在 WPF c# 中为子窗口禁用捕获屏幕

Question

我正在尝试禁用捕获屏幕，它正在与我的主窗口一起使用 -

[DllImport("user32.dll")]
        public static extern uint SetWindowDisplayAffinity(IntPtr hwnd, uint dwAffinity);

var mainWindowHandle = new WindowInteropHelper(this).Handle;
                const uint WDA_NONE = 0;
                const uint WDA_MONITOR = 1;
                SetWindowDisplayAffinity(mainWindowHandle, WDA_MONITOR);

但是当我打开一个弹出/子窗口时它不起作用。有什么方法可以与子窗口一起使用吗？ 任何帮助将不胜感激提前谢谢。

Answer 1

Deepankshee，我不确定你的问题是什么。它对我有用，而且我完全按照您的指示进行操作。

之前（两个窗口中都没有 OnLoaded 事件）：

之后（两个窗口中都有 OnLoaded 事件）：

这是我的代码...

MainWindow.xaml：

<Window x:Class="CaptureScreenTest.MainWindow"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
        xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
        xmlns:local="clr-namespace:CaptureScreenTest"
        mc:Ignorable="d"
        Title="MainWindow" Height="450" Width="800">
    <Grid>
        <StackPanel>
            <TextBlock FontSize="64" Text="What up? I'm a Main Window." HorizontalAlignment="Center" VerticalAlignment="Center" TextWrapping="Wrap" Margin="5"/>
            <Button Content="Open Child" Click="Button_Click" HorizontalAlignment="Center" VerticalAlignment="Center" Padding="15,5" Margin="15"/>
        </StackPanel>
    </Grid>
</Window>

MainWindow.xaml.cs：

using System;
using System.Runtime.InteropServices;
using System.Windows;
using System.Windows.Interop;

namespace CaptureScreenTest
{
    /// <summary>
    /// Interaction logic for MainWindow.xaml
    /// </summary>
    public partial class MainWindow : Window
    {
        [DllImport("user32.dll")]
        public static extern uint SetWindowDisplayAffinity(IntPtr hwnd, uint dwAffinity);

        public MainWindow()
        {
            InitializeComponent();
            //this.Loaded += MainWindow_Loaded;
        }

        private void MainWindow_Loaded(object sender, RoutedEventArgs e)
        {
            var mainWindowHandle = new WindowInteropHelper(this).Handle;
            const uint WDA_NONE = 0;
            const uint WDA_MONITOR = 1;
            SetWindowDisplayAffinity(mainWindowHandle, WDA_MONITOR);
        }

        private void Button_Click(object sender, RoutedEventArgs e)
        {
            var child = new ChildWindow();
            child.Show();
        }
    }
}

ChildWindow.xaml：

<Window x:Class="CaptureScreenTest.ChildWindow"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
        xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
        xmlns:local="clr-namespace:CaptureScreenTest"
        mc:Ignorable="d"
        Title="ChildWindow" Height="450" Width="800">
    <Grid>
        <TextBlock FontSize="64" Text="I'm a Child Window." HorizontalAlignment="Center" VerticalAlignment="Center" TextWrapping="Wrap"/>
    </Grid>
</Window>

ChildWindow.xaml.cs：

using System;
using System.Runtime.InteropServices;
using System.Windows;
using System.Windows.Interop;

namespace CaptureScreenTest
{
    /// <summary>
    /// Interaction logic for ChildWindow.xaml
    /// </summary>
    public partial class ChildWindow : Window
    {
        [DllImport("user32.dll")]
        public static extern uint SetWindowDisplayAffinity(IntPtr hwnd, uint dwAffinity);
        public ChildWindow()
        {
            InitializeComponent();
            //this.Loaded += ChildWindow_Loaded;
        }

        private void ChildWindow_Loaded(object sender, RoutedEventArgs e)
        {
            var mainWindowHandle = new WindowInteropHelper(this).Handle;
            const uint WDA_NONE = 0;
            const uint WDA_MONITOR = 1;
            SetWindowDisplayAffinity(mainWindowHandle, WDA_MONITOR);
        }
    }
}

Answer 2

如果您希望在应用程序的所有窗口上全局发生这种情况，您最好预先定义自己的自定义窗口类并将其放入其中。然后从您的类与默认窗口派生所有窗口......类似

using System;
using System.Windows;
using System.Runtime.InteropServices;
using System.Windows.Interop;


namespace CaptureScreenTest
{
   public class YourBaseWindow : Window
   {
      public YourBaseWindow()
      {  
         Loaded += MyWindowLoaded;
      }

      [DllImport("user32.dll")]
      public static extern uint SetWindowDisplayAffinity(IntPtr hwnd, uint dwAffinity);

      private void MyWindowLoaded(object sender, RoutedEventArgs e)
      {
         var _curWindHandle = new WindowInteropHelper(this).Handle;
         // const uint WDA_NONE = 0;
         const uint WDA_MONITOR = 1;
         SetWindowDisplayAffinity(_curWindHandle, WDA_MONITOR);
      }
   }
}

然后，在您的所有窗口中，由于“本地”引用您的 CaptureScreenTest 命名空间，因此只需从 ex 更改您的所有窗口定义：

<Window x:Class="CaptureScreenTest.MainWindow"
   … etc...
   xmlns:local="clr-namespace:CaptureScreenTest"

到

<local:YourBaseWindow x:Class="CaptureScreenTest.MainWindow"
   … etc...
   xmlns:local="clr-namespace:CaptureScreenTest"

那么您就不必在每个窗口上都这样做了。您甚至可以参数化您的窗口以确定是否要启用它。我为我的 WPF 应用程序做了类似的基线窗口声明，所有表单都从它派生。我的核心窗口类中还有其他捕获/恒定的功能，这一切都是为了顺风顺水，我永远不必忘记在其他窗口上实现。

现在，由于您未来的窗口声明继承自此，如果它们也有“已加载”事件调用，则父类仍将自己做，派生类也将继续做他们需要做的事情。

Answer 3

经过大量分析，在我的xaml文件的子窗口中找到了解决方案-

<Window x:Class="Test.Popup"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
        xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
        xmlns:local="clr-namespace:AllenClassRoom"
        xmlns:materialDesign="http://materialdesigninxaml.net/winfx/xaml/themes"
        mc:Ignorable="d"
        Title="Popup" ResizeMode="NoResize" WindowStartupLocation="CenterScreen" AllowsTransparency="True" WindowStyle="None" Loaded="Window_Loaded">
    <Grid>

AllowsTransparency="True" 此属性设置为 true。由于此属性，此函数 (SetWindowDisplayAffinity(mainWindowHandle, WDA_MONITOR);) 总是返回 false。 删除此属性并开始工作。 希望这会有所帮助