How to use class interface with a view

Question

I'm learning about interfaces in views (MVVM), and wanted to ask about a simple case:

Here's some very basic code:

public interface IInventory {
   public int Id { get; set; }
   public string Name { get; set; }
}

public class Inventory : IInventory {
    public int Id { get; set; }
    public string Name { get; set; }
}

public InventoryViewModel {
   public IEnumerable<Inventory> Inventory { get; set; }
   public PagingInfo PagingInfo { get; set; }
}

Then in the controller:

 IEnumerable<IInventory> all_inventory = from m in mydb.Inventory select m;

So here's my question: when I want to get the inventory into a VM with the PagingInfo,
I start with:

 InventoryViewModel myVM = new InventoryViewModel {
      Inventory = all_inventory.Skip(5),
      PagingInfo = new PagingInfo ........

This won't work because "Inventory" is an IEnumerable listing class instances, but "all_inventory" is an IEnumerable listing interface instances.

I know I'm using it wrong. How would I fix this?

Answer 1

Hi @Coreysan ,

 InventoryViewModel myVM = new InventoryViewModel {    

       Inventory = all_inventory.Skip(5),  
       PagingInfo = new PagingInfo ........  

Try to modify the code as below:

    InventoryViewModel myVM = new InventoryViewModel  
    {  
        Inventory = all_inventory.Select(c=> new Inventory() { Id = c.Id, Name= c.Name}).Skip(5).ToList(),  
        PagingInfo = new PagingInfo() ...  
    };  

Besides, you can also modify the controller code as below: change the all_inventory data type from IEnumerable<IInventory> to IEnumerable<Inventory>:

IEnumerable<Inventory> all_inventory = from m in mydb.Inventory select m;  

In this scenario, there is no need to use the select statement when create the InventoryViewModel model instance.

Finally, please check your view model, it should like this:

public class InventoryViewModel {  
    public IEnumerable<Inventory> Inventory { get; set; }  
    public PagingInfo PagingInfo { get; set; }  
}  

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Best regards,
Dillion