Question

我在Oncreate中设置了两个命令，并且都将数据发送到蓝牙设备。我需要第二个命令来等待从第一个命令接收数据字符串，然后再进入第二个命令。每个命令只向BT发送一个字节。我尝试了一段时间循环true但没有缝合工作并挂起while true语句。我假设while true不会让处理程序在循环中触发。只要我不发送两个命令，两个命令都可以单独工作。

这是Oncreate中的代码，包含两个命令和true语句

    looping=true;

    intByteCount =9;
    GetData(intCommand);        // (Command 1)Send byte to get data on reveiver

    while( looping) {               // Wait add data to be received before next command
        Log.d("TAG", "On Hold ?  ");
    }

     intByteCount=160;           // (command 2)
     GetTitle(intCommand);

这是蓝牙处理程序中的代码，一旦收到所有字节，就会将循环设置为false。

    Handler h = new Handler() {

    @Override
    //  public void handleMessage(android.os.Message msg) {

    public void handleMessage(android.os.Message msg) {
        byte[]readBuf = (byte[]) msg.obj;


        if (intByteCount==9){
        // Data is channel status and Master value
        byte[] encodedBytes = new byte[5];
        System.arraycopy(readBuf, 0, encodedBytes, 0, encodedBytes.length);
        looping=false;
        };

GetTitle和GetData基本相同这是GetTitle（）

    private void GetData(int FixtureNumber) {

    Log.d("TAG", "Value  " + intArrayToInt(intArray1));

    intByteCount=9;                         // set to receive 9 bytes

    byte buffer[] = new byte[6];

    buffer[0] = ((byte) 1); // Command (get data)
    buffer[1] = ((byte) Master_value);
    buffer[2] = ((byte) intArrayToInt(intArray1));
    buffer[3] = ((byte) intArrayToInt(intArray2));

    buffer[4] = ((byte) 3);
    buffer[5] = ((byte) 4);

    if (isBTConnected) {

        try {
            mmOutputStream.write(buffer);
        } catch (IOException e) {

            e.printStackTrace();
        }

    }
}

以下是获取两个控制数据的最终代码

        new Thread(new Runnable() {
        @Override
        public void run() {

            intByteCount =9;
            GetData(intCommand);        // Send byte to get data on reveiver
            //you can use a for here and check if the command was executed or just wait and execute the 2nd command
            try {
                Thread.sleep(1000); //wait 2 seconds
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            intByteCount=160;           // Sed incoming data byte count
            GetTitle(intCommand);
        }
    }).start();

Answer 1

您可以使用线程并等待x个时间，：

new Thread(new Runnable() {
    @Override
    public void run() {
        //your 1st command

        //you can use a for here and check if the command was executed or just wait and execute the 2nd command
        try {
            Thread.sleep(2000); //wait 2 seconds
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        //your 2nd command
    }
}).start();

Answer 2

谢谢你，奥古斯丁工作得很好。收到的数据很快，所以只有延迟才能在没有控制布尔值的情况下工作。这是新代码。

        new Thread(new Runnable() {
        @Override
        public void run() {

            intByteCount =9;
            GetData(intCommand);        // Send byte to get data on reveiver
            //you can use a for here and check if the command was executed or just wait and execute the 2nd command
            try {
                Thread.sleep(1000); //wait 2 seconds
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            intByteCount=160;           // Sed incoming data byte count
            GetTitle(intCommand);
        }
    }).start();

Answer 3

您目前正在举行主线程，不建议这样做。最好的选择是修改代码，以便在使用循环== false代码的函数中完成第二个蓝牙命令。

循环时为真

3 个答案: