Question

当我需要将服务器蓝牙插槽中的字符串发送到我的客户端蓝牙插槽时，我的程序出现问题。只要我一次只发送一个字符串（例如聊天），但是如果我需要在短时间内编写更多字符串（以交换信息），一切都可以正常工作，字符串将不会从客户端代码中分离出来。例如，如果我发送“FirstUser”并且在“SecondUser”之后，客户端不读取“FirstUser”，然后读取“SecondUser”。它将显示“FirstUserSecondUser”。我该如何避免这种行为？

编辑：如果我让线程在能够发送新消息之前进入休眠状态，它会读取正确的字符串，但此解决方案无法满足我的需求。

服务器代码：发送给所有客户（已编辑）

   public synchronized void sendToAll(String message)
{   
    try {
        Thread.sleep(100);
    } catch (InterruptedException e1) {
        e1.printStackTrace();
    }

    publishProgress(message);
    for(OutputStream writer:outputList) {
        try {
            writer.write(message.getBytes());
            writer.flush();
        } catch (IOException e) {
            System.out.println("Some-Error-Code");
        }
    }
}

服务器代码：从客户端读取：

   public void run() {
    String nachricht;
    int numRead;
    byte[] buffer = new byte[1024];
    while (runningFlag) 
    {
        try {
            if((numRead = inputStream.read(buffer)) >= 0) { 
                nachricht = new String(buffer, 0, numRead);
                serverThread.handleMessage(nachricht); 
            }
            }
             catch (IOException e) {
                this.cancel();
                e.printStackTrace();
            }
    } 
}

客户代码：从服务器读取（已编辑）

@Override
    protected Void doInBackground(Integer... ints) {    
        String nachricht = new String();
        byte[] buffer = new byte[1024];
        int numRead;
        while (runningFlag) 
        {
            try {
                if(((numRead = inputStream.read(buffer)) >= 0)) { 
                    nachricht = new String(buffer, 0, numRead);
                    publishProgress(nachricht);
                }
            }
             catch (IOException e) {
                clientGame.finish();
                e.printStackTrace();
            }                      
        }
        return null;
}

客户端代码：写入服务器

public synchronized void write(String nachricht)
    {
        try {
            Thread.sleep(100);
        } catch (InterruptedException e1) {
            e1.printStackTrace();
        }

        try {
            outputStream.write(nachricht.getBytes());
            outputStream.flush();
        } catch (IOException e) {
            this.cancel();
            e.printStackTrace();
        }
    }

我感谢每一点帮助:)。

Answer 1

您需要封装数据项以避免连接。这意味着您必须在继续之前编写和读取整个数据项。

您应该使用一些实用程序方法来代替直接使用OutputStream和InputStream的方法：

public static void writeItem(OutputStream out, String s) throws IOException
{
    // Get the array of bytes for the string item:
    byte[] bs = s.getBytes(); // as bytes
    // Encapsulate by sending first the total length on 4 bytes :
    //   - bits 7..0 of length
    out.write(bs.length);      // modulo 256 done by write method
    //   - bits 15..8 of length
    out.write(bs.length>>>8);  // modulo 256 done by write method
    //   - bits 23..16 of length
    out.write(bs.length>>>16); // modulo 256 done by write method
    //   - bits 31..24 of length
    out.write(bs.length>>>24); // modulo 256 done by write method
    // Write the array content now:
    out.write(bs); // Send the bytes
    out.flush();
}

public static String readItem(InputStream in) throws IOException
{
    // first, read the total length on 4 bytes
    //  - if first byte is missing, end of stream reached
    int len = in.read(); // 1 byte
    if (len<0) throw new IOException("end of stream");
    //  - the other 3 bytes of length are mandatory
    for(int i=1;i<4;i++) // need 3 more bytes:
    {
        int n = in.read();
        if (n<0) throw new IOException("partial data");
        len |= n << (i<<3); // shift by 8,16,24
    }
    // Create the array to receive len bytes:
    byte[] bs = new byte[len];
    // Read the len bytes into the created array
    int ofs = 0;
    while (len>0) // while there is some byte to read
    {
        int n = in.read(bs, ofs, len); // number of bytes actually read
        if (n<0) throw new IOException("partial data");
        ofs += n; // update offset
        len -= n; // update remaining number of bytes to read
    }
    // Transform bytes into String item:
    return new String(bs);
}

然后你将这些方法用于服务器和客户端读取和写入您的String项目。

蓝牙连接;无法正确发送字符串

1 个答案: