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我有一个摇动听众如下

public class ShakeListener implements SensorEventListener {
    private String TAG = ShakeListener.class.getSimpleName();
    private static final int FORCE_THRESHOLD = 800;
    private static final int TIME_THRESHOLD = 100;
    private static final int SHAKE_TIMEOUT = 500;
    private static final int SHAKE_DURATION = 1000;
    private static final int SHAKE_COUNT = 5;

    private SensorManager mSensorMgr;
    private float mLastX = -1.0f, mLastY = -1.0f, mLastZ = -1.0f;
    private long mLastTime;
    private OnShakeListener mShakeListener;
    private Context mContext;
    private int mShakeCount = 0;
    private long mLastShake;
    private long mLastForce;

    public interface OnShakeListener {
        public void onShake();
    }

    public ShakeListener(Context context) {

        Log.d(TAG,"ShakeListener invoked---->");
        mContext = context;
        resume();
    }

    public void setOnShakeListener(OnShakeListener listener) {

        mShakeListener = listener;
    }

    public void resume() {
        mSensorMgr = (SensorManager) mContext
                .getSystemService(Context.SENSOR_SERVICE);
        if (mSensorMgr == null) {
            throw new UnsupportedOperationException("Sensors not supported");
        }
        boolean supported = false;
        try {
            supported = mSensorMgr.registerListener(this,
                    mSensorMgr.getDefaultSensor(Sensor.TYPE_ACCELEROMETER),
                    SensorManager.SENSOR_DELAY_GAME);
        } catch (Exception e) {
            Toast.makeText(mContext, "Shaking not supported", Toast.LENGTH_LONG)
                    .show();
        }

        if ((!supported) && (mSensorMgr != null))
            mSensorMgr.unregisterListener(this);
    }

    public void pause() {
        if (mSensorMgr != null) {

            mSensorMgr.unregisterListener(this);
            mSensorMgr = null;
        }
    }

    public void onAccuracyChanged(Sensor sensor, int accuracy) {

    }

    public void onSensorChanged(SensorEvent event) {
        if (event.sensor.getType() != Sensor.TYPE_ACCELEROMETER)
            return;
        long now = System.currentTimeMillis();

        if ((now - mLastForce) > SHAKE_TIMEOUT) {
            mShakeCount = 0;
        }

        if ((now - mLastTime) > TIME_THRESHOLD) {
            long diff = now - mLastTime;
            float speed = Math.abs(event.values[SensorManager.DATA_X]
                    + event.values[SensorManager.DATA_Y]
                    + event.values[SensorManager.DATA_Z] - mLastX - mLastY
                    - mLastZ)
                    / diff * 10000;
            if (speed > FORCE_THRESHOLD) {
                if ((++mShakeCount >= SHAKE_COUNT)
                        && (now - mLastShake > SHAKE_DURATION)) {
                    mLastShake = now;
                    mShakeCount = 0;
                     Log.d(TAG,"ShakeListener mShakeListener---->"+mShakeListener);
                    if (mShakeListener != null) {
                        mShakeListener.onShake();
                    }
                }
                mLastForce = now;
            }
            mLastTime = now;
            mLastX = event.values[SensorManager.DATA_X];
            mLastY = event.values[SensorManager.DATA_Y];
            mLastZ = event.values[SensorManager.DATA_Z];
        }}}

我在我的一项活动中使用此侦听器，如下所示

mShaker = new ShakeListener(this);
        mShaker.setOnShakeListener(new ShakeListener.OnShakeListener () {
         public void onShake()
          {

              Intent myIntent = new Intent(myActivity.this, loginActivity.class);
              startActivity(myIntent);
          }

        });

仅当我在该特定活动上时才会触发此震动。但它会在所有活动中触发，更糟糕的是，即使我在震动时关闭应用程序，事件也会触发并使应用程序打开。感谢任何帮助。

4

1 回答 1

1

在您的活动中添加此方法：

    @Override
    protected void finalize() throws Throwable {
        try {
            stop();
        } catch (Exception e){
            e.printStackTrace();
        }
        super.finalize();
    }

    public void stop(){
        try {
            sensorMgr.unregisterListener(this, mAccelerometer);
        } catch (Exception e){
            e.printStackTrace();
        }
    }
于 2012-12-05T09:49:01.777 回答