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Home/ Questions/Q 8393925
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Editorial Team
  • 0
Editorial Team
Asked: June 9, 20262026-06-09T19:55:54+00:00 2026-06-09T19:55:54+00:00

I have the following code to post to an httppost method in my controller:

  • 0

I have the following code to post to an httppost method in my controller:

$QuickLoginSubmit.click(function (e) {
        e.preventDefault();
        var LoginModel = {
            'UserName': $QuickEmail.val() == "" ? null : $QuickEmail.val(),
            'Password': $QuickPassword.val() == "" ? null : $QuickPassword.val(),
            'RememberMe': $QuickRemember.val() == "on" ? true : false
        };
        $.ajax({
            url: '/Account/LogOnAjax',
            type: 'POST',
            contentType: 'application/json',
            dataType: 'json',
            data: JSON.stringify(LoginModel),
            success: function (result) {
                if (result == true) {
                    window.location = "/Dashboard";
                } else {
                    $QuickLoginErrors.text(result);
                }
            }
        });
    });

The problem is, if they haven’t typed in anything in the $QuickEmail or $QuickPassword field, the ajax call returns the following error:

The view 'LogOnAjax' or its master was not found or no view engine supports the searched locations. The following locations were searched:<br>~/Views/Account/LogOnAjax.aspx<br>~/Views/Account/LogOnAjax.ascx<br>~/Views/Shared/LogOnAjax.aspx<br>~/Views/Shared/LogOnAjax.ascx<br>~/Views/Account/LogOnAjax.cshtml<br>~/Views/Account/LogOnAjax.vbhtml<br>~/Views/Shared/LogOnAjax.cshtml<br>~/Views/Shared/LogOnAjax.vbhtml</i>

But if the the two fields are filled out my ajax method call works fine. Here is my httppost method:

[HttpPost]
        public ActionResult LogOnAjax(LogOnModel model)
        {
            if (ModelState.IsValid)
            {
                if (Membership.ValidateUser(model.UserName, model.Password))
                {
                    FormsAuthentication.SetAuthCookie(model.UserName, model.RememberMe);
                    return Json(true);                  
                }
                else
                {
                    return Json("The user name or password provided is incorrect.");
                }
            }

            // If we got this far, something failed, redisplay form
            return View(model);
        }
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-09T19:55:56+00:00Added an answer on June 9, 2026 at 7:55 pm

    The problem is with the last line of your action code.

    You actually don’t have a view named LogOnAjax. So, you should use

    return RedirectToAction(YOUR_LOGON_ACTION_NAME);
    

    or to have your model back to the form, you can use

    return View(YOUR_LOGON_VIEW_NAME, model);
    

    to get what you want.

    Update:

    Besides, it’s better to use JsonResult if you are going to return a json object in all cases.

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