i have 2 actions
public ActionResult FilesAdd(int id)
{
FillParentMenuDDL(id);
return View();
}
[HttpPost]
public ActionResult FilesAdd(int id)
{
//some logic...
FillParentMenuDDL(id);
return View();
}
but it is error because of same parameters, but i need only one parameter. first i call page /action/id and then i submit it for example with id and uploaded file, but i access to file using request.files[0]. so what the solution with controllers and same parameters? i see only declare FilesAdd(int? id) in one controller
.Net MVC has an ActionNameAttribute for this purpose. Rename your second action to something like FilesAddPost and then use ActionNameAttribute("FilesAdd")
public ActionResult FilesAdd(int id)
{
FillParentMenuDDL(id);
return View();
}
[HttpPost]
[ActionName("FilesAdd")]
public ActionResult FilesAddPost(int id)
{
//some logic...
FillParentMenuDDL(id);
return View();
}
Add an (unused) form parameter to the POST action. That will make the method signatures different.
[HttpPost]
public ActionResult FilesAdd(int id, FormCollection form)
{
//some logic...
FillParentMenuDDL(id);
return View();
}
You can control the action of the submitted form, it doesn't have to go to the same action.
// Works under MVC 2.0
<% using (Html.BeginForm("action", "controller", FormMethod.Post)) { %>
// code
<% } %>
Related
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.
In ASP.NET MVC, I have recently found out that:
This doesn't work and results in an HTTP 404.
public class TestController : Controller
{
[HttpGet]
[HttpPost]
public ActionResult Index(TestModel model)
{
return View(model);
}
}
This works fine.
public class TestController : Controller
{
public ActionResult Index(TestModel model)
{
return View(model);
}
}
This also works fine:
public class TestController : Controller
{
[HttpGet]
[ActionName("Index")]
public ActionResult GetIndex(TestModel model)
{
return View("Index", model);
}
[HttpPost]
[ActionName("Index")]
public ActionResult PostIndex(TestModel model)
{
return View("Index", model);
}
}
I would like an explanation of why the first version doesn't work, but the other two do work. I would also appreciate if someone could tell me how I can modify the first version to make it work. I like the first version because it is more concise (1 method rather than 2) and also filters out unnecessary HTTP methods.
HTTP verb attributes are mutually exclusive, a request cannot be both GET and POST at the same time. Instead, you have to do this:
[AcceptVerbs(HttpVerbs.Get | HttpVerbs.Post)]
public ActionResult Index(TestModel model) { ... }
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);
}
If you have 5 list views in a controller and in each list you can go to edit, details or delete.
On the edit, details and delete page youo have a link 'return to list'.
What's the best method to 'remember' to which list action you must return?
As a solution I've put some info like CurrentAction in the ViewModel and used that in the View. But if you want to use this with different controllers instead of one...
(You can use a Currentcontroller, CurrentArea, but that's not a 'beautifull' solution)
public class MyController : Controller
{
public ActionResult Index()
{
...
}
public ActionResult List2()
{
...
}
public ActionResult List3()
{
...
}
public ActionResult List4()
{
...
}
public ActionResult Create(...)
{
...
}
[HttpPost]
public ActionResult Create(...)
{
...
}
public ActionResult Edit(...)
{
...
}
[HttpPost]
public ActionResult Edit(...)
{
...
}
public ActionResult Delete(...)
{
...
}
[HttpPost]
public ActionResult Delete(...)
{
...
}
}
thanks
Filip
You can use Request.UrlReferrer Property to examine from where did user come to delete ot edit screens.Then bind url to return to list command.
You can set TempData["ReturnUrl"] in your caller action and then use it to set the url of return to list hyperlink.
So basically in my UserController.cs class I have an Index method that returns the ActionResult to display the dashboard for the user. On this page is a html button with a type of submit. When I hit this button I want to capture it on the server side log the user out.
How can I do this since I'm not passing information back and the method signature ends up being the same.
Thanks,
Mike
[Authorize]
public ActionResult Index()
{
return View();
}
[Authorize]
[AcceptVerbs(HttpVerbs.Post)]
public ActionResult Index()
{
FormsAuthentication.SignOut();
return RedirectToAction("Index", "Home");
}
Use ActionNameAttribute:
[AcceptVerbs(HttpVerbs.Get), ActionName("Index"), Authorize]
public ActionResult IndexGet()
{
return View();
}
[AcceptVerbs(HttpVerbs.Post), ActionName("Index"), Authorize]
public ActionResult IndexPost()
{
FormsAuthentication.SignOut();
return RedirectToAction("Index", "Home");
}
Use:
[Authorize]
[AcceptVerbs(HttpVerbs.Post)]
public ActionResult Index(FormCollection values)
{
FormsAuthentication.SignOut();
return RedirectToAction("Index", "Home");
}
Or
[Authorize]
[AcceptVerbs(HttpVerbs.Post), ActionName("Post")]
public ActionResult IndexPost()
{
FormsAuthentication.SignOut();
return RedirectToAction("Index", "Home");
}
As you know in C# you cannot have two methods with the same name and same arguments within the same class. You could either add some dummy parameter or I would advice you to rename the second action to SignOut which seems more semantically correct and better reflecting what this action actually does.