从IntentService获取位置时，在死线程上向处理程序发送消息

My app needs location fixes on regular basis, even when the phone is not awake. To do this, I am using IntentService with the pattern generously provided by Commonsware. https://github.com/commonsguy/cwac-wakeful

即使手机没有醒来，我的应用也需要定期修复定位。为此，我使用了Commonsware慷慨提供的模式的IntentService。 https://github.com/commonsguy/cwac-wakeful

To get location fixes, I rely on the following code provided by a member called Fedor. What is the simplest and most robust way to get the user's current location on Android?. I slightly modified it to return null instead of getting the last known location

为了获得位置修复，我依赖于名为Fedor的成员提供的以下代码。在Android上获取用户当前位置的最简单，最强大的方法是什么？我稍微修改它以返回null而不是获取最后的已知位置

They both work perfectly fine when not combined: I can get a location fix from Fedor's class in an Activity, and I can do some basic stuff ( like logging something) using Commonsware class.

它们在没有组合时都能很好地工作：我可以在Activity中从Fedor的类中获得一个位置修复，我可以使用Commonsware类做一些基本的东西（比如记录一些东西）。

The problem occurs when I am trying to get location fixes from the Commonsware's WakefulIntentService subclass. The LocationListeners do not react to locationUpdates and I get these warnings (I don't get yet the subtleties of threads, handlers, ...) . Can someone help me understand what the problem is? Thanks and I wish you all a very happy new year :)

当我试图从Commonsware的WakefulIntentService子类获取位置修复时，会出现问题。 LocationListeners没有对locationUpdates作出反应，我得到了这些警告（我还没有得到线程，处理程序......的细微之处）。有人可以帮我理解问题所在吗？谢谢，祝大家新年快乐:)

12-31 14:09:33.664: W/MessageQueue(3264): Handler (android.location.LocationManager$ListenerTransport$1) {41403330} sending message to a      Handler on a dead thread
12-31 14:09:33.664: W/MessageQueue(3264): java.lang.RuntimeException: Handler (android.location.LocationManager$ListenerTransport$1) {41403330} sending message to a Handler on a dead thread
12-31 14:09:33.664: W/MessageQueue(3264): at android.os.MessageQueue.enqueueMessage(MessageQueue.java:196)
12-31 14:09:33.664: W/MessageQueue(3264): at android.os.Handler.sendMessageAtTime(Handler.java:473)
12-31 14:09:33.664: W/MessageQueue(3264): at android.os.Handler.sendMessageDelayed(Handler.java:446)
12-31 14:09:33.664: W/MessageQueue(3264): at android.os.Handler.sendMessage(Handler.java:383)
12-31 14:09:33.664: W/MessageQueue(3264): at android.location.LocationManager$ListenerTransport.onLocationChanged(LocationManager.java:193)
12-31 14:09:33.664: W/MessageQueue(3264): at android.location.ILocationListener$Stub.onTransact(ILocationListener.java:58)
12-31 14:09:33.664: W/MessageQueue(3264): at android.os.Binder.execTransact(Binder.java:338)
12-31 14:09:33.664: W/MessageQueue(3264): at dalvik.system.NativeStart.run(Native Method)

This is my WakefulIntentService subclass:

这是我的WakefulIntentService子类：

public class AppService extends WakefulIntentService {
public static final String TAG = "AppService";
public AppService() {
    super("AppService");
}

public LocationResult locationResult = new LocationResult() {
    @Override
    public void gotLocation(final Location location) {
        if (location == null)
            Log.d(TAG, "location could not be retrieved");
        else
            sendMessage(location);
    }
};

@Override
protected void doWakefulWork(Intent intent) {
    PhoneLocation myLocation;
    myLocation = new PhoneLocation();
    myLocation.getLocation(getApplicationContext(), locationResult);
}

And here a copy of Fedor's class (slightly modified: no need of GPS nor last known location)

