Reputation: 35194
Lets say that i have an URL that looks something like this: localhost/userdetails/5
where 5
is the users ID. Is there any way to make use of the ID directly in the view (razor viewengine) and show the details? Or do i handle it in the default action in the controller?
Upvotes: 1
Views: 192
Reputation: 14655
To keep things simple now, focusing on getting the id
to the view, you basically want to use the id
to populate your model with data and then pass that to the view. So in your controller:
public ActionResult Index(int id)
{
UserModel model = db.Users.Where(u => u.Id == id).SingleOrDefault();
return View(model);
}
The view (very simplified) might look like this:
@model MyProject.Models.UserModel
@Html.DisplayFor(m => m.Id)
@Html.DisplayFor(m => m.Username)
This is very basic though. Eventually, you'll get to a point where you realise you should use viewmodels for your views instead of a domain model that's come directly from the data source. That in itself gives you another problem to solve in the form of mapping properties from the domain model onto your viewmodel. Typically, AutoMapper or ValueInjecter are used for that. For now though, it's probably best to just focus on passing data to a view and getting it back into a controller so that you can do something with it.
This is a simple scenario which demonstrates how to get the data back into the controller. So basically, you'd have a form which you would submit:
@using (Html.BeginForm("Index", "Home"))
{
// Form elements and submit button
}
That would post the data to this action method for you to do whatever you wish with the data:
[HttpPost]
public ActionResult Index(UserModel inputModel)
{
// Check to see if the model's data was valid.
if (ModelState.IsValid)
{
// Do something in the database here.
// Then redirect to give the user some feedback.
return RedirectToAction("Thanks");
}
// The model validation failed so redisplay the view.
return View(inputModel);
}
Upvotes: 1
Reputation: 7605
Yes you can. There is more than one way to do it, but since you've tagged your post MVC, assume you'll want to do it the 'MVC way', which means (imo) using a view model.
So you write a view model
public class MyViewModel()
{
public int ID {get; set;}
}
You populate the model in the controller and pass it to the view
public ActionResut MyView (int id)
{
var viewModel = new MyViewModel {ID = id};
return View (viewModel);
}
Then you have a strongly typed view (strongly typed to the MyViewModel, that is) and you can reference the model's properties
@Model.ID
Then to make this useful, you can add whatever other properties you're wanting to work with to your view model. Then you can populate them in your controller before rendering the view (to show user info, for example), or let the user populate them for you in the view (using textboxes and such wrapped in a form). Then you can collect the user input in the post action in the controller like so
[HttpPost]
public ActionResult MyView(MyViewModel viewModel)
{
//do stuff with the data from the viewModel
}
Upvotes: 0
Reputation: 3361
You should use the model (i.e. the model passed back to your view). A ViewBag is another option but since the ID is part of the model itself, it wouldn't make any sense to do that.
View
@model User
@{
ViewBag.Title = "User Details";
}
@Model.Id;
Controller
public ActionResult UserDetails(int id)
{
return View("UserDetails", (object)id);
}
Upvotes: 0
Reputation: 9271
You can get the id with this
@HttpContext.Current.Request.RequestContext.RouteData.Values["id"].ToString()
But I would reccomend to use a ViewMdoel to pass the value to the view and not the ViewBag or accessing directly from the view
Upvotes: 0
Reputation: 5801
you can use this in both the controller or in the View as an extension method. Example: asuming your routes id holder has the default values in global.asax
public int IdFromAdress(HttpContext httpContext)
{
RouteData rd = httpContext.Request.RequestContext.RouteData;
string stringId = (string)rd.Values["id"];
return int.Parse(stringId);
{
Upvotes: 0