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up vote 0 down vote favorite

I have controller.cs code :

public class GroupsController : Controller
    {
        public ActionResult Index()
        {
            GroupTypeRepository groupTypeRepo = new GroupTypeRepository();

            var groupTypeNames = groupTypeRepo.GetAll().ToList();

    //like GroupType I have Group also so i want to pass both GroupType and Group object to my below return View("Groups", groupTypeNames);
        // how can i do this?

            return View("Groups", groupTypeNames);
        }

I tried it in doing in /Model Folder in Model.cs but its not good approach please guide how to achieve this? thanks

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2 Answers

up vote 3 down vote accepted

You can't return multiple models from a controller action. What you do instead is define a view model that will contain all the necessary information and then return this view model.

For example:

public class MyViewModel
{
    public List<string> GroupTypeNames { get; set; }
    public string SomeOtherProperty { get; set; }
}

and then:

public ActionResult Index()
{
     GroupTypeRepository groupTypeRepo = new GroupTypeRepository();

     var model = new MyViewModel();
     model.GroupTypeNames = groupTypeRepo.GetAll().ToList();
     model.SomeOtherProperty = "some other property value";

     return View("Groups", model);
}

Now of course your view should be typed to the view model:

@model MyViewModel

and if you wanted to access the some property:

@foreach (var item in Model.GroupTypeNames)
{
    <div>@item</div>
}

or:

@model.SomeOtherProperty
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superb info can you please give me threads any link sorry to bother again as i'm new bie :( –  ashish Dec 29 '13 at 17:53
    
I don't understand what threads and links you are asking for. That's basic stuff. You should read the getting started tutorials at asp.net/mvc –  Darin Dimitrov Dec 29 '13 at 17:54
    
what if SomeOtherProperty is also like GroupTypeNames i mean List i just need to take same as in viewmodel is it :) –  ashish Dec 29 '13 at 17:55
1  
Erm what? Sorry I don't understand what you are asking. You could have arbitrary complex properties in your view model. –  Darin Dimitrov Dec 29 '13 at 17:55
    
hey related with previous question i used above code to return two model object but now getting error for previuos code for loop Operator '<' cannot be applied to operands of type 'int' and 'method group' question was stackoverflow.com/questions/20827155/… –  ashish Dec 29 '13 at 18:04
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up vote 1 down vote

Using PartialResult You can divide your complex code into multiple Partials. For example here is the code.

 public class Type1
    {
        public struct Test
        {

        }
    }

    public class Type2
    {
    }

and this is my code that I used in partial

@using MVC5.Models

@model  List<Type1.Test>

in another partial you can call type2.

Now make a action in Controller like this

   public PartialViewResult Partial1()
        {
            ViewData.Model = new List<Type1.Test>();
            return PartialView();
        }

Now in my index.cshtml 

@{
    ViewBag.Title = "Index";
}

@{
    Html.Action("Partial1","Home");   
}

I have called the partial1 in my index ActionResult. Using this implementation you can use model in every partialresult you have.

As darin say Make a call that hold all these sub-data and pass them from controller to Models.

I want to suggest you to pass multiple data on Views through ViewData or ViewBag (ViewBag is dynamic). See this post http://www.rachelappel.com/when-to-use-viewbag-viewdata-or-tempdata-in-asp.net-mvc-3-applications

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