I have Four monitors with my PC,Is there any way to make a WPF window display cross four screens and maximized?
I'd like just to complete Mark's answer. Actually the answer is no you can't and it is a windows limitation that only allows to maximize on one screen. However you can have a non maximized window stretched across multiple screens.
A more WPF approach would be something like this:
this.WindowState = WindowState.Normal;
this.Top = SystemParameters.VirtualScreenTop;
this.Left = SystemParameters.VirtualScreenLeft;
this.Width = SystemParameters.VirtualScreenWidth;
this.Height = SystemParameters.VirtualScreenHeight;
However I don't really know how it will behave when stretching across windows arranged in T or L shape.
Yes, add a project reference to System.Windows.Forms and then add this to your MainWindow class:
protected override void OnInitialized(EventArgs e)
{
base.OnInitialized(e);
// get extents of all windows
var left = System.Windows.Forms.Screen.AllScreens.Min(screen => screen.Bounds.X);
var top = System.Windows.Forms.Screen.AllScreens.Min(screen => screen.Bounds.Y);
var right = System.Windows.Forms.Screen.AllScreens.Max(screen => screen.Bounds.X + screen.Bounds.Width);
var bottom = System.Windows.Forms.Screen.AllScreens.Max(screen => screen.Bounds.Y + screen.Bounds.Height);
var width = right - left;
var height = bottom - top;
// resize main window accordingly
this.WindowState = WindowState.Normal;
this.Top = top;
this.Left = left;
this.Width = width;
this.Height = height;
}
Related
I have a WPF Image inside a Border I currently am able to click and drag to pan the image. I want to prevent users from dragging the image off the screen. Perhaps a minimum of 100 pixels should be showing at any border. That is, if a user drags the image all the way to the left, most of the image will disappear, but 100 pixels will still be hanging out toward the left boundary of the border.
Here's my current code for the MouseMove and LeftClick event:
private void img_Box_MouseMove(object sender, MouseEventArgs e)
{
if (!img_Box.IsMouseCaptured) return;
var tt = (TranslateTransform)((TransformGroup)img_Box.RenderTransform).Children.First(tr => tr is TranslateTransform);
Vector v = start - e.GetPosition(img_Border);
tt.X = origin.X - v.X;
tt.Y = origin.Y - v.Y;
}
private void img_Box_MouseLeftButtonDown(object sender, MouseButtonEventArgs e)
{
img_Box.CaptureMouse();
var tt = (TranslateTransform)((TransformGroup)img_Box.RenderTransform).Children.First(tr => tr is TranslateTransform);
start = e.GetPosition(img_Border);
origin = new System.Windows.Point(tt.X, tt.Y);
}
I think I should be able to put a pretty simple condition on that to achieve my desired result, but I can't figure out what the condition should be.
I need help in unifying size of elements in stackpanel
void MainPageLoaded(object sender, RoutedEventArgs e)
{
var random = new Random();
for (var i = 0; i < 5; i++)
{
var grid = new Grid();
var border = new Border()
{
Height = random.Next(50, 150),
Width = random.Next(50, 150),
Margin = new Thickness(10),
BorderBrush = new SolidColorBrush(Colors.White),
BorderThickness = new Thickness(1)
};
grid.Children.Add(border);
imageBoxesStackPanel.Children.Add(grid);
}
var h = imageBoxesStackPanel.Children.Max(n => n.DesiredSize.Height);
what I am trying to achieve is to find max height and max width of each grid in stackpanel and apply it to all of them. The problem is that desired size is always wrong.
You can only do this in a custom way after a measure/arrange pass, before that the sizes won't be visible.
After that (in the OnLoaded event, which you have), you can use the ActualHeight and ActualWidth of the grids.
In short:
var h = imageBoxesStackPanel.Children.Max(n => n.ActualHeight);
This is however bad for performance and will trigger another layout pass.
Remarks:
In WPF the best solution would be a SharedSizeGroup or a UniformGrid. This is not implemented in Silverlight, but there are people who have implemented it.
In WPF there is the UniformGrid to do this job, but unfortunately it's not implemented for Silverlight by default. There are several alternatives for it, e.g. this one
I have a wpf window that hosts a control.
The window's height and width are set to SizeToControl.
Now I need to position this Window relative to the position of its parent window.(basically the Top right position).
(So my windows Top = ParentWindow.Top, and Left = ParentWindow.Left + ParentWindow.ActualWidth - control.ActualWidth so that my window is positioned inside the parent window but to its right corner)
So i will need to set the Top and Left of the window. To do this I need the Actual Width of the control that is being hosted inside it....but I can only get this once I actually do,
Window.Show(control,parent)
Is there a way to get around this problem? How do I get the actual rendered width of the control before it is actually shown?
Thanks!
