ASP.Net MVC sending model to view without using URL - asp.net-mvc

I am using the same model between 2 views, but when posting the model to the second view it puts all the previously entered data in the URL. Is it possible to send the populated model to the second view without posting the data in the URL?
Controller code:
[HttpPost]
public ActionResult ViewExample1(.Models.RegisterModel model)
{
if (ModelState.IsValid)
{
return RedirectToAction("ViewExample2", model);
}
return View(model);
}
public ActionResult ViewExample2(Models.RegisterModel model)
{
return View(model);
}
Second view code where I use HiddenFor to persist the data when this view is posted back:
<% using (Html.BeginForm(null, null, FormMethod.Post, new { id="ViewExample2"})) { %>
<%: Html.HiddenFor(model => model.UserName)%>
<% } %>

When you redirect to an action with RedirectToAction(), you're doing that by GET. So the Framework passes your view model in the url to the action.
I'd suggest you to do this:
[HttpPost]
public ActionResult ViewExample1(Models.RegisterModel model)
{
if (ModelState.IsValid)
{
// Do the work you want to do in the ViewExample2 action here!
// ... and then return the ViewExample2 view
return View("ViewExample2", model);
}
return View(model);
}
// This action is not needed anymore
/*public ActionResult ViewExample2(Models.RegisterModel model)
{
return View(model);
}*/

My guess is that you're using a form tag (rather than BeginForm) and you aren't specifying a method, so it defaults to using a GET rather than a POST.
Convert to using a BeginForm, or add the method.

Related

show status after post and keep entered model as is on MVC view

I have this mvc controller that add a customer to the database called CustomerController. This Controller has one ActionResult called Add. It works as it is but I want to display a status message after the user hit submit, and I want all information added to the model be kept on the page as is. How can I keep the all the entered text in the form fields and also show a status message after the form has been posted?
public ActionResult Add()
{
// This is the empty view the user see when he is about to add new form data
return View(new CreateSupplierViewModel());
}
public ActionResult AddNew(CreateSupplierViewModel model)
{
// I post to this and need to display the status of this on the view with the entered text fields as is
return RedirectToAction("Add", "Supplier");
}
You need to refactor your code as below :
The CustomerController :
public ActionResult Add()
{
return View(new CreateSupplierViewModel());
}
public ActionResult Add(CreateSupplierViewModel model)
{
return View(model);
}
public ActionResult AddNew(CreateSupplierViewModel model)
{
return RedirectToAction("Add", "Supplier", model);
}
Your SupplierController
public ActionResult Add(CreateSupplierViewModel model)
{
//save the entity
Viewbag.Message ="submit result";
return RedirectToAction("Add", "Customer", model);
}
The Customer/Add.cshtml (show the submit result in view)
#if( Viewbag.Message != null)
{
<p> #Viewbag.Message </p>
}

Basic wizard, and passing model from one view to another?

