I'm using MVC3 razor, and I'm trying to pass an object to a partial view, and it's not working.
This works fine without sending the object model to the partial view:
Html.RenderAction("Index", "ViewName");
Trying this doesn't sent the model object, i'm getting nulls instead (the object has data, and the view expects it):'
Html.RenderAction("Index", "ViewName", objectModel);
Is this even possible using RenderAction?
Thanks!
Edit: I found the error, there was an error with the controller's action that didn't pick up the sent object. Thanks for all your help!
You can actually pass an object to a controller method using Action. This can be done on any avaialble view, for instance I have one in a shared library that gets built to project bin folders that reference my shared project (properties - Copy if newer on the view file, in Visual Studio). It is done like so:
Controller:
public class GroovyController : Controller
{
public ActionResult MyTestView(MyModel m)
{
var viewPath = #"~\bin\CommonViews\MyTestView";
return View(viewPath, m);
}
}
MVC page (using Razor syntax):
#Html.Action("MyTestView", "Groovy", new { m = Model })
or using RenderAction method:
#{ Html.RenderAction("MyTestAction", "MyTestController", new { area = "area", m = Model }); }
Note: in the #Html.Action(), the Model object must be of type MyModel and that 3rd parameter must be set to the controller variable name, of which mine is MyModel m. The m is what you must assign to, so I do m = Model.
say you want to pass foo as model, make it first
public class Foo {
public string Name { get; set; }
public int Age { get; set; }
}
now make an ActionResult
public ActionResult FooBar(Foo _foo){
return PartialView(_foo);
}
call it
#Html.RenderAction("FooBar", "Controller", new { Name = "John", Age=20 });
Usually if I have a model already available it makes more sense to use Html.Partial than trying to render an action.
#Html.Partial("Foo", Model.FooModel)
Where Foo.cshtml is a view file (perhaps in your Shared folder) strongly typed with with #model FooProject.Models.FooModel or whatever your model is called. This can be as complex a model as you need it to be. Model is your page's main model into which you must set FooModel - or just omit this parameter if the Foo view uses the same model as the parent page.
RenderAction is generally better when you have just simple parameters, because you're just simulating a request to a regular action which has routing/query string parameters - and then dumping that response into your page. It works well if you need to put something in a Layout that isn't available in your page's model such as an element in a side bar.
Related
How can we determine in which mode the view has been rendered, inside of the view (.cshtml file)?
Actually, in a view, for example Delete.cshtml, I have called the Details.cshtml as :
#Html.Partial("Details.cshtml",model)
and inside of the Details.cshtml view, I want to determine the rendering type (Full/Partial) to make a decision for some jobs.
Is there any IsPartial property or something like that?
What you are trying to achieve is not possible directly from Html.Partial but it is possible if you go to that view using an action and render it using Html.Action.
say you want to pass foo as model, make it first
public class Foo
{
public string Name { get; set; }
public int Age { get; set; }
}
Then create an action
public ActionResult Details(Foo _foo)
{
return PartialView(_foo);
}
call it like
#Html.RenderAction("FooBar", "Controller", new { Name = "John", Age=20 });
Now by doing so, You can determine weather this action is called from another action or not
if (ControllerContext.IsChildAction)
return PartialView(_foo);
Could you explain the interaction Models and ViewModels in the ASP.NET MVC?
If I need to display data on the page, but not edit, whether to create a ViewModel to display or use the Model?
I have two methods in the repository. One returns the Model and the other Model gets.In View I need to send the model. Should I convert the resulting Model to a ViewModel that would pass it to the View, and upon receipt of the submission to convert it back into the model to keep it?
For example I have a class model and class ViewModel.
public class Item
{
public int Id { get; set; }
public string Name { get; set; }
public int Price { get; set; }
}
public class EditItemItemViewModel
{
public string Name { get; set; }
public int Price { get; set; }
}
On the edit page, I clicked the edit item, and must get to the controller:
[HttpGet]
public ActionResult EditItem(int id)
{
//some code
return View();
}
Where can I get the ID if I passed in the view ViewModel in which there was no ID?
If I somehow got the ID, I need to do the following, which would save the model?
[HttpPost]
public ActionResult EditItem(EditItemItemViewModel ViewModel)
{
var Item = _dataManager.Items.GetItemById("1");
Item.Name = ViewModel.Name;
Item.Price = ViewModel.Price;
_dataManager.Items.AddItem(Item);
return View("Sucsess");
}
Could you tell me how to work with Models and ViewModels?
You can get the id a few different ways:
#Html.HiddenFor(m => m.Id)
This will include the property in the actual HTTP request.
