I have an Account controller which have:
LoginForm (get action)
Login (post action)
RegisterForm (get action)
Register (post action)
In another controller's index action i use render them as:
Html.RenderAction("RegistrationForm", "Acount");
Html.RenderAction("LoginForm ", "Acount");
Everything works ok and I can register a new user, login and validate with unobtrusive validation.
The problem is when do some back-end validation in Register/Login action and if there is an error I don't know how to transfer the error to be rendered.
I've tried with PRG pattern and it works ok. I get the error displayed on the form with the preserved data, but PRG is not the way to do it.
What is an alternative solution to this problem without using ajax to validate or move those methods in the controller where the RegistrationForm/LoginForms are used?
I want to skip using TempData due to session usage in the background.
EDIT CODE SAMPLE:
class AccountController : SomeBaseController{
[HttpGet]
public PartialViewResult RegistrationForm()
{
return PartialView(new RegisterUser());
}
[HttpPost]
public ActionResult RegisterUser(RegisterUser user)
{
if (ModelState.IsValid)
{
var _user;// create domain user from Register user Model;
var _validationOutput = _userService.DoSomeAwsomeServerSideValidation(_user);// do some custom validation
if (_validationOutput.IsFault)
{
// we preseve tempdata in base controller OnActionExecuted
_validationOutput.ErrorMessages.ForEach(x => ModelState.AddModelError(_validationOutput.ErrorCode, _validationOutput));
// redirect to home controller custom error occured
return RedirectToAction("Index", "Home", user);
}
return RedirectToAction("RegistrationInfo");
}
return RedirectToAction("SomeUserInfoAction");
}
}
class HomeController : SomeBaseController {
Index(){
return View();
}}
HomeControllerMarkup {
#{Html.RenderAction("RegistrationForm", "Acount");}
#{Html.RenderAction("LoginForm", "Acount");}
}
You can manually add errors to your ModelState within your post controller using:
ModelState.AddModelError("", #"You didn't perform task XYZ");
You should then be able to return the view and display the errors in a validation summary:
#Html.ValidationSummary(false, "Login was unsuccessful because...")
Related
I want to redirect my signIn Post methode back to the index page (Index Page has login and register form on in) but with the modelstate so i can show the errors if the sign in failed.
I have Read multiple articles about this but they either are outdate or not for asp.net core. I can't find a solution. I have tried to store the ViewData or ModelState in TempData but that doesn't work.
[AllowAnonymous]
[HttpGet]
public IActionResult Index()
{
//How to access have the modelstate from SignIn here?
return View();
}
[AllowAnonymous]
[HttpPost]
[ValidateAntiForgeryToken]
public async Task<IActionResult> Index(SignInModel model)
{
if (ModelState.IsValid)
{
....
return RedirectToAction("","");
}
// here i need to save the modelstate
return RedirectToAction(nameof(Index));
}
Don't try to pass ModelState as it is, ASP.NET will override it. But you can pass anything else. You index method has to support the state as a parameter:
public IActionResult Index(bool? IsValidAuth = null)
{
if(IsValidAuth!=true) {} // do something
}
Then you can pass the state in the second parameter of RedirectToAction:
public async Task<IActionResult> Index(SignInModel model)
{
// ...
return RedirectToAction(nameof(Index), new {IsValidAuth = false});
}
You can store your ModelState Validation error on TempData and will able to get it on your login method to show this error
you have to inject signmanager as service in configure to login
var result = await signInManager.PasswordSignInAsync(
model.Email, model.Password, model.RememberMe, true);
This will handel the user sign process through your application.
Hello everyone I would like to ask how to redirect from one view to another. Here is my view
#model IEnumerable<test.Models.contents>
#using test
#if(Request.IsAuthenticated) {
<text>Welcome<strong>#User.Identity.Name</strong>
</text>
}
else
{
???
}
Don't do any redirects inside the view. That's not its responsibility. The responsibility of a view is to display data that is passed to it from the controller action under the form of a view model.
Do this redirect inside the controller action that is rendering this view. For example you could decorate it with the [Authorize] attribute. This way if the user is not authorized he will be redirected to the loginUrl you specified in your web.config:
[Authorize]
public ActionResult SomeAction()
{
return View();
}
and if you wanted to redirect to some particular view you could simply write a custom Authorize attribute and override the HandleUnauthorizedRequest method to specify the controller and action you wish to redirect to in case of user not being authenticated:
public class MyAuthorizeAttribute : AuthorizeAttribute
{
protected override void HandleUnauthorizedRequest(AuthorizationContext filterContext)
{
var values = new RouteValueDictionary(new
{
controller = "SomeController",
action = "NotAuthorized"
});
filterContext.Result = new RedirectToRouteResult(values);
}
}
and then decorate your action with it:
[MyAuthorize]
public ActionResult SomeAction()
{
return View();
}
Now inside the corresponding view you don't need to perform any tests. It is guaranteed that if you got as far as rendering this view the user is authenticated and you could welcome him directly:
#model IEnumerable<test.Models.contents>
#using test
Welcome <strong>#User.Identity.Name</strong>
I have a controller with an Index method that has several optional parameters for filtering results that are returned to the view.
public ActionResult Index(string searchString, string location, string status) {
...
product = repository.GetProducts(string searchString, string location, string status);
return View(product);
}
I would like to implement the PRG Pattern like below but I'm not sure how to go about it.
[HttpPost]
public ActionResult Index(ViewModel model) {
...
if (ModelState.IsValid) {
product = repository.GetProducts(model);
return RedirectToAction(); // Not sure how to handle the redirect
}
return View(model);
}
My understanding is that you should not use this pattern if:
You do not need to use this pattern unless you have actually stored some data (I'm not)
You would not use this pattern to avoid the "Are you sure you want to resubmit" message from IE when refreshing the page (guilty)
Should I be trying to use this pattern? If so, how would I go about this?
Thanks!
PRG Stands for Post-Redirect-Get. that means when you post some data to the server back, you should redirect to a GET Action.
Why do we need to do this ?
Imagine you have Form where you enter the customer registration information and clicking on submit where it posts to an HttpPost action method. You are reading the data from the Form and Saving it to a database and you are not doing the redirect. Instead you are staying on the same page. Now if you refresh your browser ( just press F5 button) The browser will again do a similar form posting and your HttpPost Action method will again do the same thing. ie; It will save the same form data again. This is a problem. To avoid this problem, We use PRG pattern.
In PRG, You click on submit and The HttpPost Action method will save your data (or whatever it has to do) and Then do a Redirect to a Get Request. So the browser will send a Get Request to that Action
RedirectToAction method returns an HTTP 302 response to the browser, which causes the browser to make a GET request to the specified action.
[HttpPost]
public ActionResult SaveCustemer(string name,string age)
{
//Save the customer here
return RedirectToAction("CustomerList");
}
The above code will save data and the redirect to the Customer List action method. So your browser url will be now http://yourdomain/yourcontroller/CustomerList. Now if you refresh the browser. IT will not save the duplicate data. it will simply load the CustomerList page.
In your search Action method, You dont need to do a Redirect to a Get Action. You have the search results in the products variable. Just Pass that to the required view to show the results. You dont need to worry about duplicate form posting . So you are good with that.