这里有Fedor级别的副本（略有修改：不需要GPS也不需要上次知道位置）

public class PhoneLocation {
public static String TAG = "MyLocation";
Timer timer1;
LocationManager lm;
LocationResult locationResult;
boolean gps_enabled=false;
boolean network_enabled=false;

public boolean getLocation(Context context, LocationResult result)
{
    //I use LocationResult callback class to pass location value from MyLocation to user code.
    locationResult=result;
    if(lm==null) lm = (LocationManager) context.getSystemService(Context.LOCATION_SERVICE);

    //exceptions will be thrown if provider is not permitted.
    try{gps_enabled=lm.isProviderEnabled(LocationManager.GPS_PROVIDER);}catch(Exception ex){}
    try{network_enabled=lm.isProviderEnabled(LocationManager.NETWORK_PROVIDER);}catch(Exception ex){}

    //don't start listeners if no provider is enabled
    if(!gps_enabled && !network_enabled)
        return false;

     if(network_enabled)  lm.requestLocationUpdates(LocationManager.NETWORK_PROVIDER, 0, 0, locationListenerNetwork);

    timer1=new Timer();
    timer1.schedule(new ReturnNullLocation(), 20000);
    return true;
}

 LocationListener locationListenerNetwork = new LocationListener() {
     public void onLocationChanged(Location location) {
        timer1.cancel();
        locationResult.gotLocation(location);
        lm.removeUpdates(this);
    }
    public void onProviderDisabled(String provider) {}
    public void onProviderEnabled(String provider) {}
    public void onStatusChanged(String provider, int status, Bundle extras) {}
};

class ReturnNullLocation extends TimerTask {
    @Override
    public void run() { 
          lm.removeUpdates(locationListenerNetwork);
          locationResult.gotLocation(null);
    }
}

public static abstract class LocationResult{
    public abstract void gotLocation(Location location);

}

}

3 个解决方案

#1

You cannot safely register listeners from an IntentService, as the IntentService goes away as soon as onHandleIntent() (a.k.a., doWakefulWork()) completes. Instead, you will need to use a regular service, plus handle details like timeouts (e.g., the user is in a basement and cannot get a GPS signal).

您无法安全地从IntentService注册侦听器，因为只要onHandleIntent（）（a.k.a。，doWakefulWork（））完成，IntentService就会消失。相反，您将需要使用常规服务，以及超时等处理细节（例如，用户在地下室并且无法获得GPS信号）。

#2

I've had a similar situation, and I've used the android Looper facility to good effect: http://developer.android.com/reference/android/os/Looper.html

我有类似的情况，我已经使用android Looper工具效果很好：http：//developer.android.com/reference/android/os/Looper.html

concretely, just after calling the function setting up the the listener, I call:

具体来说，在调用设置监听器的函数之后，我调用：

Looper.loop();

That blocks the current thread so it doesn't die but at the same time lets it process events, so you'll have a chance to get notified when your listener is called. A simple loop would prevent you from getting notified when the listener is called.

这会阻塞当前线程，因此它不会死，但同时让它处理事件，因此您将有机会在调用侦听器时收到通知。一个简单的循环会阻止您在调用侦听器时收到通知。

And when the listener is called and I'm done processing its data, I put:

当监听器被调用并且我已经完成处理它的数据时，我把：

Looper.myLooper().quit();

#3

For others like me: I had a similar problem related to AsyncTask. As I understand, that class assumes that it is initialized on the UI (main) thread.

对于像我这样的人：我遇到了与AsyncTask相关的类似问题。据我所知，该类假定它是在UI（主）线程上初始化的。

A workaround (one of several possible) is to place the following in MyApplication class:

解决方法（可能的几种方法之一）是将以下内容放在MyApplication类中：

@Override
public void onCreate() {
    // workaround for http://code.google.com/p/android/issues/detail?id=20915
    try {
        Class.forName("android.os.AsyncTask");
    } catch (ClassNotFoundException e) {}
    super.onCreate();
}

(do not forget to mention it in AndroidManifest.xml:

（别忘了在AndroidManifest.xml中提及它：

<application
    android:icon="@drawable/ic_launcher"
    android:label="@string/app_name"
    android:name="MyApplication"
    >

)

）

#1

#2