My problem is that I want to access from a page to the properties of a control (button, textblock, label, or a menuitem of the window....) placed in a window. The page is placed into the window. How can I do this? Is there any method 开发者_Go百科to find controls by name in a specific window or page or entire application?
A Window object has a method FindName() (inherited from FrameworkElement). MSDN: http://msdn.microsoft.com/en-us/library/system.windows.frameworkelement.findname.aspx
If I understand correctly, you'll then need to use reflection to enumerate the control's properties.
Update:
In the XAML you'll have something like this:
ListBox o = this.FindName("myListBox") as ListBox;
In the code behind for the window you could use:
<Window>
<Grid>
<ListBox x:Name="myListBox" />
</Grid>
</Window>
There's no "new" involved in the process.
I have obtained the window reference using Reflection:
public static Window WindowByName(string strWindowName)
{
if (string.IsNullOrEmpty(strWindowName)) return null;
Assembly asm = Assembly.GetExecutingAssembly();
string strFullyQualifiedName = asm.GetName().Name + "." + strWindowName;
object obj = asm.CreateInstance(strFullyQualifiedName);
if (obj != null)
return obj as Window;
else
return null;
}
with the above method I retrieve the reference of WinMain.xaml and then I address its controls by using FindName method from PageRutas.xaml code behind like this:
// Retrieve reference WinMain.xaml reference
Window win = cWindow.WindowByName("WinMain");
// Retrieve listbox reference
System.Windows.Controls.ListBox lstbox =
(System.Windows.Controls.ListBox)win.FindName("LayoutListBox");
// Modifying IsEnabled property
lstbox.IsEnabled = false;
but now... I have another problem.... property changed to false after executing lstbox.IsEnable = false; line but something strange is happening because WinMain.xaml UI seems to ignore the change I have made (the listbox doesn't disabled in the UI). What's happening? anybody can help me?
Thanks.
var mainWindow = Application.Current.MainWindow as MainWindow;
//Where MainWindow is Window type
use mainWindow.controlName to access the control
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