我想要实现的是在方法isClose
中返回正确的结果集。问题是isClose
方法没有等待onSensorChanged
被触发,并返回isClose
字段的默认值"0"。
我这样调用Position类:Position mPosition = new Position();
boolean result = mPosition.isInPocket(this);
位置类:
public class Position implements SensorEventListener {
private SensorManager mSensorManager;
private Sensor mProximity;
private boolean isClose;
public void onAccuracyChanged(Sensor sensor, int accuracy) {
}
public void onSensorChanged(SensorEvent event) {
float[] value = event.values;
if(value[0] > 0) {
isClose = false;
} else
{
isClose = true;
}
}
public boolean isClose(Context context) {
mSensorManager = (SensorManager) context.getSystemService(Context.SENSOR_SERVICE);
mProximity = mSensorManager.getDefaultSensor(Sensor.TYPE_PROXIMITY);
mSensorManager.registerListener(this, mProximity, 0);
return isClose; // I'd like to return this with value set in onSensorChanged.
}
}
对于第一个onSensorChanged
事件,您需要在isClose
中拥有您的主线程wait
,您可以通过多种方式实现这一点,但使用Condition
变量可能是最简单的。
public class Position implements SensorEventListener {
private SensorManager mSensorManager;
private Sensor mProximity;
private boolean isClose;
private final Lock lock = new ReentrantLock();
private final Condition eventReceived = lock.newCondition();
private boolean firstEventOccurred = false;
public void onAccuracyChanged(Sensor sensor, int accuracy) {
}
public void onSensorChanged(SensorEvent event) {
float[] value = event.values;
if (value[0] > 0) {
isClose = false;
} else {
isClose = true;
}
if (!firstEventOccurred) {
lock.lock();
try {
firstEventOccurred = true;
eventReceived.signal();
} finally {
lock.unlock();
}
}
}
public boolean isClose(Context context) {
mSensorManager = (SensorManager) context.getSystemService(Context.SENSOR_SERVICE);
mProximity = mSensorManager.getDefaultSensor(Sensor.TYPE_PROXIMITY);
mSensorManager.registerListener(this, mProximity, 0);
lock.lock();
try {
while (!firstEventOccurred) {
eventReceived.await();
}
} finally {
lock.unlock();
}
return isClose; // I'd like to return this with value set in onSensorChanged.
}
我从上面的代码中省略了InterrupedException
检查,但它应该给你一个想法。