I'm new to MVC and trying to create a wizard-style series of views, passing the same model instance from one view to the next, where the user completes a little more information on each form. The controller looks something like this:-
[HttpGet]
public ActionResult Step1()
{
return View();
}
[HttpPost]
public ActionResult Step1(MyModel model)
{
if (!ModelState.IsValid)
return View(model);
return View("Step2", model);
}
[HttpPost]
public ActionResult Step2(MyModel model)
{
if (!ModelState.IsValid)
return View(model);
return View("Step3", model);
}
// etc..
Questions:-
When I submit the form from the Step1 view, it calls the Step1 POST method and results in the Step2 view being displayed in the browser. When I submit the form on this view, it calls the Step1 POST method again! I got it to work by specifying the action and controller name in Html.BeginForm(), so I'm guessing that the parameterless overload just POSTs back to the action that rendered the view?
I've noticed that the browser's address bar is out of sync with the current view - when I'm on the Step2 view it still shows the Step1 URL, and when on Step3 it shows the Step2 URL. What's going on?
Another approach I've seen for passing a model between views is to put the model in TempData then use RedirectToAction(). What are the pros and cons of this method versus what I'm currently doing?
I won't be providing any "back" buttons of my own in the wizard. Are there any pitfalls to be aware of regarding the browser's back button, and do either of the above two approaches help (or hinder)?
Edit
Prompted by #StephenMuecke's comment I've now rewritten this to use a single view. I tried this once before but had difficulties round-tripping a "step number" to keep track of where I was in the wizard. I was originally using a hidden field created with #Html.HiddenFor', but this wasn't updating as the underlying model property changed. This appears to be "by design", and the workaround is to create the hidden field using vanilla HTML (
Anyway the one-view wizard is now working. The only problem is the old chestnut of the user being able to click the back button after they have completed the wizard, make a change, and resubmit a second time (resulting in a second DB record).
I've tried adding [OutputCache(NoStore = true, Duration = 0, VaryByParam = "None")] to my POST method, but all this does is display (in my case) a Chrome error page suggesting that the user clicks refresh to resubmit the form. This isn't user friendly and doesn't prevent a second submit.
you can use RedirectToAction() in this case without worrying about TempData. Just add your model as a parameter to each action and use RedirectToAction("Step2", model);
[HttpGet]
public ActionResult Step1()
{
return View();
}
[HttpPost]
public ActionResult Step1(MyModel model)
{
if (!ModelState.IsValid)
return View(model);
return RedirectToAction("Step2", model);
}
[HttpGet]
public ActionResult Step2(MyModel model)
{
return View(model);
}
[HttpPost]
public ActionResult Step2(MyModel model)
{
if (!ModelState.IsValid)
return View(model);
return RedirectToAction("Step3", model);
}
// etc..
The answer to #1 is found in #2.. if you dont specify the Action in you Html.BeginForm() it posts to the current url.
Using TempData to avoid model displaying in url.
[HttpGet]
public ActionResult Step1()
{
return View();
}
[HttpPost]
public ActionResult Step1(MyModel model)
{
if (!ModelState.IsValid)
return View(model);
TempData["myModel"] = model;
return RedirectToAction("Step2");
}
[HttpGet]
public ActionResult Step2()
{
var model = TempData["myModel"] as MyModel;
return View(model);
}
[HttpPost]
public ActionResult Step2(MyModel model)
{
if (!ModelState.IsValid)
return View(model);
TempData["myModel"] = model;
return RedirectToAction("Step3");
}
// etc..
Another option would be to add the name of the next action to ViewBag and set your actionName in each BeginForm()
[HttpGet]
public ActionResult Step1()
{
ViewBag.NextStep = "Step1";
return View();
}
[HttpPost]
public ActionResult Step1(MyModel model)
{
if (!ModelState.IsValid)
{
ViewBag.NextStep = "Step1";
return View(model);
}
ViewBag.NextStep = "Step2";
return View("Step2", model);
}
[HttpPost]
public ActionResult Step2(MyModel model)
{
if (!ModelState.IsValid)
{
ViewBag.NextStep = "Step2";
return View(model);
}
ViewBag.NextStep = "Step3";
return View("Step3", model);
}
//View
#using (Html.BeginForm((string)ViewBag.NextStep, "ControllerName"))
{
}
I'd prefer to add NextStep as a property to MyModel and using that instead of using ViewBag though.
I understand the thought behind your approach and don't have any issues with it. Unfortunately, I don't believe that ASP.NET MVC is geared very well for passing the the same view model (with data!) between different actions.
Typically, the scaffolded actions in the controller will either create a model item or find it by identifier in the database.
I don't know if this would help, but you could try to save it to the database on every step, and then retrieve it by identifier, or you could also save it to a session and grab it that way.
One issue I do see with your approach is you have Step2 set as a get, yet you probably want to post data to it from Step1 instead of using a query string. You may need to reconcile that issue.

