如何在 WPF 中绘制可点击的矩形
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How to draw a clickable rectangle in WPF
提问by FrostyFire
I am an absolute beginner to WPF applications and need some help. All I'm trying to do is draw a rectangle from point A to point B, andbe able to detect when the rectangle is clicked. So when it is clicked it turns yellow and when clicked again, red.
我是 WPF 应用程序的绝对初学者，需要一些帮助。我想要做的就是从 A 点到 B 点绘制一个矩形，并能够检测到何时单击该矩形。所以当它被点击时它会变成黄色，再次点击时它会变成红色。
回答by McGarnagle
There are multiple ways to do this.
有多种方法可以做到这一点。
- Add a click handler to the rectangle, and toggle its color from code behind
- Bind the rectangle's color to a View Model property, and set the property on click using a Delegate Command.
- 向矩形添加单击处理程序，并从后面的代码切换其颜色
- 将矩形的颜色绑定到视图模型属性，并在单击时使用委托命令设置属性。
The first is easiest if you're just starting with XAML (although #2 is recommended if you want to adhere to MVVM).
如果您刚开始使用 XAML，第一个是最简单的（尽管如果您想坚持使用 MVVM，建议使用 #2）。
<Rectangle x:Name="rect"
Width="100" Height="100" Fill="Aquamarine"
MouseLeftButtonDown="Rectangle_MouseLeftButtonDown" />
And the code-behind handler:
以及代码隐藏处理程序：
bool toggle = false;
private void Rectangle_MouseLeftButtonDown(object sender, MouseButtonEventArgs e)
{
rect.Fill = new SolidColorBrush(toggle ? Colors.Aquamarine : Colors.DarkRed);
toggle = !toggle;
}
回答by Kirk Broadhurst
Use the Rectanglecontrol.
使用Rectangle控件。
<Rectangle
Height="100"
Width="100"
MouseLeftButtonUp="Rectangle_MouseLeftButtonUp_1"
where Rectangle_MouseLeftButtonUp_1is an event handler on the containing class.
其中Rectangle_MouseLeftButtonUp_1是包含类的事件处理程序。
Just be aware that unless the Rectangle has a background, you'll have to click the border. The background can be white, but it does need to be specified if it's to be clickable.
请注意，除非矩形有背景，否则您必须单击边框。背景可以是白色的，但如果它是可点击的，则需要指定它。
