I have a list, select a record, press the edit button and get an update view (form) via ajax. Then I modify some property and press save. This form is used both for creating and updating data. I distinguish in the view via ViewBag, like this:
<form asp-action=#(ViewBag.Mode == "new" ? "AddSender" : "UpdateSender") asp-controller="Sender" method="post">
In the ajax call I get the properties of the edited model with this url: /MasterData/Sender/UpdateSender/3 (3 is here only an example). Now, when I save and post back the data, the update action doesn't get hit, because the html getting for the update view looks like this:
<form method="post" action="/MasterData/Sender/UpdateSender/3">
The question is, why is the action method like this? Why is the earlier get url used? I thought, it should be just UpdateSender, as intended.
To be complete, here are the related actions in this order:
[Route("{SenderId}")]
public IActionResult UpdateSender(long SenderId)
{
ViewBag.Mode = "update";
return PartialView("Sender", SenderRepository.GetSender(SenderId));
}
[HttpPost]
public IActionResult UpdateSender(Sender Sender)
{
SenderRepository.UpdateSender(Sender);
return RedirectToAction(nameof(List));
}
And I have on the top of the controller class this attribute: [Route("MasterData/[controller]/[action]")]
The POST action needs to have a route as well if the intention is to use attribute routing.
Add the Route attribute on your [HttpPost] action.
[HttpPost]
[Route("UpdateSender")]
public IActionResult UpdateSender(Sender Sender)
{
SenderRepository.UpdateSender(Sender);
return RedirectToAction(nameof(List));
}
Related
I have a strongly typed razor view for a model in my MVC 3 project. Basically its for editing the model.
The model contains an Id field for the database key and some other string fields (Its a viewModel and all but thats not the point of the question).
In the view I just have a form and a submit button and nothing else. When the View is posted to the controller the model in the controller has all fields empty EXCEPT for the Id field which seems to have been auto-magically filled up.
How and where does the Id field gets populated in the model without there being a corresponding 'input' element for it in the view.
This is probably a dumb question but I would appreciate even just a link to what I should read up on. Thanks.
I bet it comes from the url as route parameter.
For example you have the following controller:
public class HomeController: Controller
{
public ActionResult Index(int id)
{
vqr model = GetModel(id);
return View(model);
}
[HttpPost]
public ActionResult Index(MyViewModel model)
{
// the model.Id property will be automatically populated here
// because the request was POST /home/index/123
...
}
}
and the following view:
#model MyViewModel
#using (Html.BeginForm())
{
<button type="submit">OK</button>
}
Now you navigate to GET /home/index/123 and you get the following markup:
<form action="/home/index/123" method="post">
<button type="submit">OK</button>
</form>
Notice the action attribute of the form? That's where the id comes from. Basically the Html.BeginForm() helper uses the current url when generating the action attribute, and since the current url is /home/index/123 it is what gets used.
And because if you have left the default routes in your Global.asax, the {id} route token is used at the end of the url, the default model binder successfully binds it to the Id property of your view model.
You are probably hitting a URL similar to the following: /MyObject/Edit/15
This is then returning the page that you have your blank form on.
What happens next is you have an HTML.BeginForm() which is posting BACK to /MyObject/Edit/15
Now because of the post back having the same format your routing rules are picking up the '15' and binding it back to your id.
Have you added the ID field as a hidden field?
e.g.
#Html.HiddenFor(x=> x.ID)
UPDATE:
My model going into the save method is PartialViewModel, which in the save method, is pushed into the index's ContactViewModel and sent back. This wasn't clear.
I am playing around with MVC3, and have a contact controller with a SaveDetails action. The index cshtml has a partial with a form whose action is pointing to this controller.
When I submit the form not having completed it fully, thereby firing the validation, the url now contains the SaveDetails action name (http://localhost:7401/Contact/SaveDetails).
The form code is:
#using (Html.BeginForm("SaveDetails", "Contact")) {
...
}
The controller action looks like this:
public ActionResult SaveDetails(Models.PartialsViewModel pvm)
{
return View("Index", new ContactViewModel{ PartialsViewModel = pvm } );
}
What am I doing wrong?