ASP.net MVC4 Clear Model Values

So I have a simple action in my controller. The project is a MVC Mobile Application.
public ActionResult Index()
{
return View();
}
this gives an form to enter data. I then handle the data in the post back.
[HttpPost]
public ActionResult Index(ScanViewModel model)
{
if (ModelState.IsValid)
{
Scan ns = new Scan();
ns.Location = model.Location;
ns.Quantity = model.Quantity;
ns.ScanCode = model.ScanCode;
ns.Scanner = User.Identity.Name;
ns.ScanTime = DateTime.Now;
_db.Scans.Add(ns);
_db.SaveChanges();
}
return View(model);
}
I want to clear the fields in the form and allow users to enter data again. However I get the exact same values back into my inputs. How can I clear them in the controller.
Just call this.ModelState.Clear()
You should follow the PRG pattern.
Just redirect to the Action method which is meant for the Create Screen. You can use the RedirectToAction method to do so.
RedirectToAction returns an HTTP 302 response to the browser, which causes the browser to make a GET request to the specified action.
[HttpPost]
public ActionResult Index(ScanViewModel model)
{
if(ModelState.IsValid)
{
//Code for save here
//..............
_db.SaveChanges();
return RedirectToAction("Index","User");
}
return View(model);
}
public ActionResult Index()
{
return View();
}
Assuming your controller name is UserController.

MVC3 Form Posted Value

Below Scenario, I think I must see the START text in my form when first loaded.
When I click send data button and submit, I was waiting to see FINISH text in my form.
Buy the START text never changes when I click the button and post the form...
Anybody can tell the problem?
MY CONTROLLER:
namespace MvcApplication1.Controllers
{
public class BuyController : Controller
{
public ActionResult Index(BuyModel model)
{
if (Request.HttpMethod == "GET")
{
model.Message= "START";
return View(model);
}
else
{
BuyModel newModel = new BuyModel();
newModel.Message= "FINISH";
return View(newModel);
}
}
}
}
MY VIEW :
#model MvcApplication1.Models.BuyModel
#using (Html.BeginForm("Index", "Buy", FormMethod.Post))
{
#Html.TextBoxFor(s => s.Message)
<button type="submit" >Send</button>
}
MY MODEL:
public class BuyModel
{
public string Message { get; set; }
}
public class BuyController : Controller
{
public ActionResult Index()
{
BuyModel model = new BuyModel();
model.Message= "START";
return View(model);
}
[HttpPost]
public ActionResult Index(BuyModel model)
{
model = new BuyModel();
model.Message= "FINISH";
ModelState.Clear(); // the fix
return View(model);
}
}
View:
#model MvcApplication1.Models.BuyModel
#using (Html.BeginForm("Index", "Buy"))
{
#Html.TextBoxFor(s => s.Message)
<button type="submit" >Send</button>
}
Your issue is because your original code, that Action Method will only be executed as an HTTP GET request. ASP.NET MVC allows you to specify a post with the [HttpPost] attribute (see above code).
I'm not sure what you are getting at with your POST desired-behavior. It seems like you are just wiping out whatever form values are pushed on the POST. So modify my above code accordingly, but it should give you the general idea.
Edit: it seems to be that the text box is retaining its value after the POST. It's not just with "START", but if you type anything into that text box and hit submit, you'll have a POST with the exact same text in the text box that was there when you submitted the form.
Edit Edit: see the changed code. Call ModelState.Clear() in your POST action method and you'll have the right value reflected.
If you are posting, and not returning a RedirectResult, by default the helpers will use the value from ModelState. You either need to clear ModelState or have a different approach.
The PRG (post redirect get) pattern in MVC is very important. So if its a post, and you aren't redirecting, the helpers assume there is an error that needs to be corrected and the value is pulled from ModelState.

