WPF,窗口拖动事件与窗口内控件点击事件 - 教程

帮手程序(assister.exe)在运行鼠标自动操作脚本时,可以弹出置顶的悬浮窗口,实时显示程序运行状态。悬浮窗口上一个小按钮可以快速切换回帮手脚本主窗口。如图:

这个悬浮窗口许可自由拖动到屏幕任意位置,避免遮挡其他操作窗口。实现该特性时,通常只需在窗口控件中处理PreviewMouseLeftButtonDown事件,并执行以下代码:

private void MouseLeftButtonDown(object sender, MouseButtonEventArgs e)
{
if (e.LeftButton == MouseButtonState.Pressed)
{
this.DragMove();
}
}

然而,由于窗口内部包含按钮控件需响应点击操作,上述代码会导致按钮点击事件无法触发。

修改后的解决方案是让窗口处理以下三个关键事件:


<Window
...
PreviewMouseLeftButtonDown="Window_PreviewMouseLeftButtonDown"
PreviewMouseMove="Window_PreviewMouseMove"
PreviewMouseLeftButtonUp="Window_PreviewMouseLeftButtonUp"

...
Topmost="True"
WindowStyle="None"
AllowsTransparency="True"
Background="Transparent"
>

private voidWindow_PreviewMouseLeftButtonDown(object sender, MouseButtonEventArgs e)
{
_mousePressedPosition = e.GetPosition(this);
}

private voidWindow_PreviewMouseMove(object sender, MouseEventArgs e)
{
if (Mouse.LeftButton == MouseButtonState.Pressed && _mousePressedPosition != e.GetPosition(this))
{
_isMouseDragMove = true;
DragMove();
}
}

private voidWindow_PreviewMouseLeftButtonUp(object sender, MouseButtonEventArgs e)
{
if (_isMouseDragMove)
{
_isMouseDragMove = false;
e.Handled = true;
}
}

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posted @ 2025-07-30 12:04  yfceshi  阅读(7)  评论(0)    收藏  举报