Question

我正在尝试构建一个弹出窗口，当用户将鼠标悬停在元素上时会显示该弹出窗口。如果用户将鼠标移到弹出窗口上，我希望它保持显示状态。但是，如果用户同时保留owner元素和popup元素，则弹出元素应该消失。

我尝试了以下但不起作用：

<UserControl x:Class="SilverlightApplication1.MainPage"
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"
mc:Ignorable="d"
d:DesignHeight="300" d:DesignWidth="400">

<StackPanel MouseEnter="OnMouseEnter" MouseMove="OnMouseMove" MouseLeave="OnMouseLeave">
    <HyperlinkButton Content="Root" HorizontalAlignment="Left"/>
    <Popup x:Name="popup" MouseEnter="OnMouseEnter" MouseMove="OnMouseMove"  MouseLeave="OnMouseLeave">
        <StackPanel x:Name="leaves" HorizontalAlignment="Left">
            <HyperlinkButton Content="Leaf1" />
            <HyperlinkButton Content="Leaf2" />
        </StackPanel>
    </Popup>
</StackPanel>

代码背后：

public partial class MainPage : UserControl
{
    public MainPage()
    {
        InitializeComponent();
    }

    private void OnMouseMove(object sender, MouseEventArgs e)
    {
        this.popup.IsOpen = true;
    }

    private void OnMouseEnter(object sender, MouseEventArgs e)
    {
        this.popup.IsOpen = true;
    }

    private void OnMouseLeave(object sender, MouseEventArgs e)
    {
        this.popup.IsOpen = false;
    }
}

触发所有者元素（Root）的MouseLeave并弹出Popup会发生什么。

有什么想法吗？

Answer 1

使用DispatcherTimer将IsOpen实际设置为false，如下所示： -

public partial class MainPage: UserControl
{
    DispatcherTimer popupTimer = new DispatcherTimer();

    public MainPage()
    {
        InitializeComponent();

        popupTimer.Interval = TimeSpan.FromMilliseconds(100);
        popupTimer.Tick += new EventHandler(popupTimer_Tick);
    }

    void popupTimer_Tick(object sender, EventArgs e)
    {
        popupTimer.Stop();
        popup.IsOpen = false;
    }

    private void OnMouseEnter(object sender, MouseEventArgs e)
    {
        popupTimer.Stop();
        popup.IsOpen = true;
    }

    private void OnMouseLeave(object sender, MouseEventArgs e)
    {
        popupTimer.Start();
    }
}

还将MouseEnter MouseMove事件从Popup移到弹出窗口内的StackPanel。

Silverlight：只要鼠标位于所有者元素上方或弹出窗口本身，就会显示弹出窗口

1 个答案: