如何将 WPF UserControl 从一个窗口移动到另一个窗口？

Question

假设我在 MainWindow.xaml 中有一个名为 MyVideoControl 的用户控件：

<Window Name="_mainWindow">
    <Grid>
        <MyVideoControl Name="_localVideo"/>
    </Grid>
</Window>

现在用户单击一个按钮，我希望 UserControl 浮动在 MainWindow.xaml 的顶部，在一个名为 PopUp.xaml 的新创建的窗口内。

<Window Name="_popUpWindow">
    <Grid>
        <MyVideoControl Name="_localVideo"/>
    </Grid>
</Window>

我如何做到这一点，以便整个对象被移动？目前我使用 XAML 以声明方式将 MyVideoControl 放置在我的窗口中，但我猜我需要以编程方式完成所有操作？

Answer 1

是的，您可以通过从Mainwindow 中删除userControl 并将其作为逻辑子项添加到PopupWin 窗口中的任何控件来实现。

UserControl.xaml：

<UserControl x:Class="WpfApplication1.UserControl1"
             xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
             xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
             xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006" 
             xmlns:d="http://schemas.microsoft.com/expression/blend/2008" 
             mc:Ignorable="d" 
             d:DesignHeight="100" d:DesignWidth="100">
    <Grid>
        <TextBox x:Name="txtBlock1" Text="hai"/>
    </Grid>
</UserControl>

MainWindow.xaml：

<Window x:Class="WpfApplication1.MainWindow"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml" xmlns:WpfApplication1="clr-namespace:WpfApplication1" Title="MainWindow" Height="550" Width="555">
    <Grid>
        <StackPanel x:Name="mainPanel" Orientation="Vertical ">
            <Button Content="Button" Height="23" Name="button1" Width="75" Click="button1_Click" />
            <WpfApplication1:UserControl1 x:Name="myUserControl" />
        </StackPanel>
    </Grid>
</Window>

PopupWin.xaml：

<Window x:Class="WpfApplication1.PopupWin"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        Title="PopupWin" Height="300" Width="300">

    <StackPanel x:Name="mainPanel"/>

</Window>

PopupWin.xaml.cs: 公开一个新的构造函数来接受userControl 并将其作为子级添加到mainPanel

public partial class PopupWin : Window
{
    public PopupWin()
    {
        InitializeComponent();
    }

    private UserControl control;

    public PopupWin(UserControl control)
        : this()
    {
        this.control = control;

        this.mainPanel.Children.Add(this.control);
    }
}

MainWindow.xaml.cs 在 Button_Click 上，从当前 MainWindow 中移除 userControl，并将其传递给 PopupWin，在本例中是通过构造函数。

    private void button1_Click(object sender, RoutedEventArgs e)
    {
        this.mainPanel.Children.Remove(this.myUserControl);

        var wind = new PopupWin(this.myUserControl);

        wind.ShowDialog();
    }

注意：userControl 实例在任何时候都应该是仅one 元素的逻辑子元素。

Answer 2

如果您将UserControl 设为资源，您可以这样做。

例子：

App.Xaml

<Application x:Class="WpfApplication10.App"
             xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
             xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
             StartupUri="MainWindow.xaml">
    <Application.Resources>
        <ContentControl x:Key="sharedContent">
            <Label Content="StackOverFlow" />
        </ContentControl>
    </Application.Resources>
</Application>

主窗口

<Window x:Class="WpfApplication10.MainWindow"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        Title="MainWindow" Height="325" Width="422" Name="UI">

    <StackPanel>
        <Button Content="Open PopUp" Click="Button_Click" />
        <ContentControl Content="{DynamicResource sharedContent}" />
    </StackPanel>
</Window>

弹出窗口

<Window x:Class="WpfApplication10.PopupWindow"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        Title="PopupWindow" Height="300" Width="300" xmlns:my="clr-namespace:WpfApplication10">
    <Grid>
        <ContentControl Content="{DynamicResource sharedContent}" />
    </Grid>
</Window>

结果：

Answer 3

另一种实现方式，使用动态布局...

 public partial class MainWindow : Window
{
    public Button ControlToMove { get; set; }

    public MainWindow()
    {
        InitializeComponent();

        //create button
        ControlToMove = new Button();

        //add text to button
        ControlToMove.Content = "Click to move me to pop up window";

        //add a new routed event to the buttons click property
        ControlToMove.Click += new RoutedEventHandler(MoveControlToPopUp);

        //add control to the layout grid
        MainGrid.Children.Add(ControlToMove);
    }

    //This method moves the button to a popup window
    private void MoveControlToPopUp(object sender, RoutedEventArgs e)
    {
        //get the name of the control from sender
        var control = sender as Button;
        var controlName = control.Name;

        //checks to see if this is the control we want moved
        //if its not, method exits
        if (controlName != ControlToMove.Name) return;

        //create copy of the control
        var copiedControl = control;

        //remove control from existing window
        MainGrid.Children.Remove(control);

        //create pop up window
        var popUpWindow = new PopUpWindow(copiedControl);
        popUpWindow.Show();
    }
}

public class PopUpWindow : Window
{
    public Grid Layout { get; set; }
    public PopUpWindow(Button button)
    {
        //create a grid for the new window
        Layout = new Grid();

        //add control to grid
        Layout.Children.Add(button);

        //add grid to window
        this.AddChild(Layout);
    }
}

}

我唯一没有做的就是添加在弹出窗口中单击按钮时将按钮移回主窗口的功能。