Detect The Device Is Vibrating?

I have used the below code to vibrate the device. public void vibrator() { try { Vibrator vibrator = (Vibrator) getSystemService(Context.VIBRATOR_SERVICE); vibr

Solution 1:

No, you can't.

Same question is here.

You can check only if the device vibrating is supported:

Vibrator mVibrator = (Vibrator) getSystemService(Context.VIBRATOR_SERVICE);
boolean hasVibrator = mVibrator.hasVibrator();

See more:

• Android Vibrator

Solution 2:

Introduction

The Vibrator class does not have the isVibrating() method that you are looking for. It uses services, so you cannot easily override Vibrator and add in the extra functionality.

ManagedVibrator

Below, is a ManagedVibrator class that is a wrapper for the Vibrator class. All Vibrator methods are included, with the additional isVibrating() method.

The constant vibration methods with signatures that accept long[] pattern are easy to track because cancel() needs to be called to end the vibration. However, the one time vibration methods with signatures that accept long millseconds are much harder to track.

This implementation uses a ScheduledThreadPoolExecutor to track one time validation methods. It sets the mIsVibrating flag to false just after a one time vibration method finishes.

public class ManagedVibrator {

    public static final String TAG = ManagedVibrator.class.getSimpleName();

    private Context mContext;
    private Vibrator mVibrator;
    private boolean mIsVibrating = false;
    private ScheduledThreadPoolExecutor mExecutor;

    private Runnable mVibrationEndRunnable = new Runnable() {
        @Override
        public void run() {
            setVibrating(false);
        }
    };

    public ManagedVibrator(Context context) {
        this.mContext = context;
        mVibrator = (Vibrator) mContext.getSystemService(Context.VIBRATOR_SERVICE);
        mExecutor = new ScheduledThreadPoolExecutor(1);
    }

    public boolean hasVibrator() {
        return mVibrator.hasVibrator();
    }

    public void vibrate(long milliseconds) {
        setVibrating(true);
        mVibrator.vibrate(milliseconds);
        notifyOnVibrationEnd(milliseconds);
    }

    // Requires API v21
    public void vibrate(long milliseconds, AudioAttributes attributes) {
        setVibrating(true);
        mVibrator.vibrate(milliseconds, attributes);
        notifyOnVibrationEnd(milliseconds);
    }

    public void vibrate(long[] pattern, int repeat) {
        setVibrating(true);
        mVibrator.vibrate(pattern, repeat);
    }

    // Requires API v21
    public void vibrate(long[] pattern, int repeat, AudioAttributes attributes) {
        setVibrating(true);
        mVibrator.vibrate(pattern, repeat, attributes);
    }

    public void cancel() {
        mVibrator.cancel();
        setVibrating(false);
    }

    public boolean isVibrating() {
        return mIsVibrating;
    }

    private void setVibrating(boolean isVibrating) {
        mIsVibrating = isVibrating;
    }

    private void notifyOnVibrationEnd(long milliseconds) {
        try {
            mExecutor.schedule(mVibrationEndRunnable, milliseconds, TimeUnit.MILLISECONDS);
        } catch (RejectedExecutionException e) {
            Log.e(TAG, e.getMessage());
        }
    }

}

Usage

ManagedVibrator vibrator = new ManagedVibrator(this);
vibrator.vibrate(5000);
...
if (vibrator.isVibrating()) {
    // Do something
}

Limitations

1. You need to use one ManagedVibrator instance in your application
2. ManagedVibrator can only tell you about vibrations initiated by your application. It does not know anything about vibrations caused by other applications of services.
3. Very long vibration times or very frequent one time vibrations may cause issues

Solution 3:

There's a bad way for rooted phones but you will get atleast something.

You can read the file at:

"/sys/class/timed_output/vibrator/enable"

It stores the time left as microseconds when the device is vibrating. (tried on 5.1)

For unrooted phones, you can just check "dumpsys vibrator" Process output with a BufferedReader. It's updated once the vibrator finishes vibrating.