I have an HTML select in View, for the select options i have to retrieve value from database. I have a function in model that returns LIST of options. How can I call the model's function from view.
Something like this:
public class XXXXViewModel {
//properties...
public List<Option> Options {get;set;}
}
and in your controller:
public class XXXXController : Controller {
public ActionResult SomeAction(){
var model = GetModelFromRepository();
var viewModel = new XXXXViewModel{
//Properties...
Options = model.GetOptions();
};
return View(viewModel);
}
}
So, you controller takes care of providing the options to the view using a ViewModel class containing everything your view needs.
Hope it helps.
create an event which is going to call that function and by the help of json and Ajax javascript you can easily do it
its a small concept that model canot be call from html after rendering on client side you have to take care of it by controller action and ajax is a good practice for it in maximum of mvc application
Related
I'm trying to do that:
Create a Model, add it on a session and send it to the view.
Change Model fields on my view
Get the Model from session updated on my controller
The problem is that my model is never updated when I'm changing values on textboxes, I'm sure that I'm missing something with razor,
View:
#model MvcTestApp.Models.Car
<div class="b1">
<div class="b2">#Html.EditorFor(e => e.KM)</div>
<div class="b2">#Html.EditorFor(e => e.RegistrationNumber)</div>
</div>
#Html.ActionLink("Car", "sendCar")
Controller:
On SendCar, I would like to get the model updated.
namespace MvcTestApp.Controllers
{
public class CarController : Controller
{
public ActionResult Show()
{
var model = new MvcTestApp.Models.Car()
{
RegistrationNumber ="12345",
KM = "12345"
};
Session["temp"] = model;
return View("Show",Session["temp"]);
}
public ActionResult sendCar()
{
return View("Show", Session["temp"]);
}
}
}
Model:
namespace MvcTestApp.Models
{
public class Car
{
[DataType(DataType.Text)]
public string KM { get; set;}
[DataType(DataType.Text)]
public string RegistrationNumber { get; set;}
}
}
You need to make your sendCar controller to update the model. Currently, all the changes you do will only persist locally until you navigate away from the page. You need to post the changed model back to the server.
Take a look at the "Handling edits" part of this example to see how it can be done:
Asp.net tutorials
The way to do this is by wrapping your model details in a form with a submit function. Then in your sendCar method take in a Car object and the model binding will take care of setting everything on the new object.
If you're wanting to persist this (I assume this is just for testing purposes?) then perhaps make your car that you're returning in your show method a class variable.
You should read a beginner tutorial about ASP.NET MVC, which will explain you how to send data from a form to a controller, as it seems you are absolutely not aware of how to do this.
You are not missing 'something', you are missing all about sending data from forms to controllers.
I have a situation where i have to take input(form) from user. After continue button is pressed next view page is displayed. But after continue is pressed i don't want to store the model in the DB. I have to display some details(combining some tables) according to input given by the user earlier and again get some data from user. Only then i want to store the model in the respective tables.
How can i perform this? I tried getting Model from user and passing to the function that generates next page. Is this is way to do it? or there is other way around?
Store the model submitted by the first form in session.
[HttpPost]
public ActionResult ContinueForm1(Model1 model1)
{
if(ModelState.IsValid)
{
Session["Model1"] = model1;
return View("Form2");
}
return View();
}
[HttpPost]
public ActionResult ContinueForm2(Model2 model2)
{
if(ModelState.IsValid)
{
... model2 is already here, get the model1 from session
... and save to datatbase finally return a different view or redirect to some
... other action
}
return View();
}
You are heading down the right track.
You need to grab the model that is passed back from the first view - preferably you are using ViewModels here rather than binding directly to your db models. Have a look at http://lostechies.com/jimmybogard/2009/06/30/how-we-do-mvc-view-models/ and Why should I use view models? as to why these are good things.
The easiest way to do this is to pass the model in as an argument to your method e.g.
Assuming that your views are using the same ViewModel ( which may or may not be true) then you can send the viewmodel straight to your new view - else you can copy the elements into a new viewModel and send that.
e.g.
