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Home/ Questions/Q 8188061
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Editorial Team
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Editorial Team
Asked: June 7, 20262026-06-07T02:46:39+00:00 2026-06-07T02:46:39+00:00

My two links to filter: @Html.ActionLink(Customer 1, Index, new { customer = 1 })

  • 0

My two links to filter:

@Html.ActionLink("Customer 1", "Index", new { customer = 1  })
@Html.ActionLink("Project A", "Index", new { project = "A"  })

My Controller with filtering:

public ViewResult Index(int? customer, int? project) {
        var query = ...

        if (customer != null) {
            query = query.Where(o => o.CustomerID == customer);
        }

        if (project != null) {
            query = query.Where(o => o.ProjectID == project);
        }
        return View(query.ToList());
}

I can now filter on either customer or project but not on both at the same time!

If I click Customer 1, url = Object?customer=1

If I click Project A, url = Object?project=a

I would like to be able to first click Customer 1 and then Project A and get url = Object?customer=1&project=a

Is this possible or should I do it in another way?

Thanks!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-07T02:46:40+00:00Added an answer on June 7, 2026 at 2:46 am

    The correct way to do this is to return a model with your various parameters to your view.

    Model

    public class TestModel {
       public int? Customer { get; set; }
       public int? Project { get; set; }
       public List<YourType> QueryResults { get; set; }
    }
    

    View

    @model Your.Namespace.TestModel
    
    ...
    
    @Html.ActionLink("Project A", "Index", new { customer = Model.Customer, project = Model.Project })
    

    Controller

    public ViewResult Index(TestModel model) {
        var query = ...
    
        if (model.Customer != null) {
            query = query.Where(o => o.CustomerID == model.Customer);
        }
    
        if (model.Project != null) {
            query = query.Where(o => o.ProjectID == model.Project);
        }
    
        model.QueryResults = query.ToList();
    
        return View(model);
    }
    
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