I have a simple problem:
In my application i have a "mainview" / Mainwindow.
From this the user can controll all the things like DB querys, searching and so on.

But i need to open a property window on a click on the menuebutton.

So i bound it to my **SearchWindowCommand**:

```
private ICommand searchwindowcommand;

       public ICommand SearchWindowCommand
       {
           get
           {
               if (searchwindowcommand == null)
               {
                   searchwindowcommand = new RelayCommand(p => ExcecuteSearchwindowcommand());
               }
               return searchwindowcommand;
           }
       }

       public void ExcecuteSearchwindowcommand()
       {

       }

```

Usualy i would open it like this: (i had the application working, but i didnt used mvvm and now i have to redo the application and figure mvvm out :) )

```
 public void auswahl_click(object sender, RoutedEventArgs e)
        {
            Einstellungen suchwindow = new Einstellungen();
            app_config_load(suchwindow);
            suchwindow.Show();

        }

```

What do i have to write into my Executecommand to show up the other window?
And should i create a new viewmodel for the other view? (i guess yes?)

Edit:
My Startapplikationcode:

```
 public partial class MainWindow : Window

{
    public MainWindow()
    {
        InitializeComponent();
        SetupBindings();
    }

    private void SetupBindings()
    {
        pViewModelList viewModel = new pViewModelList();

        personlistview.DataContext = viewModel;
    }

}

```