[HttpPost]
public ViewResult Step1(MyViewModel viewModel)
{
//Do some validation here perhaps
MySecondViewModel secondViewModel = new MySecondViewModel{
Id = viewModel.Id,
// etc. etc.
};
return View("Step2", secondViewModel);
}
Then you can carry on as you need until you have to persist the entity to the database.
NB as you do not need to do anything special in the form to make it post the model as an argument as long as the view is strongly typed to that ViewModel.
I have this post method:
[HttpPost]
[ValidateAntiForgeryToken]
public ActionResult Invitations(SuperInvitationsEditModel model)
{
...
var newmodel = new SuperInvitationsEditModel();
if (hasErrors)
{
SuperInvitationsErrorModel newErrorModel = new SuperInvitationsErrorModel();
newErrorModel.Errors = model.Errors;
return View(newErrorModel);
}
return View(newmodel);
}
When this code in the if(hasErrors) executes I get this error.
The model item passed into the dictionary is of type 'MyProject.Models.SuperInvitationsErrorModel', but this dictionary requires a model item of type 'MyProject.Models.SuperInvitationsEditModel'.
I thought I can do this since the return value of the method is a generic ActionResult. Can anyone tell me why is this not working?
because your current view is strongly typed. change the code as
return View("yourviewname",newErrorModel);
It has nothing to do with casting ViewResult to ActionResult. The problem is, that you have strongly typed view that expects the model of type SuperInvitationsEditModel (see #model on the top of Invitations.cshtml), but you are passing the model of type SuperInvitationsErrorModel to it.
You should merge the two view model classes (SuperInvitationsEditModel and SuperInvitationsErrorModel) into one, or create a standalone view for each of them.
I have a view that is strongly typed. Inside this view i have jqueryui tabs, that when clicked call my Controller and return a partial view
("#tab0").load('#Url.Action("ProfileImage", "User")');
public ActionResult ProfileImage()
{
return PartialView("_ProfileImage");
}
What I'd like to do is pass the model from the "parent" view to the controller which can then bind it to the partial when it is returned:
("#tab0").load('#Url.Action("ProfileImage", "User", new {model=model})');
public ActionResult ProfileImage(UserViewModel model)
{
return PartialView("_ProfileImage", model);
}
Is this possible? how is this normally done? Where you have the model data in one view and you'd like to pass it to a asynchronously loaded partial view?
You could create a ToJson method on your viewmodel, which could be something like this:
public IHtmlString ToJson()
{
return MvcHtmlString.Create(Json.Encode(this));
}
It just serializes the viewmodel into a json. The IHtmlString returntype makes sure the output isn't encoded in your view.
The call to your controller would be something like this:
("#tab0").load('#Url.Action("ProfileImage", "User", new {model=model.ToJson()})');
The json modelbinder can recreate the viewmodel on the serverside. You probably run into some problems along the way, but nothing unsolvable I guess.
Can anyone help me with the subject? I'm using Razor view engine and I need to pass some data to _Layout. How can I do it?
As usual you start by creating a view model representing the data:
public class MyViewModel
{
public string SomeData { get; set; }
}
then a controller which will fetch the data from somewhere:
public class MyDataController: Controller
{
public ActionResult Index()
{
var model = new MyViewModel
{
SomeData = "some data"
};
return PartialView(model);
}
}
then a corresponding view (~/Views/MyData/Index.cshtml) to represent the data:
#{
Layout = null;
}
<h2>#Model.SomeData</h2>
and finally inside your _Layout.cshtml include this data somewhere:
#Html.Action("index", "mydata")
You could use the ViewBag to pass data.
In your controller:
ViewBag.LayoutModel = myData;
Access in you layout:
#ViewBag.LayoutModel
It is a dynamic object, so you can use any property name you want.
The ViewBag method is the easiest. However if you need advanced and typed features, you can also try taking that part to a partial view (the part that'll render the dependent section) with a common controller (if the value can be calculated on it's own and doesn't need input from other controllers), and call RenderPartial on it from _Layout.
If you'd like I can give you some more info about it...