mvc 用jquery ajax成功后页面跳转。

 ?$.ajax({ ????????????//url: ‘http://192.168.0.102/webApiDemo/api/WebApiTest/PostStudentInforOnePara‘, ????????????//url: "http://localhost/webapidemo/api/WebApiTest/PostStudentInforOnePara", ????????????url: "UserLogin/Login", ????????????type: ‘GET‘, ????????????data: { bname: txtbname, psw: txtpsw },//一个参数时，必须这样写，webapi中http的请求体里面去取参数才能取到。 ????????????contentType: "json", ????????????async: true,//异步 ????????????success: function (data) //成功后的回调方法 ????????????{ ????????????????if (data =="suscess") ?????????? ?????window.location.href = "@Url.Action("Index","Home")";//可以跳转. ????????????}, ????????????error: function () { ????????????????alert("失败！"); ????????????????//window.location.href = "EasyUILoutMain.aspx";//可以跳转. ????????????}

 ??public IActionResult Login(string bname,string psw) ???????{ ???????????if (!ModelState.IsValid) ???????????{ ???????????????return View(); ???????????} ???????????if (string.IsNullOrEmpty(bname)) ???????????{ ???????????????bname = Request.Query["bname"].ToString(); ????????????} ???????????if(string.IsNullOrEmpty(psw)) ???????????{ ???????????????psw = Request.Query["psw"].ToString(); ???????????} ???????????if (bname.Equals("longdb") && psw.Equals("123")) ???????????{ ???????????????return Json("suscess"); ???????????} ???????????return View(); ???????}