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How to tell if a ChildWindow is the 'top most' window

I'm working in Silverlight, but potentially a WPF solution would work as well.
My problem is very simple. I have lots of modal Child Windows that can be open, and in their generic menu is a home button. This button is supposed to close all of the child windows and return to the base screen. I have a few different types of 'generic child windows' that host lots of different UserControls, so by far the easiest way to implement this is to, when the window comes into focus, check if the global ReturnToHome bool is true, and if it is, just close it.
I've tried all of these
private void ChildWindow_GotFocus(object sender, RoutedEventArgs e)
{
if (CommonResources.ReturnToHome) DialogResult = false;
}
private void ChildWindow_MouseEnter(object sender, MouseEventArgs e)
{
if (CommonResources.ReturnToHome) DialogResult = false;
}
private void ChildWindow_IsEnabledChanged(object sender, DependencyPropertyChangedEventArgs e)
{
if (CommonResources.ReturnToHome) DialogResult = false;
}
The issue is, GotFocus doesn't fire until I actually click on the window. MouseEnter is a little better, but doesn't fire until I move the mouse. IsEnabledChanged never fires because the Child Window doesn't disable anything. Checking every child window when it closes to see if Home has been clicked isn't easy because of the sheer number of places where you can open child windows, and several of them are nested within User Controls where I couldn't even easily access DialogResult. Any idea how I could do this?
Also I should note that I want each of the windows to close one by one, from top down, because each window that closes does its own verification to see if it should warn the user before closing (giving the user the option to cancel closing)

TopMost is a bool property which is either set to true or false and as far as I'm aware, there is no public property like Z-Index that will tell you the order in which your Windows were set to TopMost. However, there is a simple solution... just maintain a static int variable that will register this order. Each time you add a new Window, set the number into its Tag property:
Window childWindow = new Window();
childWindow.Tag = currentWindowNumber++;
...
childWindow.ShowDialog();
Then, when you want to close them in order, you can just do something like this:
foreach (Window window in Application.Current.Windows.OfType<YourWindowType>()
.OrderBy(w => (int)w.Tag))
{
((AnimationWindow)window).CloseWindow();
}

How to bring the main form back to the front after the file open dialog is closed

My C# application starts by opening a form. In the constructor for that form I "showDialog" an openfiledialog. After selecting a file to open, the openfile dialog closes, the file is loaded and the contents displayed in the main form but the main form is buried behind every other open window on my desktop.
I have to find it in the task bar and bring it to focus. I just started the application, I want the form to have focus.
I have written other applications that do not use the openfiledialog and when I start them the main form opens with focus as you would expect.
How do I make the main form get focus after the openfiledialog closes?
I have tried
this.focus(),
this.activate(),
this.bringtofront();
and this.TopMost = true;
None of them make any apparent difference at all.
I have research this problem extensively and this are the things everyone suggests and say work, but they don't work for me. Some have insinuated that I am violating all that is holy by trying to make my form topmost. However, I don't think very many people would like to open an application and have the main form for it show up behind everything else.
Any one have any other ideas about how to make sure my form is "in front", topmost, has focus?

When you do it this way, your application will have a brief moment where no window is available to receive the focus after the dialog closes. Windows is forced to find another window to give the focus to, that will be a window of another app. Your main window eventually appears, now behind that other's app window.
Display the dialog in an event handler of the Shown event instead. Or use the boilerplate File + Open command.

SOLUTION: this.Activate(); works but if called from the form Load event.

This will set the window on top:
[DllImport("user32.dll")]
static extern bool SetForegroundWindow(IntPtr hWnd);
private void Form1_Load(object sender, EventArgs e)
{
....
//after your code place the call to the function at the end.
SetForegroundWindow(this.Handle);
}
Althought as Mr. hans said and very well you are better off with another design.

If you can, move the ShowDialog out of the constructor, or try putting this in the constructor:
this.Shown += OnShown;
and move your ShowDialog to here:
private void OnShown(object sender, EventArgs eventArgs)
{
var result = new OpenFileDialog().ShowDialog();
}

new window appears under main window

I am writing an application using WPF. I would like such an action: when I press down mouse button on a button, another window appears, when I release mouse button wherever, this window hides. That is my code so far:
XAML:
<Button Margin="0,0,0,0" Name="button_wykres" PreviewMouseUp="button_wykres_PreviewMouseUp" PreviewMouseDown="button_wykres_PreviewMouseDown">
C#:
private void button_wykres_PreviewMouseDown(object sender, MouseButtonEventArgs e)
{
omww.Top = this.Top+50;
omww.Left = this.Left +180;
omww.Show();
}
private void button_wykres_PreviewMouseUp(object sender, MouseButtonEventArgs e)
{
omww.Hide();
}
The problem is that new window (omww) appears under the main one, but I need it on top. When I tried other events, i.e. button.Click window is showed above the old one, as I want, and that confuses me. However, Click event doesn't meet my needs. I'd be grateful if anybody help me.

As I see it, you have two possible options for making the child Window appear on top of the parent Window. The first was mentioned by #Viv in a comment and that is to set the TopMost property on the child Window to true:
omww.Topmost = true;
The second option would be to set the Owner property of the child Window to the parent Window (if there is a direct relationship between them):
omww.Owner = this; // if called from the parent Window code behind

C# WPF button navigation to a usercontrol page returns null

i have this code here in my mainwindow :
private void btnHome_Click(object sender, RoutedEventArgs e)
{
NavigationService.GetNavigationService(this).Navigate("home.xaml");
}
home.xaml is my user control page which i want the button to navigate to , but when i click it , the whole thing crashes and i debug it , it shows NullException.
What actually gone wrong?
I am able to run the MainWindow and home.xaml separately well.

Click to new window.xaml

im learning wpf for the first time,
i have made this far
private void Button_Click(object sender, System.Windows.RoutedEventArgs e)
{
// TODO: Add event handler implementation here.
}
lets say its my click button 'home' some how i have made a new window store.xaml at the same product.
how can i connect them ?
heres a sc

Your question seems a bit vague to me, if you simply want to display the store inside the same window you should not implement the content of the window directly but only use the window as a shell for your content, if your store is a window as well you should refactor it into a UserControl which then can be added to the window.
You can also use Pages, see the Navigation Overview for more info on that.