The form has the action attribute set to SaveDetails action, so after submit it redirects the browser to this action.
I don' think you are doing anything wrong but I don't think you are able to do what you are tying to achieve. A request has to go somewhere and in mvc the url is used to identify which action you want to perform. If you are not submitting a post back then the url is going to change.
One way to submit the forms to different actions would be using some ajax.
Submitting the form is a POST. You can use an attribute to identify what request method an action should respond to. This means that you can create another action also called Index but give it the [HttpPost] attribute.
[HttpPost]
public ActionResult Index(Models.ContactViewModel cvm)
{
return View();
}
This way it won't display the action in the url.
I'm looking into using partial views in MVC3 using Razor, and I get my partial view to render and it works fine.
What I'd like to do, though, is refresh the parent view when the partial view is submitted.
Code in my parent view to render partial view
<div id="mydiv">
#{ Html.RenderAction("Add", "Request"); }
</div>
Action for parent view is simple,
public ActionResult Index()
{
List<obj> reqs = //some query
return View(reqs);
}
In my partial view's get action I have:
public ActionResult Add()
{
AddRequestViewModel vm = new AddRequestViewModel();
//set some stuff on the VM here
return PartialView(vm);
}
In the post action called by the partial view, if modelstate isn't valid, return PartialView(vm)
If it is valid, I'd like the parent and partial views to refresh.
I tried RedirectToAction, but this can't be called in an action called by a partial, apparently, and I tried return Index();, but this causes an issue with the code used to render the partial view,
Exception Details: System.InvalidOperationException: The model item passed into the dictionary is of type 'System.Collections.Generic.List'1[DatRequests.Models.ReqRequest]', but this dictionary requires a model item of type 'DatRequests.ViewModels.AddRequestViewModel'.
Any suggestions on how to do this would be appreciated. The purpose of the page is to show a list of elements, and the partial contains a form to add a new element to the list.
Edit: The partial's model is different, as it contains data for selection, which is from a db, which is why I tried RenderAction, but I'm not sure if there are other ways of doing this.
When the partial view is submitted normally you submit it to some controller action. You could either submit it using a normal request or an AJAX request. If you use a normal request you could perform a standard redirect to the Index inside the POST controller action that will handle the form submission. If you use AJAX, you could return a JSON result pointing to the url that you want to redirect:
[HttpPost]
public ActionResult Foo(MyViewModel model)
{
if (!ModelState.IsValid)
{
return PartialView(model);
}
return Json(new { url = Url.Action("Index") });
}
and inside your AJAX success callback:
success: function(result) {
if (result.url) {
// we have a success
window.location.href = result.url;
} else {
// invalid modelstate => refresh the partial
$('#mydiv').html(result);
}
}
Probably RenderAction should not be used this way.
When using Html.RenderAction, a new/seperate request would be sent to the server. And you got another chance to load some data from db or somewhere else to display to the client. Also, you could apply OutputCache to this action. this is usually the way doing global cache.
Here you are doing a POST to the server. Either directly put a element here or using a partial view to do the Post. And in the corresponding action, do a RedirectToAction.
Do it with ajax or not isn't the point. my opinion is more about the right way using RenderAction
I have layout with login partial view (username, password and submit button) and models ( some controls and submit button)with validation(server side and client side) displayed in normal views (#RenderBody() in layout).
My problem is when do server side validation in any of my views it also validate the login partial view because it execute the httppost function of the login. how can I stop that??
login view controller
[HttpGet]
public ActionResult LogOn()
{
return PartialView();
}
//
// POST: /Account/LogOn
[HttpPut]
public ActionResult LogOn(LogOnModel model)
{
if (ModelState.IsValid)
{
if (MembershipService.ValidateUser(model.UserName, model.Password))
{
FormsService.SignIn(model.UserName, model.RememberMe);
ViewBag.UserName = model.UserName;
}
else
{
ModelState.AddModelError("", Resources.Account.Account.LoginFailureText);
}
}
return PartialView(model);
}
and model controller
public ActionResult MyModel()
{
ViewBag.DisplayThxMsg = false;
return View();
}
[HttpPost]
public ActionResult MyModel(Models.FeedbacksModel feedback)
{
if (ModelState.IsValid)
{
//do something
}
else{
//do another thing
}
return View(feedback);
}
I find your question very difficult to understand. Im guessing your problem is you have a login partial control displayed as part of site layout and is shown on all pages. So while submitting any page, the username password validation kicks in, and you want to prevent that.
Understand that all validation # server - side happens while model binding, As properties are bound to the posted fields, the attributes on the fields are looked at and honored / catered to. So to prevent server side validation simply put the login partial view in it's own form so it is not sent while submitting other forms on the page.
In short have 2 forms - one form for login and one for feedback. Don't put all input fields in the same form.
If you still have validation errors after that, then it is because of other reasons like, type conversion problems. The default model binder will add some errors for basic type conversion issues (for example, passing a non-number for something which is an "int").The sample DataAnnotations model binder will fill model state with validation errors taken from the DataAnnotations attributes on your model.
EDIT
If you look at line number 125
#using (Html.BeginForm()){Html.RenderAction("LogOn", "Account");}
You have the above code which will render the login form.
It will do so inside the other form at line 45
<form id="form1" runat="server" method="post">
This has no end tag therefore it will encompass the whole document till </html>
You should change the structure from
<form id="form1" runat="server" method="post">
#using (Html.BeginForm()){Html.RenderAction("LogOn", "Account");}
</form
to
<form id="form1" runat="server" method="post">
</form>
#using (Html.BeginForm()){Html.RenderAction("LogOn", "Account");}
This line #using (Html.BeginForm()){Html.RenderAction("LogOn", "Account");} will render this form <form id="LoginView1" action="LogOn"> and all child elements of it.
LATEST EDIT
In your layout page use this :
#Html.Partial("~/Views/Shared/LogOnPartial.cshtml", new LogOnModel())
instead of this :
#Html.Action("LogOnPartial", "Account")
The reason why it all works is, the LogOnPartial method marked with [HttpPost] is called because the request was in a POST context. What you want is, You just need the view without the action executing even when POSTing. The above code does that. It renders the view without calling the action method. MVC3 is sort of a stupid servent : It only knows that it should call the Action method marked with [HttpPost] when the request is in a post context. It doesn't know that the request is in a post context for another action (index) and not this one (logonpartial). So now you can remove this method
public ActionResult LogOnPartial()
{
return PartialView();
}
which will no longer be used.
Note that you need to change the account controller's LogOnPartial Method to return
return RedirectToAction("Index","Home"); instead of return PartialView(model); on successful login. And on FAILURE you cannot render a partialview as you have coded. You must return an entirely new View. It must neither be index nor LogonPartails - just return return View("Login_Error_View"); which has its own layout. Otherwise it will be difficult to control the workflow.
I am using ASP.NET MVC 3 for my website.
I have created a partial view that has "Previous, Next and Save" buttons. I am calling this partial view on my master page.
My requirement is that on what ever View I am I must be able to call different Save methods in different controllers by passing respective Model data to controller actions.
Example
I have 4 step data input, I have a different controller for each step.
If i am on step 1 and i click Save the form Values should go to Step1Controller's action method,
If I am on step 2 then post should call Step2Controller
Something like this this:
public ActionResult Save(GenericModel model)
{
//use reflection to find out model type
//call appropriate controller action with model
return RedirectToAction("Create", new { Controller = "Conference", Action = "Create" });
}
This save method will be called for Save button on the Master page. How can I achieve this?
Are the forms on separate actions on the controllers?
If so, just set the form action on each page to point to the relevant controller. So form 1 is
<form method="post" action="/step1controller/action">
Form 2 is:
<form method="post" action="/step2controller/action">
Does that solve your problem?
I would create a private method in the controller and call in every part of the first steps.
private bool Save(GenericModel model)
{
......
}
[HttpPost]
public bool SaveStep1(GenericModel model)
{
this.Save(model);
}
[HttpPost]
public bool SaveStep2(GenericModel model)
{
this.Save(model);
}