13

我正在尝试启动service然后打开socket以与服务器建立连接。

在按钮上单击我创建新Thread的然后启动服务。

Thread t = new Thread(){
        public void run(){
            mIntent= new Intent(MainActivity.this, ConnectonService.class);
            mIntent.putExtra("KEY1", "Value used by the service");
            context.startService(mIntent);
        }
    };
t.start();

然后service，我尝试打开socket并与服务器建立连接

@Override
public int onStartCommand(Intent intent, int flags, int startId) {
    //TODO do something useful


    try {
        InetAddress serverAddr = InetAddress.getByName(SERVER_IP);
        socket = new Socket(serverAddr, SERVERPORT);
        Scanner scanner = new Scanner(socket.getInputStream());
        message = scanner.nextLine();
    } catch (IOException e) {
        e.printStackTrace();
    }

    return Service.START_NOT_STICKY;
}

但是当我调用它时，我有错误

08-30 08:56:49.268: E/AndroidRuntime(3751): java.lang.RuntimeException: Unable to start service com.example.testofconnection.ConnectonService@40ef02a8 with Intent { cmp=com.example.testofconnection/.ConnectonService (has extras) }: android.os.NetworkOnMainThreadException*

我认为问题在于service，main thread但我找不到我应该如何在新（独立）线程上启动服务以保持connection alive？

4

2 回答 2

20

您可以为此使用 IntentService。只需使用主线程中的 Intent 正常启动它。onHandleIntent()方法在后台线程中执行。把你的套接字代码放在那里。这是一个示例代码。

public class MyIntentService extends IntentService {

    public MyIntentService() {
        super("MyIntentService");
    }

    @Override
    protected void onHandleIntent(Intent intent) {
        // this method is called in background thread
    }

    @Override
    public IBinder onBind(Intent intent) {
        return null;
    }    
}

在您的活动中，您按以下方式启动服务。

startService(new Intent(this, MyIntentService.class));

如果你需要一个持久的服务，你可以创建一个普通的服务并在那里启动一个线程。这是一个例子。确保将其作为“前台”服务启动。这将允许服务运行更长时间而不会被 Android 杀死。

public class MyAsyncService extends Service {

    private AtomicBoolean working = new AtomicBoolean(true)

    private Runnable runnable = new Runnable() {
        @Override
        public void run() {
            while(working.get()) {
                // put your socket-code here
                ...
            }
        }
    }

    @Override
    public void onCreate() {

        // start new thread and you your work there
        new Thread(runnable).start();

        // prepare a notification for user and start service foreground
        Notification notification = ...
        // this will ensure your service won't be killed by Android
        startForeground(R.id.notification, notification);
    }

    @Override
    public onDestroy() {
        working.set(false)
    }
}
于 2013-08-30T06:21:22.787 回答
6

将此代码移动到您的线程：

try {
    InetAddress serverAddr = InetAddress.getByName(SERVER_IP);
    socket = new Socket(serverAddr, SERVERPORT);
    Scanner scanner = new Scanner(socket.getInputStream());
    message = scanner.nextLine();
} catch (IOException e) {
    e.printStackTrace();
}

举个例子（我不确定这是否适合您的任务）：

Thread t = new Thread(){
        public void run(){
            try {
                InetAddress serverAddr = InetAddress.getByName(SERVER_IP);
                socket = new Socket(serverAddr, SERVERPORT);
                Scanner scanner = new Scanner(socket.getInputStream());
                message = scanner.nextLine();
            } catch (IOException e) {
                e.printStackTrace();
            }

            mIntent= new Intent(MainActivity.this, ConnectonService.class);
            mIntent.putExtra("KEY1", "Value used by the service");
            context.startService(mIntent);
        }
    };
t.start();

你应该知道一个服务正在 UI 线程上运行，所以你得到了这个错误。查看这个不错的站点以获取有关 Android 中各种线程方法的更多信息。

于 2013-08-30T06:21:42.077 回答