返回新视图时如何保留网址

Question

A beginner question - I have a HomeController, HomeModel and HomeView. When the user comes to the http://page/Home page, then Index method is executed and he can fill out some controls. After he clicks on a button (postback), the Process action is executed and in the case of an error, the application calls the ModelState.AddModelError method. Then the Index action is called again and I can display the error on the page.

This works OK, but the problem is that after the postback the new url is http://page/Home/Index instead of http://page/Home . Any idea how to prevent this?

PS - I tried this solution but then the new url was something like http://page/Home?...long string of serialized ModelState data...

My Controller:

[HttpGet]
public ActionResult Index(MyModel model)
{
    return View(model);
}

[HttpPost]
public ActionResult Process(MyModel model)
{
    if (...error...)
    {
        model.SetErrorState();
        ModelState.AddModelError("ProcessError", "error message");
        return View("Index", model);
    }
    else
    {
        // do something...
        model.SetSuccessState();
        return View("Index", model);
    }
}

Answer 1

The problem is you're pushing to a new URL for the HttpPost action. If you change this to a HttpPost version of your Home Action you can neatly return to the page without the URL changing on error.

eg

[HttpGet]
public ActionResult Index(ImportData model)
{
    return View(model);
}

[HttpPost]
public ActionResult Index(MyModel model, FormCollection data)
{
    if (...error...)
    {
        model.SetErrorState();
        ModelState.AddModelError("ProcessError", "error message");
        return View(model);
    }
    else
    {
        // do something...
        model.SetSuccessState();
        return View(model);
    }
}