Returning multiple object properties in a list

Question

I'm able to return the CategoryName field into a list via ViewBag.Categories. I also want to return the CategoryId field and use it to dynamically build the category Details view URL

This is the controller action:

public ActionResult Details(Item item)
{
    var db = new appContext();
    ViewBag.Item = item.ItemTitle;
    ViewBag.ItemLink = "http://localhost:4444/Items/Details/" + item.ItemId;
    ViewBag.Categories = new List<string>(item.Categories.Select(c => c.CategoryName));

    return View();
}

This is the view where the list of categories is returned:

@foreach (string category in ViewBag.Categories)
{
    <li>
        <a href="http://localhost:4444/Categories/Details/???">@category</a>
    </li>
}

How can I return CategoryIdas shown above?

I recommend you use @Html.Action("Details", "Categories", new { ID = category.ID }) (based on dotnetoms' answer) to create the link. As soon as you publish your hard coded string above (..//localhost:4444/..) to a server it will fail. Similarly `ViewBag.ItemLink should just be the ID. — user3559349
– user3559349, Commented Jul 18, 2014 at 8:28

krivtom · Accepted Answer · 2014-07-18 08:21:11Z

3

Use custom class for storing categories:

public class CategoryViewModel
{
    public int CategoryId { get; set; }
    public string CategoryName { get; set; }
}

Assign it like this:

ViewBag.Categories = item.Categories
    .Select(c => 
        new CategoryViewModel 
        { 
            CategoryId = c.CategoryId, 
            CategoryName = c.CategoryName
        }).ToList();

And in your view use it like this:

@foreach (CategoryViewModel category in ViewBag.Categories)
{
    <li>
        <a href="http://localhost:4444/Categories/Details/@category.CategoryId">@category.CategoryName</a>
    </li>
}

answered Jul 18, 2014 at 8:21

krivtom

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Comments

CodingIntrigue · Accepted Answer · 2014-07-18 08:21:46Z

1

You can use anonymous types to achieve this:

public ActionResult Details(Item item)
{
    var db = new appContext();
    ViewBag.Item = item.ItemTitle;
    ViewBag.ItemLink = "http://.../Items/Details/" + item.ItemId;
    ViewBag.Categories = item.Categories.Select(c => new { c.CategoryName, c.CategoryId });

    return View();
}

Then in your view:

@foreach (var category in ViewBag.Categories)
{
    <li>
        <a href="http://localhost:4444/Categories/Details/@(category.CategoryId)">@category.CategoryName</a>
    </li>
}

However, you may want to consider replacing the ViewBag with strongly-typed models. It offers a lot of great benefits - including IntelliSense support.

answered Jul 18, 2014 at 8:21

CodingIntrigue

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Comments

DGibbs · Accepted Answer · 2014-07-18 08:22:34Z

1

Instead of using a List<string> you could use a:

List<Tuple<int, string>>

E.g.

ViewBag.Categories = item.Categories.Select(cat => 
       Tuple.Create(cat.CategoryID, cat.CategoryName)).ToList();

answered Jul 18, 2014 at 8:22

DGibbs

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Comments

Sandeep · Accepted Answer · 2014-07-18 08:42:53Z

Define a view-model like this:

 public class CustomViewModel
    {
        public string Item { get; set; }
        public string Title { get; set; }
        public Categories Categories { get; set; }
    }

    public class Categories
    {
        public int Id { get; set; }
        public string CategoryName { get; set; }
    }

    public ActionResult Details(Item item)
    {
        var db = new appContext();
        var model = new CustomViewModel{
            Item = item.ItemTitle,
            ItemLink = "http://localhost:4444/Items/Details/" + item.ItemId,
            Categories = (item.Categories.Select(c => 
                new Categories{ Id = c.Id,CategoryName= c.CategoryName})).ToList()
        };
        return View(model);
    }

and view like:

   @model Categories
   @foreach (Categories category in Model.Categories)
    {
      <li>
        <a href="http://localhost:4444/Categories/Details/@category.CategoryId">@category.CategoryName</a>
     </li>
    }

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