IntentService 失去与互联网的连接

IntentService loses connection with internet

我的 IntentService 有问题。我需要我的服务在后台每 1 分钟访问网站中的 JSON 文件,而不打开应用程序(发出通知,我在 JSON 中收到)。它会这样做,但 3-4 分钟后,它会抛出“java.net.SocketTimeoutException:连接超时”错误,我仍然可以将域解析为 IP 地址,但我无法连接到网站,并且 phone 仍然有移动连接。所有网站都会发生同样的事情。

我的意图服务:

public class BackgroundService extends IntentService {

Runnable runnable = () -> new bgJob().execute();
Handler handler = new Handler();
DatabaseHandler db = new DatabaseHandler(this);

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

@Override
public void onCreate() {
    super.onCreate();
    runnable.run();
}

@Override
public int onStartCommand(final Intent intent, final int flags, final int startId) {
    return START_STICKY;
}
@Override
public void onTaskRemoved(Intent rootIntent){
    Log.d("BG_SERVICE","TASK REMOVED!");
    Intent restartServiceTask = new Intent(getApplicationContext(),this.getClass());
    restartServiceTask.setPackage(getPackageName());
    PendingIntent restartPendingIntent = PendingIntent.getService(getApplicationContext(), 1,restartServiceTask, PendingIntent.FLAG_ONE_SHOT);
    AlarmManager myAlarmService = (AlarmManager) getApplicationContext().getSystemService(Context.ALARM_SERVICE);
    myAlarmService.set(
            AlarmManager.ELAPSED_REALTIME,
            SystemClock.elapsedRealtime() + 1000,
            restartPendingIntent);
}

@Override
public void onDestroy() {
    super.onDestroy();
    Log.d("BG_SERVICE","TASK DESTROYED!");
    Intent restartServiceTask = new Intent(getApplicationContext(),this.getClass());
    restartServiceTask.setPackage(getPackageName());
    PendingIntent restartPendingIntent = PendingIntent.getService(getApplicationContext(), 1,restartServiceTask, PendingIntent.FLAG_ONE_SHOT);
    AlarmManager myAlarmService = (AlarmManager) getApplicationContext().getSystemService(Context.ALARM_SERVICE);
    myAlarmService.set(
            AlarmManager.ELAPSED_REALTIME,
            SystemClock.elapsedRealtime() + 1000,
            restartPendingIntent);
}

@Override
protected void onHandleIntent(Intent intent) {
    WakefulBroadcastReceiver.completeWakefulIntent(intent);
    PowerManager powerManager = (PowerManager) getSystemService(POWER_SERVICE);
    PowerManager.WakeLock wakeLock = powerManager.newWakeLock(PowerManager.PARTIAL_WAKE_LOCK,
            "MyWakelockTag");
    wakeLock.acquire();
    runnable.run();
}



private class bgJob extends AsyncTask<Void, Void, Void> {
    @Override
    protected Void doInBackground(Void... params) {
            OkHttpClient client = new OkHttpClient().newBuilder()
                    .build();
            Request request = new Request.Builder()
                    .url("https://google.com/")
                    .build();
            Response response = null;
            try {
                response = client.newCall(request).execute();
                int webresponse = response.code();
                Log.d("Response","CODE: "+webresponse);
            } catch (IOException e) {
                e.printStackTrace();
            }
        return null;
    }

    @Override
    protected void onPostExecute(Void aVoid) {
        super.onPostExecute(aVoid);
        handler.postDelayed(runnable, 60000);
    }
}}

我在 phone 上通过移动连接使用 Android 7.0。

您必须按照给出的说明进行操作here 据我所知只有红米手机有此问题。因此,像这样在 activity 中获取品牌名称。

String manufacturer = Build.MANUFACTURER;
String model = Build.MODEL;

并且使用这个你可以只通知红米用户开启后台服务访问。