How Return to View after ajax post to controller?

Question

I'm trying post data from View to controller and then return back to view to get feedback from controller. This is my script in view.

<script type="text/javascript">
$(document).ready(function () { 
    $("Submit").click(function (e) {
        e.preventDefault();

        var token = $('input[name=__RequestVerificationToken]').val();
        console.log(token);

        var dataToPost = $(this).serialize();
        $.ajax({
            url: "Process/Payment"
            , method: "POST"
            , data: {
                __RequestVerificationToken: token
                , page: dataToPost
            }
            , done: function (data) {
                window.location.replace (data.newUrl);
                alert(data);
            }
            , error: function () {
                alert("Ajax call failed");

            }
        })
    })
})

and this is my controller. All data that wrap in 'Payment' model is in good shape, pass well from view to controller. And, I want to return back to view (url: "Process/Payment")

[HttpPost]
public ActionResult Allowance (Allowance allowance) {
    return Json(new {
            newUrl = Url.Action("Payment", "Process")
        }
    );

but when the page return to localhost, I just get this page.

How should I do to return the page that look like before I hit submit button as POST ?

why "window.location.replace ()" can't redirect page to the original page ?

Thx's

Why in the world are you making an ajax call when you want to redirect (it makes no sense at all - just make a normal submit) — , May 07 '18 at 10:51
And there is no such html element as `` so none of the code in you function will ever be executed (you should not be writing scripts without knowing at least the basics of how to debug them) - I assume you mean `$("#Submit")` assuming you have an element with `id="Submit"` — , May 07 '18 at 10:54
Hi all, I made mistake by putting this url: "Process/Payment". This should be the action method only without controller. Thank's All. — Daleman, May 08 '18 at 01:47
Why have you accepted an answer that has nothing to do with your issue (it just misleads other users) — , May 08 '18 at 08:52

score -1 · Answer 1 · answered May 07 '18 at 10:35

-1

Just add a window.location.href=="Your Desired Page"; on the success event of your ajax call. That'll do the thing.

Happy Coding.

answered May 07 '18 at 10:35

Maaz Ali

83
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How Return to View after ajax post to controller?

1 Answers1