Another option is to simply include the id as part of your route (this is what I usually do):
#Html.BeginForm("EditItem", "SomeController", new { Id = Model.Id }, FormMethod.Post) {
...
}
In both instances, make sure to validate that the user should be able to update the record that corresponds to that id! There's nothing stopping a user from tampering with the id and sending you a different value.
As for whether or not to display the database model or the view model, that's really up to you. I would always advocate building a view model and keep your database models out of the picture except for in your controller. The convention I use at work is for every database object that I need to send to users I will create a corresponding view model. So if I have a database object called Product I will build another class for the view called ProductModel.
An advantage of following that pattern is something I actually explained to another user earlier in regards to model binding.
If I need to display data on the page, but not edit, whether to create a ViewModel to display or use the Model?
If it's very simple (like your example) and the properties map 1-1 like they do now. I would just use the Model as a view model, it's easier. Though if you want you could create a view model with the exact same properties and populate that and display it..it's a bit more work, but makes it so your domain models aren't necessarily tied to your views.
If you do that you may want to look into something like AutoMapper which would allow you to do something like
//simply copies the properties from one type to another
var viewModel = Mapper.Map<Item, EditItemItemViewModel>();
I have two methods in the repository. One returns the Model and the other Model gets.In View I need to send the model. Should I convert the resulting Model to a ViewModel that would pass it to the View, and upon receipt of the submission to convert it back into the model to keep it?
If you go the view model route then yes, you will end up doing a lot of converting between model and viewmodel, that's where something like AutoMapper can help out, or you can just create a couple extension methods that convert between the two types.
It looks like Justin Helgerson's answer explaining a way to handle models/viewmodels is pretty good.
Your posting your viewmodel, so i'm going to assume that you've referenced your model within the page
#model MyNameSpace.EditItemItemViewModel
Is there a reason the id is not included in your view model? The easiest method would be to include that in the model and pass the instantiated model when creating the view.
[HttpGet]
public ActionResult EditItem(int id)
{
var myViewModel = new EditItemItemViewModel() { Id = id };
return View(myViewModel);
}
Like Justin said, it is easier to put it in a hidden field somewhere inside the #Html.BeginForm as the id should just be for reference.
[HttpPost]
public ActionResult EditItem(EditItemItemViewModel viewModel)
{
var Item = _dataManager.Items.GetItemById(viewModel.Id);
Item.Name = viewModel.Name;
Item.Price = viewModel.Price;
// Should this be an Add?
_dataManager.Items.AddItem(Item);
return View("Success");
}
My _Layout page uses #RenderSection at a couple of points, for example to render the sidebars (if there are any). I have about 30 different views that use 10 different models where they get the content from. However there are only 4 different sidebars at the moment, so I out them into partial views that are called like this:
#section SBLeft {
#Html.Partial("_SidebarTopics)
}
This works fine on my frontpage because the sidebar partial _SidebarTopics that is called from the Frontpage\Index.cshtml view uses the same model (WebsiteStructureModel) that is called at the start of the Index view:
#model Web.Areas.Public.Models.WebsiteStructureModel
Now I run into problems when I want to use a sidebar that uses Model A, if the "parent" view uses Model B. It results in an error like this:
The model item passed into the dictionary is of type 'Web.Areas.Public.Models.ProjectDetailsModel', but this dictionary requires a model item of type 'Web.Areas.Public.Models.WebsiteStructureModel'.
Using two #model statements at the beginning of the Index view does not work so I can't pass the second model explicitly to the sidebar as second parameter of the #Html.Partial command. And using a #model statement at the beginning of the partial view is ignored.
There must be some way to call a partial view and have that partial view use a specified model that may not neccessarily be the one used by the calling/parent view - Please help me understand how this can be done!
There are 2 ways to do this.
First way:
You could combine your 2 models into a view model:
public class ViewModel
{
public WebsiteStructureModel WebModel { get; set; }
public ProjectDetailsModel ProjectModel { get; set; }
}
You'd obviously populate this in your Action and then pass that to the View
Second way:
Rather than calling #Html.Partial("_SidebarTopics") you could create an Action in your controller which will return PartialView("_SidebarTopics", model); where model is the model passed into the partial view, for example:
#section SBLeft {
#Html.Action("SidebarTopics", new { /* route params */ });
}
Controller:
public ActionResult SidebarTopics(/* route params */)
{
var model = new ProjectDetailsModel();
return PartialView("_SiderbarTopics", model);
}
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?
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.
