如何通过蓝牙一起读取所有字节?

时间:2012-02-10 16:56:46

标签: java android stream bluetooth bytearray

我有一个使用蓝牙从其他设备接收一些数据(字节)的应用程序。一切进展顺利,但我在接收所有字节时遇到了一个小问题。收到字节后,我在Toast上显示它们只是为了测试它们。当另一个设备一起发送10个字节时(例如:“ABCDEFGHIJ”),程序将仅采用第一个字节“A”并在Toast上显示,然后转到第二个迭代并读取其他9个字节并显示“ BCDEFGHIJ“在吐司上。这是我的代码:

byte[] buffer = new byte[1024]; // Read 1K character at a time.
int bytes = 0; // Number of bytes.

while(true)
{
    try
    {
        // Read from the InputStream.
        bytes = bInStream.read(buffer);

        // Send the obtained bytes to the MainActivity.
        mainActivityHandler.obtainMessage(MainActivity.MESSAGE_READ, bytes, -1, buffer).sendToTarget();
    }
    catch(IOException e)
    {
        connectionLost();
        break;
    }
}

在MainActivity中,我有:

// The Handler that gets information back from the BluetoothManager.
private final Handler handler = new Handler()
{
    @Override
    public void handleMessage(Message msg)
    {
        switch(msg.what)
        {
            case MESSAGE_READ:
                byte[] readBuf = (byte[]) msg.obj;

                // construct a string from the valid bytes in the buffer.
                String readMessage = new String(readBuf, 0, msg.arg1);
                Toast.makeText(MainActivity.this, readMessage, Toast.LENGTH_SHORT).show();
                break;

            // ...
        }
    }
};

如何一起收到所有字节?!

6 个答案:

答案 0 :(得分:7)

嗯,最有可能的罪魁祸首就在于你发送邮件的方式。您的接收没有问题,它将收到与写入的字节数(最多1024个)。

如果您无法控制消息的发送方式,您可以一次读取一个字节,然后在达到预定义的终结符时发送处理程序消息。例如:“ABCDEFGHIJ#”,其中#是终止符。

String msg = "";
byte ch;
while((ch=mInStream.read())!='#') {
    bytes++;
    msg+=ch;
}

答案 1 :(得分:5)

蓝牙连接是基于流的,而不是基于数据包的。无法保证或尝试保留打包。因此,任何数量的写入都可以导致任意数量的读取,只保证字节流是正确的。如果需要检测数据包,则需要提供自己的数据包结构来包装数据。例如,在每个数据包之前添加一个长度字段,以便您可以在接收方重建。

答案 2 :(得分:4)

@broody接受的回答是正确的。但如果数据本身包含“#”,则可能很难获取数据。因此,根据我的最佳方法是在设备中附加'\ n'后跟'\ n'(或不太可能作为数据Refer ASCII Table提供的任何其他字符),将数据发送到您的Android应用。它只是作为换行符并标记数据结束。

例如:ABCDEFGH \ n \ r

那么您的代码可以是这样的:

byte[] buffer = new byte[1024];
 while (true) {

         // Read from the InputStream
          buffer[bytes] = (byte) mmInStream.read();                 
         // Send the obtained bytes to the UI Activity
 if ((buffer[bytes] == '\n')||(buffer[bytes]=='\r'))
 {
   mHandler.obtainMessage(MainActivity.MESSAGE_READ, bytes, -1, buffer).sendToTarget();
   bytes=0;
 }
 else
 bytes++;
}

希望它有所帮助 此致

答案 3 :(得分:1)

要一起读取所有字节,您需要在代码中用“\ n”或“\ r”或“\ r \ n”分隔数据。

例如:如果您想通过蓝牙将数据从Arduino发送到Android应用程序:

(Arduino代码):

    int count =0;
    int sensorPin = 0;

    void setup()
    {
      Serial.begin(9600);
    }

    void loop()
    {
    int val= analogRead(sensorPin);
      if(val<threshold){ 
         a++; 
      }
      else{
        delay(2);
        count = count + 1;
        Serial.print(count);
        Serial.print("\n");
          }

现在,要读取发送的数据(变量'count'的值),这里是Android Studio的代码:

 private class ConnectedThread extends Thread {
        private final InputStream mmInStream;

        public ConnectedThread(BluetoothSocket socket) {
            InputStream tmpIn = null;

            // Get the input streams, using temp objects because
            // member streams are final
            try {
                tmpIn = socket.getInputStream(); // opens the input stream in order to retrieve InputStream objects
            } catch (IOException e) {
            }

            mmInStream = tmpIn;
        }

     public void run() {
        int bytes; // bytes returned from read()
        int availableBytes = 0;

        // Keep listening to the InputStream until an exception occurs
        while (true) {
            try {
                availableBytes = mmInStream.available();
                if(availableBytes>0){
                    byte[] buffer = new byte[availableBytes];  // buffer store for the stream
                    // Read from InputStream
                    bytes = mmInStream.read(buffer); // Get number of bytes and message in "buffer"
                    if (bytes>0){
                        h.obtainMessage(RECEIVE_MESSAGE, bytes, -1, buffer).sendToTarget();     // Send to message queue Handler
                    }
                   }
                } catch (IOException e) {
                break;
                        }
        }
    }

这个:(在onCreate()方法中):

mConnectedThread = new ConnectedThread(btSocket);
mConnectedThread.start();
// The Handler that gets information back
 h = new Handler() { // Handler-->used to communicate b/w UI & BG Thread
            public void handleMessage(android.os.Message msg) {
                switch (msg.what) {
                    case RECEIVE_MESSAGE:// if receive message
                        byte[] readBuf = (byte[]) msg.obj;
                       String strIncom = new String(readBuf, 0, msg.arg1); // create string from bytes array

                        sb.append(strIncom);                                                // append string
                        int endOfLineIndex = sb.indexOf("\n");                            // determine the end-of-line
                        if (endOfLineIndex > 0) {                                            // if end-of-line,
                            String sbprint = sb.substring(0, endOfLineIndex);               // extract string
                            sb.delete(0, sb.length());// and clear
                            Toast.makeText(ledControl.this,sbprint,Toast.LENGTH_SHORT).show();
                            footSteps.setText(sbprint);            // update TextView
                        }
                        break;
                }}};

答案 4 :(得分:0)

以上答案对我都不起作用。 Thread.sleep(1000)给了我解决方案

答案 5 :(得分:-1)

一种处理长于缓冲区消息的方法(可能很危险,因为如果从未清除就会占用你的内存)是以块的形式处理它们:

String inString = "";
byte[] buffer = new byte[1024];  // buffer store for the stream
int bytes; // bytes returned from read()

while (true) {
    try {
        bytes = mmInStream.read(buffer);
        String chunk = new String(buffer, 0, bytes);
        if(chunk.contains(";"))
        {
            inString += chunk.substring(0,chunk.indexOf(';'));
            Message msg = new Message();
            msg.obj  = inString;
            handler.sendMessage(msg);
            inString =  chunk.substring(chunk.indexOf(';'));
        }
        else
        {
            inString += chunk;
        }
    } catch (IOException e) {
        break;
    }
}