[HttpPost]
public ActionResult Index(ViewModel model) {
if (ModelState.IsValid) {
var products = repository.GetProducts(model);
return View(products)
}
return View(model);
}
A redirect is just an ActionResult that is another action. So if you had an action called SearchResults you would simply say
return RedirectToAction("SearchResults");
If the action is in another controller...
return RedirectToAction("SearchResults", "ControllerName");
With parameter...
return RedirectToAction("SearchResults", "ControllerName", new { parameterName = model.PropertyName });
Update
It occurred to me that you might also want the option to send a complex object to the next action, in which case you have limited options, TempData is the preferred method
Using your method
[HttpPost]
public ActionResult Index(ViewModel model) {
...
if (ModelState.IsValid) {
product = repository.GetProducts(model);
TempData["Product"] = product;
return RedirectToAction("NextAction");
}
return View(model);
}
public ActionResult NextAction() {
var model = new Product();
if(TempData["Product"] != null)
model = (Product)TempData["Product"];
Return View(model);
}
I got an action List
//[HttpGet] (will come back to that!)
public ViewResult List(int page = 1)
{
//blah blah blah
return View(viewModel);
}
In its view we render action:
#{
Html.RenderAction("UpdateSearch");
}
Action definitions:
[ChildActionOnly]
[HttpGet]
public PartialViewResult UpdateSearch()
{
// do something and display a form in view
return PartialView(so);
}
[HttpPost]
public RedirectToRouteResult UpdateSearch(Options searchOptions)
{
// do something and redirect to List
return RedirectToAction("List");
}
and I'm getting: Child actions are not allowed to perform redirect actions exception every time someone submits the form. I'm new to MVC3, but it looks like the redirection is also a POST, because if [HttpGet] above List method is uncommented "the resource cannot be found" happens.
How do I change Http method on redirection or am I doing something wrong? I did try to Bing it, but no success.
The redirect info is stored in response header. However, the response is already being sent when child action is run so headers can't be written.
In short, there's no way of performing a redirect from child action other than through the use of javascript on client side.
I have simple controller:
public class TestController : Controller
{
public ActionResult Test(string r)
{
return View();
}
}
I have simple View Test.cshtml:
<h2>#ViewContext.RouteData.Values["r"]</h2>
#using (Html.BeginForm("Test", "Test"))
{
<input type="text" name="r" />
<button>Submit</button>
}
I have route rule in Global.asax:
routes.MapRoute(
null,
"Test/{r}",
new { action = "Test", controller = "Test",
r = UrlParameter.Optional }
);
I want to make such thing: user types route value in input, press submit and controller redirects him to page Test/value. But controller show just page with name Test everytime. ViewContext.RouteData.Values["r"] is empty too. I check in debug, Test action recieves user value of r correctly.
How can I realize my idea?
Thanks.
I'm super late to the party, but just wanted to post a solution for reference. Let's assume that this form has more than just a strong as it's input. Assuming there are other inputs, we can wrap up the inputs of the form into a class in our model, called TestModel whose properties maps to the id's of the form's inputs.
In our post, we redirect to the get, passing in the route values we need in the URL. Any other data can then be shuttled to the get using a TempData.
public class TestController : Controller
{
[HttpGet]
public ActionResult Test(string r)
{
TestModel model = TempData["TestModel"] as TestModel;
return View(model);
}
[HttpPost]
public ActionResult Test(string r,TestModel model) //some strongly typed class to contain form inputs
{
TempData["TestModel"] = model; //pass any other form inputs to the other action
return RedirectToAction("Test", new{r = r}); //preserve route value
}
}
You cannot do this without javascript. There are two types of methods that exist when submitting a <form>: GET and POST. When you use POST (which is the default), the form is POSTed to the url but all data entered in input fields is part of the POST body, so it is not part of the url. When you use GET, the input fields data is part of the query string but of the form /Test?r=somevalue.
I wouldn't recommend you trying to send user input as part of the path but if you decide to go that route you could subscribe to the submit event of the form and rewrite the url:
$('form').submit(function() {
var data = $('input[name="r"]', this).val();
window.location.href = this.action + '/' + encodeURIComponent(data);
return false;
});
As far as you are saying to post the form to Html.BeginForm("Test", "Test") you will be always posted back to the same page.
A solution could be to use an explicit Redirect to the action using 'RedirectToAction' (in view) or you can use javascript to change the form's action:
<input type="text" name="r" onchange="this.parent.action = '\/Test\/'+this.value"/>