意图服务失去与互联网的连接

Question

我的意图服务有问题。我需要我的服务在后台每 1 分钟访问一次网站的 JSON 文件，而无需打开应用程序（发出通知，我以 JSON 格式收到）。它这样做，但是在3-4分钟后，它抛出"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，并且手机仍然具有移动连接。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);
    }
}}

我在手机上使用安卓 7.0 和移动连接。

Answer 1

您必须按照此处给出的说明进行操作 据我所知，只有红米手机有这个问题。因此，像这样在您的活动中获取品牌名称。

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

使用此功能，您只能通知Redmi用户启用后台服务访问。