Passing a variable from [HttpPost] method to [HttpGet] method

I am redirecting the view from [HttpPost] method to [HttpGet] method. I have gotten it to work, but want to know if this is the best way to do this.
Here is my code:
[HttpPost]
public ActionResult SubmitStudent()
{
StudentViewModel model = TempData["model"] as StudentResponseViewModel;
TempData["id"] = model.Id;
TempData["name"] = model.Name;
return RedirectToAction("DisplayStudent");
}
[HttpGet]
public ActionResult DisplayStudent()
{
ViewData["id"] = TempData["id"];
ViewData["name"] = TempData["name"];
return View();
}
View:
<%# Page
Language="C#"
Inherits="System.Web.Mvc.ViewPage"
%>
<html>
<head runat="server">
<title>DisplayStudent</title>
</head>
<body>
<div>
<%= ViewData["id"]%> <br />
<%= ViewData["name"]%>
</div>
</body>
</html>
There are basically 3 techniques in ASP.NET MVC to implement the PRG pattern.
TempData
Using TempData is indeed one way of passing information for a single redirect. The drawback I see with this approach is that if the user hits F5 on the final redirected page he will no longer be able to fetch the data as it will be removed from TempData for subsequent requests:
[HttpPost]
public ActionResult SubmitStudent(StudentResponseViewModel model)
{
if (!ModelState.IsValid)
{
// The user did some mistakes when filling the form => redisplay it
return View(model);
}
// TODO: the model is valid => do some processing on it
TempData["model"] = model;
return RedirectToAction("DisplayStudent");
}
[HttpGet]
public ActionResult DisplayStudent()
{
var model = TempData["model"] as StudentResponseViewModel;
return View(model);
}
Query string parameters
Another approach if you don't have many data to send is to send them as query string parameters, like this:
[HttpPost]
public ActionResult SubmitStudent(StudentResponseViewModel model)
{
if (!ModelState.IsValid)
{
// The user did some mistakes when filling the form => redisplay it
return View(model);
}
// TODO: the model is valid => do some processing on it
// redirect by passing the properties of the model as query string parameters
return RedirectToAction("DisplayStudent", new
{
Id = model.Id,
Name = model.Name
});
}
[HttpGet]
public ActionResult DisplayStudent(StudentResponseViewModel model)
{
return View(model);
}
Persistence
Yet another approach and IMHO the best consists into persisting this model into some data store (like a database or something and then when you want to redirect to the GET action send only an id allowing for it to fetch the model from wherever you persisted it). Here's the pattern:
[HttpPost]
public ActionResult SubmitStudent(StudentResponseViewModel model)
{
if (!ModelState.IsValid)
{
// The user did some mistakes when filling the form => redisplay it
return View(model);
}
// TODO: the model is valid => do some processing on it
// persist the model
int id = PersistTheModel(model);
// redirect by passing the properties of the model as query string parameters
return RedirectToAction("DisplayStudent", new { Id = id });
}
[HttpGet]
public ActionResult DisplayStudent(int id)
{
StudentResponseViewModel model = FetchTheModelFromSomewhere(id);
return View(model);
}
Each method has its pros and cons. Up to you to choose which one suits best to your scenario.
If you are inserting this data into a database then you should redirect them to a controller action that has this data in the route:
/Students/View/1
You can then write code in the controller to retrieve the data back from the database for display:
public ActionResult View(int id) {
// retrieve from the database
// create your view model
return View(model);
}
One of the overrides of RedirectToAction() looks like that:
RedirectToAction(string actionName, object routeValues)
You can use this one as:
[HttpPost]
public ActionResult SubmitStudent()
{
StudentViewModel model = TempData["model"] as StudentResponseViewModel;
return RedirectToAction("DisplayStudent", new {id = model.ID, name = model.Name});
}
[HttpGet]
public ActionResult DisplayStudent(string id, string name)
{
ViewData["id"] = TempData["id"];
ViewData["name"] = TempData["name"];
return View();
}
Hope that works.
This is the classic Post-Redirect-Get pattern (PRG) and it looks fine but I would add one bit of code. In the DisplayStudent method check if your TempData variables are not null otherwise do a redirect to some default Index action. This is in case a user presses F5 to refresh the page.
public ActionResult DisplayStudent()
{
if(TempData["model"] == null)
{
return RedirectToAction("Index");
}
var model = (StudentResponseViewModel)TempData["model"];
return View(model);
}
public ViewResult Index()
{
IEnumerable<StudentResponseViewModel> students = GetAllStudents();
return View(students);
}

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