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Home/ Questions/Q 8232547
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Editorial Team
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Editorial Team
Asked: June 7, 20262026-06-07T17:52:12+00:00 2026-06-07T17:52:12+00:00

So I’m just starting out with ASP.NET MVC, and I’ve run into an issue

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So I’m just starting out with ASP.NET MVC, and I’ve run into an issue I just can’t find an answer to.

I’m grabbing some data from an SQL database, and send it to the view with this code:

var listingsQuery = (from s in db.tblListings select s).OrderBy(s => s.listingID).Skip(100).Take(25);
var listings = listingsQuery.ToList();

return View(listings);

This works great, but I want to add a value in the list of results. Basically what I’m trying to do is something like this:

foreach (var item in listings)
{
    this.Add("propertyType", "Home");
}

But obviously that doesn’t work. I’ve tried doing ToArray() instead of ToList() and that got me nowhere.

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  1. Editorial Team
    Editorial Team
    2026-06-07T17:52:14+00:00Added an answer on June 7, 2026 at 5:52 pm

    Do you want to add a new property to each object in the List collection? If so, you can create a new type that inherits from whatever object type is in the list (hover over the listingsQuery variable – the type is inside the <> symbols) and add the new property to it:

    public class MyNewType : ExistingTableType
    {
        public string PropertyType { get; set; }
    }
    

    Then inside of your query, project the properties into this new type:

    var listingsQuery = (from s in db.tblListings  
                         orderby s.listingID
                         select new NewType
                         {
                             listingID = s.listingID,
                             someOtherField = s.someOtherField
                         }
                        ).Skip(100).Take(25);
    

    Now you can cycle through each record and assign a value:

    foreach(var record in listings)
    {
       record.ProperyType = "Home";
    }
    

    There are other ways of doing this as well (assuming you’re using EF), for example, if you used raw SQL you can cast the result type directly into the new type (without having the map each field), but this should get you started.

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