Question

我使用MouseEnter事件进行简单的边框控制，打开弹出控件。弹出窗口比边界更大，因此它超出了边界的范围。弹出窗口还有一些按钮和文本块，所以我需要对它进行一些控制。现在，我想要获得：鼠标进入边框 - 弹出窗口显示。鼠标离开边框和弹出窗口 - 然后弹出窗口关闭。因此，当鼠标不在边界上但超过弹出窗口时 - 弹出窗口保持打开状态。我不想点击关闭。你能帮助我吗？ WPF中的弹出控件对我来说很困惑。

编辑：对，代码：

<Window x:Class="Sandbox.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:Sandbox"
    mc:Ignorable="d"
    Title="MainWindow" Height="350" Width="525">
<Grid Margin="20">
    <Popup PopupAnimation="Fade" Placement="Top" PlacementTarget="{Binding ElementName=OpenPopupBorder}" AllowsTransparency="True" StaysOpen="True" x:Name="DeviceStatusSnippetPopup">
        <Border Background="White" Padding="5" BorderBrush="Black" BorderThickness="1">
            <Grid>
                <StackPanel Orientation="Vertical">
                    <TextBlock HorizontalAlignment="Center" Text="Device status snippet" Margin="0 10 0 10"/>
                    <StackPanel Orientation="Horizontal">
                        <Button Style="{StaticResource btn-default}" Margin="0 0 5 0" Content="Go to device"/>
                        <Button Style="{StaticResource btn-default}" Content="Refresh device"/>
                    </StackPanel>
                </StackPanel>
            </Grid>
        </Border>
    </Popup>
    <Border x:Name="OpenPopupBorder" Width="90" Height="30" Background="LightGray" MouseEnter="Border_MouseEnter"/>
</Grid>

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

    private void Border_MouseEnter(object sender, MouseEventArgs e)
    {
        DeviceStatusSnippetPopup.IsOpen = true;
    }
}

Answer 1

代码隐藏：

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

    private void Border_MouseEnter(object sender, MouseEventArgs e)
    {
        DeviceStatusSnippetPopup.IsOpen = true;
    }

    private void Popup_MouseLeave(object sender, MouseEventArgs e)
    {
        if (!OpenPopupBorder.IsMouseOver)
            DeviceStatusSnippetPopup.IsOpen = false;
    }


    private void Border_MouseLeave(object sender, MouseEventArgs e)
    {
        if (!DeviceStatusSnippetPopup.IsMouseOver)
            DeviceStatusSnippetPopup.IsOpen = false;
    }
}

XAML（减去窗口）：

<Grid Margin="20">
    <Popup MouseLeave="Popup_MouseLeave" PopupAnimation="Fade" Placement="Top" PlacementTarget="{Binding ElementName=OpenPopupBorder}" AllowsTransparency="True" StaysOpen="True" x:Name="DeviceStatusSnippetPopup">
        <Border Background="White" Padding="5" BorderBrush="Black" BorderThickness="1">
            <Grid>
                <StackPanel Orientation="Vertical">
                    <TextBlock HorizontalAlignment="Center" Text="Device status snippet" Margin="0 10 0 10"/>
                    <StackPanel Orientation="Horizontal">
                        <Button Margin="0 0 5 0" Content="Go to device"/>
                        <Button Content="Refresh device"/>
                    </StackPanel>
                </StackPanel>
            </Grid>
        </Border>
    </Popup>
    <Border x:Name="OpenPopupBorder" Width="90" Height="30" Background="LightGray" MouseEnter="Border_MouseEnter" MouseLeave="Border_MouseLeave"/>
</Grid>

基本上，你为边框和弹出窗口实现MouseLeave的事件处理程序，并在关闭弹出窗口之前检查鼠标是否在另一个上面。

WPF弹出窗口 - 仅在鼠标上隐藏离开控件和弹出窗口

1 个答案: