javaNIO中的缓冲区写入/发送消息问题

Question

我的问题是关于JAVANIO客户端服务器消息传递，我不确定从技术上定义这个问题，但是:看起来buffer正在缓存数据，当它完成后，它会一起发送，这是令人不安的逻辑：

private void sendCreate(String line,SocketChannel from)
 /* A new client wishes to join the world.

      This requires the client to find out about the existing
      clients, and to add itself to the other clients' worlds.

      Message format: create name xPosn zPosn

      Store the user's name, extracted from the "create" message
  */
 { StringTokenizer st = new StringTokenizer(line);
 st.nextToken();                  // skip 'create' word
 userName = st.nextToken();
 String xPosn = st.nextToken();   // don't parse
 String zPosn = st.nextToken();   // don't parse

 // request details from other clients
 sendBroadcastMessage( "wantDetails " + achannel.socket().getInetAddress() + " " + port,from);

 // tell other clients about the new one
 sendBroadcastMessage( "create " + userName + " "+xPosn+" "+zPosn,from);

 } // end of sendCreate()

负责从服务器广播消息的方法：

private void sendBroadcastMessage(String mesg, SocketChannel from) {
  prepWriteBuffer(mesg);
  Iterator i = clients.iterator();
  while (i.hasNext()) {
   SocketChannel channel = (SocketChannel) i.next();
   if (channel != from)
    channelWrite(channel, writeBuffer);
  }
 }

我假设这应该发送第一条消息，即sendBroadcastMessage( "wantDetails“+ achannel.socket().getInetAddress() +”“+ port，from)；但这不是，它似乎在等待其他方法调用，即sendBroadcastMessage( "create”+ userName + "+xPosn+“"+zPosn，from)；然后将这两条消息作为一条消息发送，这是影响应用程序logic.ideally的一条消息，或者它应该在第一次调用sendBroadcastMessage之后发送第一条消息，然后当客户端收到第一个调用时，应该处理其他调用。

以下是在sendBroadcastMessage()中使用的方法：

private void prepWriteBuffer(String mesg) {
  // fills the buffer from the given string
  // and prepares it for a channel write
  writeBuffer.clear();
  writeBuffer.put(mesg.getBytes());
  writeBuffer.putChar('\n');
  writeBuffer.flip();
 }

 private void channelWrite(SocketChannel channel, ByteBuffer writeBuffer) {
  long nbytes = 0;
  long toWrite = writeBuffer.remaining();

  // loop on the channel.write() call since it will not necessarily
  // write all bytes in one shot
  try {
    nbytes += channel.write(writeBuffer);

  } catch (ClosedChannelException cce) {
   cce.printStackTrace();
  } catch (Exception e) {
   e.printStackTrace();
  }
  // get ready for another write if needed
  writeBuffer.rewind();
 }

请建议一些解决方案。

谢谢,

摇摆拉拉

编辑:这个怎么样，我从一些聊天应用上得到了这个补丁：

private void prepWriteBuffer(String mesg) {
        // fills the buffer from the given string
        // and prepares it for a channel write
        writeBuffer.clear();
        writeBuffer.put(mesg.getBytes());
        writeBuffer.putChar('\n');
        writeBuffer.flip();
    }


// called needs to remove the channel if it fails, otherwise it will fail forever.
        private void channelWrite(SocketChannel channel, ByteBuffer writeBuffer)  {    
            long nbytes = 0;
            long toWrite = writeBuffer.remaining();
            // loop on the channel.write() call since it will not necessarily
            // write all bytes in one shot
            try {
            while (nbytes != toWrite) {
                nbytes += channel.write(writeBuffer);

                try {
                    Thread.sleep(CHANNEL_WRITE_SLEEP);
                } catch (InterruptedException e) {
                }
            }
        } catch (ClosedChannelException cce) {
        } catch (Exception e) {
        }
        // get ready for another write if needed
        writeBuffer.rewind();
    }

Answer 1

也许你是有意

 while(writeBuffer.remaining()>0)
      channel.write(writeBuffer);

然而，您的问题似乎是您假设在消息之间存在某种类型的魔法标记。但是，并不存在这样的分隔符。一个流就是一个字节流。当你在阻塞模式下读取时，你会得到至少一个字节，你可能会得到更多，这可能会跨越多次写入，但除非你在流中包含你希望消息开始和结束的地方，否则你将无法知道。

一种简单的方法是在消息的开头写入消息的长度，并读取最多一条消息，直到您获得所有消息。就像这样。

private void prepWriteBuffer(String mesg) {    
  // fills the buffer from the given string    
  // and prepares it for a channel write    
  writeBuffer.clear();
  byte[] bytes = mesg.getBytes());
  writeBuffer.putInt(bytes.length);    
  writeBuffer.put(bytes);
  writeBuffer.flip();    
 } 


// called needs to remove the channel if it fails, otherwise it will fail forever.
private void channelWrite(SocketChannel channel, ByteBuffer writeBuffer) throws IOException {    
 while(writeBuffer.remaining()>0)
      channel.write(writeBuffer);
 writeBuffer.rewind();
}