Changing the start up location of a WPF window

Solution 1:

Just set WindowStartupLocation, Height, Width, Left, and Top in xaml:

<Window x:Class="WpfApplication1.Window1" 
    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation" 
    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml" 
    Title="Window1" 
    Height="500" Width="500"
    WindowStartupLocation="Manual" 
    Left="0" Top="0">
</Window>

Solution 2:

For people who like me wanted to set the position of the window to the current mouse position, you can do it like this:

myWindow.WindowStartupLocation = WindowStartupLocation.Manual;
myWindow.Left = PointToScreen(Mouse.GetPosition(null)).X;
myWindow.Top = PointToScreen(Mouse.GetPosition(null)).Y;

Solution 3:

I like to use WindowStartupLocation="CenterOwner" (MSDN docs for it)

The caller needs to specify itself as owner for this to work though, such as:

new MyWindow() { Owner = this }.ShowDialog();

Then just define the window height and width, e.g:

<Window ...
     Height="400" Width="600"
     WindowStartupLocation="CenterOwner"
>
...

Solution 4:

There is a property for Window, called "WindowStartupLocation" You can find that in properties window. Simply just select Window in constructor, then go to properties list. Search for "Startup" or smth similar and you can find that property. Change it to "CenterScreen" and it will make the deal. NOTE! Make sure, that you did not select grid instead of window! Otherwise you`ll fail.

Or you just can done it via XAML editing as some guys wrote before.