Update
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);
}
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);
{
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
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);
}
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
}
We are using a friendly Name URL route scheme. Basically using a combination of the Principal Identity and a friendly name this can be mapped back to an identity internally (Person ID). So a route like this:
routes.MapRoute(string.Empty, "{friendlyname}/Products/{action}", new {controller = "Products", action = "List"});
Would map to a URL like this:
Adam/Products/List
This all works fine and abstracts away the internal Id of the named person which is required as well.
The problem is our Views are comprised of many partial views. When there are rendered by using the #Html.Action method they ultimately need the PersonID but from the URL we only have the 'friendly name'.
I have thought about this for a while and there are two solutions to my mind:
Pass the 'friendly name' into each of the controller action methods that return the partial views and internally the method will have to do a lookup on the currently logged in identity and the friendly name. This will give the PersonID to me and I can then efficiently query from then on. The only problem with this apporach is that due to the multiple partial views I will be querying on the currently logged in identity and friendly name for each partial view call which is innefficeint and I feel I should only have to write this code once.
Somehow query in the view and get the PersonID so it can be passed to each #Html.Action call so the partial view controller methods will not have to do that lookup themselves saving round trips to the database for the same shared informtion. The problem with this is that I am not sure of a way of doing this cleanley in the view using the DI that we use through the rest of the application.
Any thoughts on approach to would be greatly appreciated.
Thank you,
Adam
You could add the Id to the session variables and access it from within the views with:
#{var personId = int.Parse(Session["PersonId"])}
Then you can pass it directly to partial views from the Parent without it hitting the client or having to pass parameters to any controllers.
Update
You could also access the session variable from the Controller if you wanted to do the work there instead without roundtripping to the database.
EDIT
If you put the property in a model and pass it to a page that post's back then the model will not persist between posts.
If for example your controller does:
[HttpPost]
public ActionResult DoSomething(ViewModel model)
{
if(ModelState.IsValid)
{
// Logic Here
}
return View(model)
}
when the page is reloaded, the model will have forgotten about the ID.
There are a couple of ways around this. Either use #Html.HiddenFor(m => m.ID)
which will put the property in the rendered HTML, which if it is truely a sensitive piece of information, is bad.
Or you can rebuild the view model on each subsequent postback.
Hope this helps
As Marc states I could you the Session to deal with this but I have gone with using the Model as he has stated in his update. If the parent View Controller action takes in the friendly name it can do the lookup, put the PersonID into the model and then any Partial Renders can have the models value passed into them in the parent View Controllers Action's view. An example is shown below (this is demo code but it hopefully gets the point across, I would never use a static data context in real code)
Home controller
public class HomeController : Controller
{
public ActionResult Index(string friendlyName)
{
int pupilId = Data.People.Single(x => x.Name == friendlyName).PersonId;
HomeIndexViewModel homeIndexViewModel = new HomeIndexViewModel {PupilId = pupilId};
return View(homeIndexViewModel);
}
}
Home Index View
#model SharingInformationBetweenPartials.Web.Models.HomeIndexViewModel
#{
ViewBag.Title = "Index";
}
<h2>Index</h2>
#Html.Action("DisplayPersonDetail", "Person", new {Model.PersonId})
The PersonController's DisplayPersonDetail method can then present the respective data it wishes using the passed in PersonId:
public class PupilController : Controller
{
[ChildActionOnly]
public ActionResult DisplayPupilDetail(int pupilId)
{
Person person = Data.People.Single(x => x.PersonId == pupilId);
return View(person);
}
}
I did try what I thought was this before but I must have got something wrong as the ViewModels properties were getting shown in the URL which is what I was trying to get away from. Anyway, I hope this helps anyone else who may be looking to do something similar. If you have any questions then let me know.
Thanks,
Adam
You can use [Bind] attribute to specify the exact properties a model binder should include in binding or use Exclude parameter on the attribute to exclude PersonId
[HttpPost]
public ActionResult EditPerson([Bind(Exclude = "PersonId")] Person person)
{
//do domething
}
You can also use [ReadOnly] attribute that model binder will understand and not assign to that property.
[ReadOnly(true)]
public int PersonId{ get; set; }
But the best approach is to use separate ViewModels: one only for viewing and one for editing.
public abstract class Person
{
public string Name { get; set; }
public string Surname { get; set; }
}
public class PersonCreateVM : Person
{
//no PersonId here
}
public class PersonEditVM : Person
{
public int PersonId{ get; set; }
}
This approach is maybe a "overkill" but when used properly and with AutoMapper http://automapper.codeplex.com/ it's an ease to work with.