告诉ThreadPoolExecutor何时应该继续运行

Question

我必须通过某个端口向一些计算机发送一组文件。事实是，每次调用发送文件的方法时，都会计算目标数据（地址和端口）。因此，使用为每个方法调用创建一个线程的循环，并用BindException的try-catch语句包围方法调用，以处理尝试使用已使用端口的程序的情况（不同的目标地址可能通过相同的端口接收消息）告诉线程等待几秒钟，然后重新启动重试，并继续尝试，直到不会抛出异常（运输成功执行）。 我不知道为什么（虽然我第一次看到它的时候可以猜到），Netbeans警告我说，在循环中睡眠一个Thread对象并不是最好的选择。然后我搜索了一下以获取更多信息，并找到this link to another stackoverflow post, which looked so interesting（我从来没有听说过the ThreadPoolExecutor class）。我一直在阅读这个链接和API以改进我的程序，但我还不确定我应该如何将它应用到我的程序中。请有人帮忙吗？

编辑：重要的代码：

 for (Iterator<String> it = ConnectionsPanel.list.getSelectedValuesList().iterator(); it.hasNext();) { 
     final String x = it.next(); 
     new Thread() { 

      @Override 
      public void run() { 
       ConnectionsPanel.singleAddVideos(x); 
      } 
     }.start(); 
    } 

    private static void singleAddVideos(String connName) { 
    String newVideosInfo = ""; 

    for (Iterator<Video> it = ConnectionsPanel.videosToSend.iterator(); it.hasNext();) { 
     newVideosInfo = newVideosInfo.concat(it.next().toString()); 
    } 

    try { 
     MassiveDesktopClient.sendMessage("hi", connName); 
     if (MassiveDesktopClient.receiveMessage(connName).matches("hello")) { 
      MassiveDesktopClient.sendMessage(newVideosInfo, connName); 
     } 
    } catch (BindException ex) { 
     MassiveDesktopClient.println("Attempted to use a port which is already being used. Waiting and retrying...", new Exception().getStackTrace()[0].getLineNumber()); 
     try { 
      Thread.sleep(MassiveDesktopClient.PORT_BUSY_DELAY_SECONDS * 1000); 
     } catch (InterruptedException ex1) { 
      JOptionPane.showMessageDialog(null, ex1.toString(), "Error", JOptionPane.ERROR_MESSAGE); 
     } 
     ConnectionsPanel.singleAddVideos(connName); 
     return; 
    } 

    for (Iterator<Video> it = ConnectionsPanel.videosToSend.iterator(); it.hasNext();) { 
     try { 
      MassiveDesktopClient.sendFile(it.next().getAttribute("name"), connName); 
     } catch (BindException ex) { 
      MassiveDesktopClient.println("Attempted to use a port which is already being used. Waiting and retrying...", new Exception().getStackTrace()[0].getLineNumber()); 
      try { 
       Thread.sleep(MassiveDesktopClient.PORT_BUSY_DELAY_SECONDS * 1000); 
      } catch (InterruptedException ex1) { 
       JOptionPane.showMessageDialog(null, ex1.toString(), "Error", JOptionPane.ERROR_MESSAGE); 
      } 
      ConnectionsPanel.singleAddVideos(connName); 
      return; 
     } 
    } 
} 

Answer 1

你的问题不是很清楚 - 我知道你想重新运行你的任务，直到成功为止（没有的BindException）。要做到这一点，你可以：

• 尽量不捕捉异常
• 捕获来自未来
• 例外重新安排任务稍后运行你的代码，如果它失败

一简化的代码将如下 - 添加错误消息，并根据需要进行优化：

public static void main(String[] args) throws Exception { 
    ScheduledExecutorService scheduler = Executors.newScheduledThreadPool(corePoolSize); 
    final String x = "video"; 
    Callable<Void> yourTask = new Callable<Void>() { 
     @Override 
     public Void call() throws BindException { 
      ConnectionsPanel.singleAddVideos(x); 
      return null; 
     } 
    }; 
    Future f = scheduler.submit(yourTask); 
    boolean added = false; //it will retry until success 
          //you might use an int instead to retry 
          //n times only and avoid the risk of infinite loop 
    while (!added) { 
     try { 
      f.get(); 
      added = true; //added set to true if no exception caught 
     } catch (ExecutionException e) { 
      if (e.getCause() instanceof BindException) { 
       scheduler.schedule(yourTask, 3, TimeUnit.SECONDS); //reschedule in 3 seconds 
      } else { 
       //another exception was thrown => handle it 
      } 
     } 
    } 
} 

public static class ConnectionsPanel { 

    private static void singleAddVideos(String connName) throws BindException { 
     String newVideosInfo = ""; 

     for (Iterator<Video> it = ConnectionsPanel.videosToSend.iterator(); it.hasNext();) { 
      newVideosInfo = newVideosInfo.concat(it.next().toString()); 
     } 

     MassiveDesktopClient.sendMessage("hi", connName); 
     if (MassiveDesktopClient.receiveMessage(connName).matches("hello")) { 
      MassiveDesktopClient.sendMessage(newVideosInfo, connName); 
     } 

     for (Iterator<Video> it = ConnectionsPanel.videosToSend.iterator(); it.hasNext();) { 
      MassiveDesktopClient.sendFile(it.next().getAttribute("name"), connName); 
     } 
    } 
}