Refer to active Window in WPF?

How can I refer to active Window of WPF application in C#, using something like ActiveForm property in WinForms?

One possible way would be to scan the list of open windows in the application and check which one of them has IsActive = true:

Application.Current.Windows.OfType<Window>().SingleOrDefault(x => x.IsActive);

Not sure if there may be more than one active window if, for example, there's a modal dialog showing, in which case, the owner of the dialog and the dialog itself might be active.

There is better way to do this using PInvoke. Aviads answer is not working all the time (there are some edge cases with dialogs).

IntPtr active = GetActiveWindow();

ActiveWindow = Application.Current.Windows.OfType<Window>()
    .SingleOrDefault(window => new WindowInteropHelper(window).Handle == active);

One must include following import first:

[DllImport("user32.dll")]
static extern IntPtr GetActiveWindow();

I know this is a bit old question but I think my answer could help someone.

My problem was this: I had a WPF MVVM application and I needed to get my MainWindow instance in the second view, i.e. second view model, in order to set the visibility of title bar button to visible.

This is my solution:

MainWindow window = (MyApp.MainWindow)App.Current.MainWindow;
window.btnSearch.Visibility = System.Windows.Visibility.Visible;

Hope this would help someone.

I have problems With this way "Application.Current.Windows.OfType().SingleOrDefault(x => x.IsActive);" specialy because I was building an aplication with a main Window then i had problems when the main window was selected. I resolve it creating this:

In some base class or App.xaml.cs create this:

       public static Window ActivatedWindow {get;set;}

Then put in your base class deriving Window or all of your Window's Activate Event:

First Option - personal Window Base Class:

       public class MetroToolWindowBase
       {
         public MetroToolWindowBase()
         {   
            Activated += new EventHandler(MakeActive); 
         }   
         private void MakeActive(object sender, EventArgs e)
         {
        App.ActivatedWindow= this;
         }
       }

Second Option- In Activated Event of Windows:

   private void XWindow_Activated(object sender,EventArgs e)
    {
     App.ActivatedWindow= this;
    }
       

Another way to do it is to use the native GetActiveWindow function from user32.dll.

[DllImport("user32.dll", CharSet = CharSet.Auto)]
public static extern IntPtr GetActiveWindow();

To convert it to an actual WPF Window:

IntPtr handle = GetActiveWindow();

HwndSource hwndSource = HwndSource.FromHwnd(handle);
var window = hwndSource?.RootVisual as Window;

If hosting a WPF Window in a WinForms app, WindowInteropHelper should be used. This makes for example the Window owner work correctly:

var wih = new WindowInteropHelper(window)
{
    Owner = GetActiveWindow()
};

I edited my old answer because the edge case I encountered disappeared after a Visual Studio update, but it can be checked from answer history. I encountered an issue there where I was getting null for active window in certain circumstances while debugging.