Question

我有一项需要知道手机位置的服务。该服务的主要活动在一个主题中执行如下（删除处理内容：

Semaphore locationAcquired = new Semaphore(1);
LocationFinder finder;

...
public void run() {
            delaySeconds = 60;
            Looper.prepare();
            while (true) {
                try {

                    finder.StartFinder();

                locationAcquired.acquire();
                // do some stuff...
                } catch (InterruptedException e) {
                    if (isDestroy) {
                        Log.d(TAG, "Closing Monitor Thread");
                        break;
                    } // else just wake up and process the location 
                } catch (Exception e) {
                    e.printStackTrace();

                }
            } // end while
        } // end run

LocationFinder类实现（再次，略微简化）：

package com.ksdagile.opengate;

import...

public class LocationFinder {

public static final int ONE_SECOND = 1000;

LocationListener locationListener;
String provider = LocationManager.PASSIVE_PROVIDER; // passive by default
LocationManager locationManager;
public Location currentLocation;
long updateSeconds;
private boolean isLooking = false;
OpenGateService openGateService;

public LocationFinder(LocationManager _lm, OpenGateService _openGateService) {

    openGateService = _openGateService;
    locationManager = _lm;
     // initialize with whatever location might be available
    currentLocation = locationManager.getLastKnownLocation(LocationManager.PASSIVE_PROVIDER); 


    locationListener = new LocationListener() {
        @Override
        public void onLocationChanged(Location location) {
            // A new location update is received.  Do something useful with it.  In this case,
            // we're sending the update to a handler which then updates the UI with the new
            // location.
            currentLocation = location;
            String newLoc = String.format("Found new Location lat:%.2f long:%.2f", currentLocation.getLatitude(), currentLocation.getLongitude());
            Log.d(getClass().getName(), newLoc);
            openGateService.locationAcquired.release();
        }
                    // simple implementations of onProvider<> etc.
    };

}

public void SetProvider(boolean isActive) {
    if (isActive)
        provider = LocationManager.GPS_PROVIDER;
    else
        provider = LocationManager.PASSIVE_PROVIDER;
}

public void SetFrequency(long delay) {
    updateSeconds = delay;
}

public void StartFinder() {
    if (!isLooking) {
        isLooking = true;
        locationManager.requestLocationUpdates(provider, updateSeconds*ONE_SECOND, 10, locationListener);
        Log.d(getClass().getName(), String.format("Request location from %s provider, every %d sec.", provider, updateSeconds));            
    } else
        Log.d(getClass().getName(), "Location request running");
}

public void StopFinder() {
    locationManager.removeUpdates(locationListener);
    isLooking = false;
}

public boolean IsLocating() {
    return isLooking;
}
}

我的问题是没有调用onLocationChanged例程，即使我知道有新的读数。例如，当配置为在被动模式下读取时，我运行Waze并看到自己移动。是否有可能信号量阻止对onLocationChanged的调用？如果是这样，我该如何解决这个问题呢？我希望能够动态更改requestLocationUpdate的参数。

Answer 1

我认为获得风滚草徽章是一种非常可疑的荣誉，但这里的答案是：是的，.acquire（）调用会阻塞该线程，并且不会调用onLocationChanged（）方法。

解决方案是让处理程序运行runnable，例如：

public void onLocationChanged(Location location) {
            xxxService.locHandler.post(xxxService.locationRun);
        }

其中

Runnable locationRun = new Runnable() {

    @Override
    public void run() {
        // handle stuff for new location
    }
};

一般原则是您发送和处理事件，而不是运行轮询循环。如果您需要处理非事件，例如没有读取新位置，然后设置倒数计时器。

服务中的位置查找器被信号量阻止？

1 个答案: