Question

我创建了一个简单的WPF应用程序，并在默认窗口中添加了一个按钮。当我点击按钮时，会调用一个模拟的长工作方法（使用Thread.Sleep（15000）进行模拟）。我试图让按钮异步执行，但是尽管有以下在线示例，按钮和整个窗口会立即锁定单击并保持不变，直到Thread.Sleep（...）结束。

为什么会发生这种情况？

以下是代码：

private void button1_Click(object sender, RoutedEventArgs e)
{
   DoSomeAsyncWork();
}

private void DoSomeAsyncWork()
{
     System.Windows.Threading.Dispatcher.Run();
     Thread thread = new System.Threading.Thread(
         new System.Threading.ThreadStart(
          delegate()
          {
               Application.Current.Dispatcher.BeginInvoke(DispatcherPriority.Normal, new Action(() => Thread.Sleep(15000)));
          }
        ));
     thread.Start();
}

Answer 1

您将长操作放回UI线程中。让我评论你的例子：

Thread thread = new System.Threading.Thread( 
    new System.Threading.ThreadStart( 
        delegate() { 
            // here we are in the background thread

            Application.Current.Dispatcher.BeginInvoke(DispatcherPriority.Normal, 
                new Action(() => {
                    // here we are back in the UI thread
                    Thread.Sleep(15000);
                })); 
      } 
    ));

所以，你应该像这样修改你的例子：

Thread thread = new System.Threading.Thread( 
    new System.Threading.ThreadStart( 
        delegate() { 
            // here we are in the background thread

            Thread.Sleep(15000);  // <-- do the long operation here

            Application.Current.Dispatcher.BeginInvoke(DispatcherPriority.Normal, 
                new Action(() => {
                    // here we are back in the UI thread

                    // do stuff here that needs to update the UI after the operation finished
                })); 
      } 
    ));

正如其他人所提到的，使用BackgroundWorker类更容易。这是一个例子：

private void DoSomeAsyncWork()   
{   
    BackgroundWorker bw = new BackgroundWorker();

    bw.DoWork += (sender, args) => {
        // do your lengthy stuff here -- this will happen in a separate thread
        Thread.Sleep(15000);
    }

    bw.RunWorkerCompleted += (sender, args) => {
        if (args.Error != null)  // if an exception occurred during DoWork,
            MessageBox.Show(args.Error.ToString());  // do your error handling here

        // do any UI stuff after the long operation here
        ...
    }

    bw.RunWorkerAsync(); // start the background worker
}

Answer 2

通过使用BeginInvoke，您实际上是在UI线程上执行代码。

您只需在从后台工作线程更新UI时使用此功能。如果您只是：

 Thread thread = new System.Threading.Thread(
     new System.Threading.ThreadStart(
      delegate()
      {
           Thread.Sleep(15000);
      }
    ));

我认为它会奏效。

但是，您没有提出“已完成工作”事件，因此您无法知道线程何时（或确实如此）已完成。查看BackgroundWorker class。这对你来说很重要。您只需将代码插入DoWork方法。

Answer 3

您应该使用 BackgroundWorker 。

UI线程块

3 个答案: