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Question

I have a basic Edit method in my controller that redirects back to a top level listing     Index     when the edit succeeds  Standard behavior after MVC scaffolding   I am trying to change this Edit method to redirect back to the previous page  not Index    Since my Edit method wasn t using the default mapped input parameter    id     I first tried using that to pass in the previous URL   In my Edit    get    method  I used this line to grab the previous URL and it worked fine   ViewBag ReturnUrl   Request UrlReferrer    I then sent this return URL to the Edit    post    method by using my form tag like this    using  Html BeginForm new   id   ViewBag ReturnUrl       Now this is where the wheels fell off   I couldn t get the URL parsed from the id parameter properly       UPDATE  SOLVED      Using Garry s example as a guide  I changed my parameter from  id  to  returnUrl  and used a hidden field to pass my parameter  instead of the form tag    Lesson learned  Only use the  id  parameter how it was intended to be used and keep it simple   It works now   Here is my updated code with notes     First  I grab the previous URL using Request UrlReferrer as I did the first time                 GET   Question Edit 5      public ActionResult Edit int id                Question question   db Questions Find id           ViewBag DomainId   new SelectList db Domains   DomainId    Name   question DomainId           ViewBag Answers   db Questions                              AsEnumerable                                Select d   gt  new SelectListItem                                                               Text   d Text                                  Value   d QuestionId ToString                                    Selected   question QuestionId    d QuestionId                                            Grab the previous URL and add it to the Model using ViewData or ViewBag         ViewBag returnUrl   Request UrlReferrer          ViewBag ExamId   db Domains Find question DomainId  ExamId          ViewBag IndexByQuestion   string Format  IndexByQuestion  0    question QuestionId           return View question           and I now pass the returnUrl parameter from the Model to the  HttpPost  method using a hidden field in the form    using  Html BeginForm           lt input type  hidden  name  returnUrl  value   ViewBag returnUrl    gt            In the  HttpPost  method we pull the parameter from the hidden field and Redirect to it                    POST   Question Edit 5       HttpPost      public ActionResult Edit Question question  string returnUrl     Add parameter               int ExamId   db Domains Find question DomainId  ExamId          if  ModelState IsValid                        db Entry question  State   EntityState Modified              db SaveChanges                  return RedirectToAction  Index                return Redirect returnUrl                     ViewBag DomainId   new SelectList db Domains   DomainId    Name   question DomainId           return View question

User · Answer

Here is just another option you couold apply for ASP NET MVC.

Normally you shoud use BaseController class for each Controller class.

So inside of it's constructor method do following.

public class BaseController : Controller
{
        public BaseController()
        {
            // get the previous url and store it with view model
            ViewBag.PreviousUrl = System.Web.HttpContext.Current.Request.UrlReferrer;
        }
}

And now in ANY view you can do like

<button class="btn btn-success mr-auto" onclick="  window.location.href = '@ViewBag.PreviousUrl'; " style="width:2.5em;"><i class="fa fa-angle-left"></i></button>

Enjoy!

User · Answer

For ASP NET Core You can use asp-route-  attribute    lt form asp-action  Login  asp-route-previous   Model ReturnUrl  gt    An 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  lt a asp-action  Edit  asp-route-previous  Details  asp-route-id   Model CarId  gt Edit lt  a gt   Index cshtml  lt a asp-action  Edit  asp-route-previous  Index  asp-route-id   item CarId  gt Edit lt  a gt   Edit cshtml  lt form asp-action  Edit  asp-route-previous   Model ReturnUrl  class  form-horizontal  gt           lt div class  box-footer  gt               lt a asp-action   Model ReturnUrl  class  btn btn-default  gt Back to List lt  a gt               lt button type  submit  value  Save  class  btn btn-warning pull-right  gt Save lt  button gt           lt  div gt       lt  form gt    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 lt VehicleViewModel gt  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

User · Answer

I am assuming  please correct me if I am wrong  that you want to re-display the edit page if the edit fails and to do this you are using a redirect   You may have more luck by just returning the view again rather than trying to redirect the user  this way you will be able to use the ModelState to output any errors too   Edit   Updated based on feedback   You can place the previous URL in the viewModel  add it to a hidden field then use it again in the action that saves the edits   For instance   public ActionResult Index         return View        HttpGet     This isn t required public ActionResult Edit int id          load object and return in view    ViewModel viewModel   Load id          get the previous url and store it with view model    viewModel PreviousUrl   System Web HttpContext Current Request UrlReferrer      return View viewModel       HttpPost  public ActionResult Edit ViewModel viewModel          Attempt to save the posted object if it works  return index if not return the Edit view again     bool success   Save viewModel      if  success              return Redirect viewModel PreviousUrl           else            ModelState AddModelError  There was an error          return View viewModel            The BeginForm method for your view doesn t need to use this return URL either  you should be able to get away with    model ViewModel   using  Html BeginForm                   lt input type  hidden  name  PreviousUrl  value   Model PreviousUrl    gt      Going back to your form action posting to an incorrect URL  this is because you are passing a URL as the  id  parameter  so the routing automatically formats your URL with the return path   This won t work because your form will be posting to an controller action that won t know how to save the edits   You need to post to your save action first  then handle the redirect within it

User · Answer

I know this is very late  but maybe this will help someone else  I use a Cancel button to return to the referring url  In the View  try adding this       ViewBag Title    quot Page title quot     Layout    quot   Views Shared  Layout cshtml quot      if  Request UrlReferrer    null          string returnURL   Request UrlReferrer ToString        ViewBag ReturnURL   returnURL         Then you can set your buttons href like this   lt a href  quot  ViewBag ReturnURL quot  class  quot btn btn-danger quot  gt Cancel lt  a gt   Other than that  the update by Jason Enochs works great

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