使用Window类中的ShowDialog方法显示WPF窗口对话框,就像在主窗口上按下按钮一样,如下所示.
private void button1_Click(object sender, RoutedEventArgs e)
{
try
{
var window = new Window1();
window.ShowDialog();
}
catch (ApplicationException ex)
{
MessageBox.Show("I am not shown.");
}
}
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该窗口在xaml中订阅了一个Loaded事件,如下所示:
<Window x:Class="Stackoverflow.Window1"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
Title="Window1" Loaded="Window_Loaded">
<Grid />
</Window>
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Window_Loaded事件中抛出异常
private void Window_Loaded(object sender, RoutedEventArgs e)
{
throw new ApplicationException();
}
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但是,ShowDialog调用周围没有捕获异常,调用也没有返回.吞下异常并仍然显示窗口.
为什么会发生这种情况?如何处理WPF窗口的Window_Loaded事件中的异常?我是否必须在事件处理程序中捕获它并手动处理窗口?
在WinForms中,您需要打电话 Application.SetUnhandledExceptionMode(UnhandledExceptionMode.ThrowException)
为了让异常通过ShowDialog调用冒泡.是否需要在WPF上设置类似的开关?
我只在x64机器上看到过这个问题,代码用Any Cpu编译.将程序更改为编译为x84可能会修复它,但我自己也有问题,具体取决于我们的程序集.
我唯一的代码建议如下,即便如此,也不保证能够提取它.捕获异常,并将其重新抛出到Background worker中.
private void Window_Loaded(object sender, RoutedEventArgs e)
{
try
{
/// your code here...
throw new ApplicationException();
/// your code here...
}
catch (Exception ex)
{
if (IntPtr.Size == 8) // 64bit machines are unable to properly throw the errors during a Page_Loaded event.
{
BackgroundWorker loaderExceptionWorker = new BackgroundWorker();
loaderExceptionWorker.DoWork += ((exceptionWorkerSender, runWorkerCompletedEventArgs) => { runWorkerCompletedEventArgs.Result = runWorkerCompletedEventArgs.Argument; });
loaderExceptionWorker.RunWorkerCompleted += ((exceptionWorkerSender, runWorkerCompletedEventArgs) => { throw (Exception)runWorkerCompletedEventArgs.Result; });
loaderExceptionWorker.RunWorkerAsync(ex);
}
else
throw;
}
}
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