Have you tried this approach?
public partial class ShellWindow : Window
{
public ShellWindow(IShellPresentationModel model)
{
InitializeComponent();
this.Loaded += ShellWindow_Loaded;
}
void ShellWindow_Loaded(object sender, RoutedEventArgs e)
{
var innerWindow = new InnerWindow();
innerWindow.Owner = this;
innerWindow.Loaded += InnerWindow_Loaded;
innerWindow.Show();
}
void InnerWindow_Loaded(object sender, RoutedEventArgs e)
{
var w = (InnerWindow)sender;
w.Owner = this;
w.Top = this.Top;
w.Left = this.Left + this.ActualWidth - w.ActualWidth;
}
}
I would like to make a small silverlight app which displays one fairly large image which can be zoomed in by scrolling the mouse and then panned with the mouse.
it's similar to the function in google maps and i do not want to use deepzoom.
here is what i have at the moment. please keep in mind that this is my first silverlight app:
this app is just for me to see it's a good way to build in a website. so it's a demo app and therefor has bad variable names.
the initial image is 1800px width.
private void sc_MouseWheel(object sender, MouseWheelEventArgs e)
{
var st = (ScaleTransform)plaatje.RenderTransform;
double zoom = e.Delta > 0 ? .1 : -.1;
st.ScaleX += zoom;
st.ScaleY += zoom;
}
this works, but could use some smoothing and it's positioned top left and not centered.
the panning is like this:
found it # Pan & Zoom Image
and converted it to this below to work in silverlight
Point start;
Point origin;
bool captured = false;
private void plaatje_MouseLeftButtonDown(object sender, MouseButtonEventArgs e)
{
plaatje.CaptureMouse();
captured = true;
var tt = (TranslateTransform)((TransformGroup)plaatje.RenderTransform)
.Children.First(tr => tr is TranslateTransform);
start = e.GetPosition(canvasje);
origin = new Point(tt.X, tt.Y);
}
private void plaatje_MouseLeftButtonUp(object sender, MouseButtonEventArgs e)
{
plaatje.ReleaseMouseCapture();
captured = false;
}
private void plaatje_MouseMove(object sender, MouseEventArgs e)
{
if (!captured) return;
var tt = (TranslateTransform)((TransformGroup)plaatje.RenderTransform).Children.First(tr => tr is TranslateTransform);
double xVerschuiving = start.X - e.GetPosition(canvasje).X;
double yVerschuiving = start.Y - e.GetPosition(canvasje).Y;
tt.X = origin.X - xVerschuiving;
tt.Y = origin.Y - yVerschuiving;
}
so the scaling isn't smooth and the panning isn't working, because when i click it, the image disappears.
first thing I noticed was this:
var st = (ScaleTransform)plaatje.RenderTransform;
and
(TransformGroup)plaatje.RenderTransform
. So in one handler, you're casting "plaatje.RenderTransform" to ScaleTransform, in the other you're casting to TransformGroup?
This is probably causing an exception, making your image disappear.
For the zooming part, you might want to try setting the RenderTransformOrigin of the object you want to scale ("plaatje"?) to "0.5,0.5", meaning the center of the UI element. So the image will be scaled around its center point instead of its top left corner.
Cheers, Alex
I'm just creating my own AboutBox and I'm calling it using Window.ShowDialog()
How do I get it to position relative to the main window, i.e. 20px from the top and centered?
You can simply use the Window.Left and Window.Top properties. Read them from your main window and assign the values (plus 20 px or whatever) to the AboutBox before calling the ShowDialog() method.
AboutBox dialog = new AboutBox();
dialog.Top = mainWindow.Top + 20;
To have it centered, you can also simply use the WindowStartupLocation property. Set this to WindowStartupLocation.CenterOwner
AboutBox dialog = new AboutBox();
dialog.Owner = Application.Current.MainWindow; // We must also set the owner for this to work.
dialog.WindowStartupLocation = WindowStartupLocation.CenterOwner;
If you want it to be centered horizontally, but not vertically (i.e. fixed vertical location), you will have to do that in an EventHandler after the AboutBox has been loaded because you will need to calculate the horizontal position depending on the Width of the AboutBox, and this is only known after it has been loaded.
protected override void OnInitialized(...)
{
this.Left = this.Owner.Left + (this.Owner.Width - this.ActualWidth) / 2;
this.Top = this.Owner.Top + 20;
}
gehho.
I would go the manual way, instead of count on WPF to make the calculation for me..
System.Windows.Point positionFromScreen = this.ABC.PointToScreen(new System.Windows.Point(0, 0));
PresentationSource source = PresentationSource.FromVisual(this);
System.Windows.Point targetPoints = source.CompositionTarget.TransformFromDevice.Transform(positionFromScreen);
AboutBox.Top = targetPoints.Y - this.ABC.ActualHeight + 15;
AboutBox.Left = targetPoints.X - 55;
Where ABC is some UIElement within the parent window (could be Owner if you like..) , And could also be the window itself (top left point)..
Good luck