如何使用按钮单击从另一个 xaml 窗口打开一个 xaml 窗口?

问题描述：

private void Button_Click(object sender, System.Windows.RoutedEventArgs e)
        {
            Win1 OP= new Win1();
            OP.show();
        }

OP.show() 抛出错误.

OP.show() is throwing an error.

这是一个用户控件表单.

It is a usercontrol form.

答

你说 Win1 是它是一个 usercontrol 表单."(重点是我的).

You say that Win1 is "It is a usercontrol form." (emphasis is mine).

如果 Win1 实际上是 UserControl 类型，问题是 UserControl 类型没有定义 Show()代码>方法.所以它不能作为一个窗口打开".

If Win1 is actually of type UserControl, the issues is that the type UserControl doesn't define Show() method. So it cannot be "opened" as a window.

要解决这个问题，您需要打开一个窗口并将 UC 作为该窗口的内容:

To solve this you need to open a window and have the UC as the content for that window:

private void Button_Click(object sender, System.Windows.RoutedEventArgs e)
{
    Win1 OP= new Win1();
    var host = new Window();
    host.Content = OP;
    host.Show();
}

附带说明，您可以在 App.xaml 中使用 UserControl 作为 StartupUri 并且它会起作用，因为框架识别出它不是窗口并为

As a side note, you can use UserControl as StartupUri in App.xaml and it will work since the framework recognizes that it's not a window and creates a window for it.

如何使用按钮单击从另一个 xaml 窗口打开一个 xaml 窗口?

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