Question

我正在尝试在android中实现REST接口，我需要在后台发送一个线程，将“我还活着”的消息发送到ip地址。为此，我创建了一个线程调用的RestPostThread，它在后台运行，而我在UI线程中执行操作。

问题是，在将第一条消息发送到RestPostThread之后，我无法退出looper或使用其他IP或其他内容向其发送不同的消息。

以下是UI和RestPostThread的代码：

public class MainActivity extends AppCompatActivity{

Handler workerThreadHandler;


protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    Toolbar toolbar = (Toolbar) findViewById(R.id.toolbar);
    final TextView text1 = (TextView) findViewById(R.id.text1);
    final TextView text2 = (TextView) findViewById(R.id.text2);
    setSupportActionBar(toolbar);


    final RestPostThread RPT = new RestPostThread();
    RPT.start();

    while(workerThreadHandler == null ) {
        workerThreadHandler = RPT.getThreadHandler();
    }

    Button buttonStop = (Button) findViewById(R.id.buttonStop);
    buttonStop.setOnClickListener(new View.OnClickListener(){
        public void onClick(View view) {
            try {



                workerThreadHandler.getLooper().quit();
            }catch(Exception e){
                text1.setText(e.getMessage());
                text2.setText( "Exception!");
            }

        }
    });

    Button buttonSend = (Button) findViewById(R.id.buttonSend);
    buttonSend.setOnClickListener(new View.OnClickListener(){
        public void onClick(View view) {
            try {
                text1.setText(new RestGet().execute(editText.getText().toString()).get());
                text2.setText("everything went well!");
            }catch(Exception e){
                text1.setText(e.getMessage());
                text2.setText( "Exception!");
            }

        }
    });
}

这是RestPostThread的代码：

public class RestPostThread extends Thread  {
public Handler mHandler;

@Override
public void run(){

    Looper.prepare();
    mHandler = new Handler() {
        public void handleMessage(Message msg) {
            Log.d("MYASDASDPOASODAPO", "dentro mensaje");
            while (!msg.obj.equals(null)) {
                try {
                    Thread.sleep(1000);
                    URL url = new URL(msg.obj.toString());
                    HttpURLConnection conn = (HttpURLConnection) url.openConnection();
                    conn.setDoOutput(true);
                    conn.setRequestMethod("POST");
                    String input = "<Instruction><type>put_me_in</type><room>Room 1</room></Instruction>";

                    OutputStream os = conn.getOutputStream();
                    os.write(input.getBytes());
                    os.flush();

                    if (conn.getResponseCode() != HttpURLConnection.HTTP_CREATED) {
                        //  throw new RuntimeException("Failed : HTTP error code : " + conn.getResponseCode());
                    }
                    BufferedReader br = new BufferedReader(new InputStreamReader((conn.getInputStream())));
                    String output;
                    String aux = new String();
                    while ((output = br.readLine()) != null) {
                        aux = aux + output;
                    }
                    conn.disconnect();
                    //return aux;
                } catch(MalformedURLException e) {
                    e.printStackTrace();
                    //return null;
                } catch(IOException e) {
                    e.printStackTrace();
                    //return null;
                } catch(Exception e) {
                }
            }
            Log.d("CLOSING MESSAGE", "Closing thread");
        }
    };
    Looper.loop();
}

public Handler getThreadHandler() {
    return this.mHandler;
}

Answer 1

看看HandlerThread处理一个线程来处理消息。你的Handler不应该像这样的消息循环，它将无法正常工作。这是Looper的工作，处理发送到Message的新Runnable或Handler个对象，这些对象绑定到Looper。

无论如何，您应该仔细查看使用Loader来处理REST类型API;或者，探索第三方库，例如改造，以处理REST。

Answer 2

我设法解决了这个问题。问题是我把所有东西都包裹在里面：

while (!msg.obj.equals(null)) {}

我在这个线程和UI线程中实现了处理程序，现在我在两者之间来回通信，我的RestPostThread现在看起来像这样：

public class RestPostThread extends Thread  {

public Handler mHandler,uiHandler;


public RestPostThread(Handler handler) {
    uiHandler = handler;
}

@Override
public void run(){
    Looper.prepare();
    mHandler = new Handler() {
        public void handleMessage(Message msg) {
                try {
                    //Thread.sleep(1000);
                    URL url = new URL(msg.obj.toString());
                    HttpURLConnection conn = (HttpURLConnection) url.openConnection();
                    conn.setDoOutput(true);
                    conn.setRequestMethod("POST");
                    String input = "<Instruction><type>put_me_in</type><room>Room 1</room></Instruction>";

                    OutputStream os = conn.getOutputStream();
                    os.write(input.getBytes());
                    os.flush();

                    if (conn.getResponseCode() != HttpURLConnection.HTTP_CREATED) {
                        //  throw new RuntimeException("Failed : HTTP error code : " + conn.getResponseCode());
                    }
                    BufferedReader br = new BufferedReader(new InputStreamReader((conn.getInputStream())));
                    String output;
                    String aux = new String();
                    while ((output = br.readLine()) != null) {
                        aux = aux + output;
                    }
                    conn.disconnect();
                    Message msg2 = uiHandler.obtainMessage();
                    msg2.obj = aux;
                    uiHandler.sendMessage(msg2);
                }catch(MalformedURLException e){
                    e.printStackTrace();
                }catch(IOException e){
                    e.printStackTrace();
                }catch(Exception e){
                }
            }
    };
    Looper.loop();
}


public Handler getThreadHandler() {
    return this.mHandler;
}

}

在我的MainActivity中，我有这个处理程序，允许我循环＃34; （基本上只是在RestPostThread和UIThread之间来回）我的Post消息，直到我决定停止改变布尔循环的MainActivity：

 public Handler uiHandler = new Handler() {
    public void handleMessage(Message inputMessage) {
        Log.d("FROM UI THREAD",inputMessage.obj.toString());
        if(loop) {
            Message msg = workerThreadHandler.obtainMessage();
            String url = "http://192.168.1.224:9000/xml/android_reply";
            msg.obj = url;
            workerThreadHandler.sendMessageDelayed(msg,1000);
        }
    }
};

Android处理程序只发送一条消息

2 个答案: