Question

在我团队的Android应用程序中，我有一个从启动运行的服务，它与服务器通信以执行诸如登录，注册，在电话之间聊天和更新电话数据库等操作。

我需要让我的服务与双向活动进行通信：例如，我正在处理登录活动，用户名和密码是从应用程序屏幕上的文本字段中获取的字符串，我已经能够将它们传递给服务，以便它向服务器发送授权命令。

public void loginPressed(View v){
    usernameStr = usernameField.getText().toString();
    passwordStr = passwordField.getText().toString();

    if (!bound) return;
    Bundle b = new Bundle();
    Message msg = Message.obtain(null, ChatService.LOGIN);
    try {
        b.putString("username", usernameStr);
        b.putString("password", passwordStr);
        msg.setData(b);
        messenger.send(msg);
    }
    catch (RemoteException e) {

    }

这可以像我预期的那样工作。当服务器响应一条消息，说明登录是否成功时，我需要它将消息传递回活动，以便我可以成功启动主活动，或者如果没有则提示重新进入。

我尝试使用msg.replyTo字段来获取返回信使以发回信息，但是当我运行应用程序时，它会以空指针异常关闭，我不知道为什么会发生这种情况。以下是似乎是罪魁祸首的代码：

private class IncomingHandler extends Handler {
    @Override
    public void handleMessage(Message msg) {
        switch(msg.what) {
        case LOGIN:

            Bundle b = msg.getData();
            String username = b.getString("username");
            String password = b.getString("password");

            String loginMessage = TCPCall.login(username, password);
            connection.sendMessage(loginMessage);

            String loginReturn = connection.retrieveMessage();
            Message m;

            Scanner s = new Scanner(loginReturn);
            s.useDelimiter(",");
            String c = s.next();
            String status = s.next();
            String message = s.next();

            if (status.equals("OK")) {
                m = Message.obtain(null, LoginActivity.OK);
                try {
                    msg.replyTo.send(m);
                } catch (RemoteException e) {}
            }
            else {
                m = Message.obtain(null, LoginActivity.ERR);
                try {
                    msg.replyTo.send(m);
                } catch (RemoteException e) {}
            }
            break;

空指针似乎来自

msg.replyTo.send(m);

两种情况下的代码行（登录成功和登录失败）

任何帮助解决这个问题都将非常感激：）

Answer 1

正如格雷格在评论中指出的那样。您需要在发送原始邮件的位置设置if int。

可在此处找到一个示例：http://www.survivingwithandroid.com/2014/01/android-bound-service-ipc-with-messenger.html

Answer 2

我认为您忘记了通过Service捆绑发送对Login Activity的响应。因此，我在Messenger Service中进行了一些更改

定义一个全局变量并在传入处理程序中进行一些更改

static final int LOGIN_STATUS = 1;

private class IncomingHandler extends Handler {
@Override
public void handleMessage(Message msg) {
    switch(msg.what) {
    case LOGIN:

        Bundle b = msg.getData();
        String username = b.getString("username");
        String password = b.getString("password");

        String loginMessage = TCPCall.login(username, password);
        connection.sendMessage(loginMessage);

        String loginReturn = connection.retrieveMessage();
        Message m = Message.obtain(null, LOGIN_STATUS);

        Scanner s = new Scanner(loginReturn);
        s.useDelimiter(",");
        String c = s.next();
        String status = s.next();
        String message = s.next();

        if (status.equals("OK")) {
            b.putString("responseC",c);
            b.putString("responseStatus",status);
            b.putString("responseMessage",message)

            m.setData(b);
            try {
                msg.replyTo.send(m);
            } catch (RemoteException e) {}
        }
        else {
           /*if something is wrong with username and password you can put 
           a toast*/

            }
        break;

现在，我们必须在LoginActivity和在登录活动中也使用IncomingHandler

class IncomingHandler extends Handler{

        @Override
        public void handleMessage(Message msg) {
            switch (msg.what){
                case ChatService.LOGIN_STATUS:
                    String C = msg.getData().getString("responseC");
                    String Status = msg.getData().getString("responseStatus");
                    String Message = msg.getData().getString("responseMessage");

                    //Here is your response in LoginActivity, enjoy!!! 

                    break;

                default:
                    super.handleMessage(msg);
            }
        }
    }

final Messenger mMessenger = new Messenger(new IncomingHandler());

public void loginPressed(View v){
usernameStr = usernameField.getText().toString();
passwordStr = passwordField.getText().toString();

if (!bound) return;
Bundle b = new Bundle();
Message msg = Message.obtain(null, ChatService.LOGIN_SATUS,0,0);
try {
    b.putString("username", usernameStr);
    b.putString("password", passwordStr);
    msg.setData(b);
    msg.replyTo = mMessenger;
    messenger.send(msg);
}
catch (RemoteException e) {
    // In this case the service has crashed before we could even
    // do anything with it; we can count on soon being
    // disconnected (and then reconnected if it can be restarted)
    // so there is no need to do anything here.

}

此代码运行良好，希望对您有所帮助，谢谢

Android Service-Activity 2方式沟通

2 个答案: