How do I redirect to the previous action in ASP.NET MVC?

try:

public ActionResult MyNextAction()
{
    return Redirect(Request.UrlReferrer.ToString());
}

alternatively, touching on what darin said, try this:

public ActionResult MyFirstAction()
{
    return RedirectToAction("MyNextAction",
        new { r = Request.Url.ToString() });
}

then:

public ActionResult MyNextAction()
{
    return Redirect(Request.QueryString["r"]);
}

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If you want to redirect from a button in the View you could use:

@Html.ActionLink("Back to previous page", null, null, null, new { href = Request.UrlReferrer})

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If you are not concerned with unit testing then you can simply write:

return Redirect(ControllerContext.HttpContext.Request.UrlReferrer.ToString());

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A suggestion for how to do this such that:

the return url survives a form's POST request (and any failed validations) the return url is determined from the initial referral url without using TempData[] or other server-side state handles direct navigation to the action (by providing a default redirect)

.

public ActionResult Create(string returnUrl)
{
    // If no return url supplied, use referrer url.
    // Protect against endless loop by checking for empty referrer.
    if (String.IsNullOrEmpty(returnUrl)
        && Request.UrlReferrer != null
        && Request.UrlReferrer.ToString().Length > 0)
    {
        return RedirectToAction("Create",
            new { returnUrl = Request.UrlReferrer.ToString() });
    }

    // Do stuff...
    MyEntity entity = GetNewEntity();

    return View(entity);
}

[AcceptVerbs(HttpVerbs.Post)]
public ActionResult Create(MyEntity entity, string returnUrl)
{
    try
    {
        // TODO: add create logic here

        // If redirect supplied, then do it, otherwise use a default
        if (!String.IsNullOrEmpty(returnUrl))
            return Redirect(returnUrl);
        else
            return RedirectToAction("Index");
    }
    catch
    {
        return View();  // Reshow this view, with errors
    }
}

You could use the redirect within the view like this:

<% if (!String.IsNullOrEmpty(Request.QueryString["returnUrl"])) %>
<% { %>
    <a href="<%= Request.QueryString["returnUrl"] %>">Return</a>
<% } %>

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In Mvc using plain html in View Page with java script onclick

<input type="button" value="GO BACK" class="btn btn-primary" 
onclick="location.href='@Request.UrlReferrer'" />

This works great. hope helps someone.

@JuanPieterse has already answered using @Html.ActionLink so if possible someone can comment or answer using @Url.Action

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I'm using .Net Core 2 MVC , and this one worked for me, in the controller use HttpContext.Request.Headers["Referer"];

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Pass a returnUrl parameter (url encoded) to the change and login actions and inside redirect to this given returnUrl. Your login action might look something like this:

public ActionResult Login(string returnUrl) 
{
    // Do something...
    return Redirect(returnUrl);
}

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You could return to the previous page by using ViewBag.ReturnUrl property.

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To dynamically construct the returnUrl in any View, try this:

@{
    var formCollection =
        new FormCollection
            {
                new FormCollection(Request.Form),
                new FormCollection(Request.QueryString)
            };

    var parameters = new RouteValueDictionary();

    formCollection.AllKeys
        .Select(k => new KeyValuePair<string, string>(k, formCollection[k])).ToList()
        .ForEach(p => parameters.Add(p.Key, p.Value));
}

<!-- Option #1 -->
@Html.ActionLink("Option #1", "Action", "Controller", parameters, null)

<!-- Option #2 -->
<a href="/Controller/Action/@object.ID?returnUrl=@Url.Action(ViewContext.RouteData.Values["action"].ToString(), ViewContext.RouteData.Values["controller"].ToString(), parameters)">Option #2</a>

<!-- Option #3 -->
<a href="@Url.Action("Action", "Controller", new { object.ID, returnUrl = Url.Action(ViewContext.RouteData.Values["action"].ToString(), ViewContext.RouteData.Values["controller"].ToString(), parameters) }, null)">Option #3</a>

This also works in Layout Pages, Partial Views and Html Helpers

Related: MVC3 Dynamic Return URL (Same but from within any Controller/Action)

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For ASP.NET Core You can use asp-route-* attribute:

<form asp-action="Login" asp-route-previous="@Model.ReturnUrl">

Other in details example: Imagine that you have a Vehicle Controller with actions

Index

Details

Edit

and you can edit any vehicle from Index or from Details, so if you clicked edit from index you must return to index after edit and if you clicked edit from details you must return to details after edit.

//In your viewmodel add the ReturnUrl Property
public class VehicleViewModel
{
     ..............
     ..............
     public string ReturnUrl {get;set;}
}



Details.cshtml
<a asp-action="Edit" asp-route-previous="Details" asp-route-id="@Model.CarId">Edit</a>

Index.cshtml
<a asp-action="Edit" asp-route-previous="Index" asp-route-id="@item.CarId">Edit</a>

Edit.cshtml
<form asp-action="Edit" asp-route-previous="@Model.ReturnUrl" class="form-horizontal">
        <div class="box-footer">
            <a asp-action="@Model.ReturnUrl" class="btn btn-default">Back to List</a>
            <button type="submit" value="Save" class="btn btn-warning pull-right">Save</button>
        </div>
    </form>

In your controller:

// GET: Vehicle/Edit/5
    public ActionResult Edit(int id,string previous)
    {
            var model = this.UnitOfWork.CarsRepository.GetAllByCarId(id).FirstOrDefault();
            var viewModel = this.Mapper.Map<VehicleViewModel>(model);//if you using automapper
    //or by this code if you are not use automapper
    var viewModel = new VehicleViewModel();

    if (!string.IsNullOrWhiteSpace(previous)
                viewModel.ReturnUrl = previous;
            else
                viewModel.ReturnUrl = "Index";
            return View(viewModel);
        }



[HttpPost]
    public IActionResult Edit(VehicleViewModel model, string previous)
    {
            if (!string.IsNullOrWhiteSpace(previous))
                model.ReturnUrl = previous;
            else
                model.ReturnUrl = "Index";
            ............. 
            .............
            return RedirectToAction(model.ReturnUrl);
    }

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