检查服务是否正在运行的正确方法是简单地询问它。在您的服务中实现一个BroadcastReceiver来响应来自活动的ping。在服务启动时注册BroadcastReceiver，在服务被销毁时注销它。从您的活动(或任何组件)发送一个本地广播意图到服务，如果它响应，您就知道它正在运行。注意下面代码中ACTION_PING和ACTION_PONG之间的细微差别。

public class PingableService extends Service {
    public static final String ACTION_PING = PingableService.class.getName() + ".PING";
    public static final String ACTION_PONG = PingableService.class.getName() + ".PONG";

    public int onStartCommand (Intent intent, int flags, int startId) {
        LocalBroadcastManager.getInstance(this).registerReceiver(mReceiver, new IntentFilter(ACTION_PING));
        return super.onStartCommand(intent, flags, startId);
    }

    @Override
    public void onDestroy () {
        LocalBroadcastManager.getInstance(this).unregisterReceiver(mReceiver);
        super.onDestroy();
    }

    private BroadcastReceiver mReceiver = new BroadcastReceiver() {
        @Override
        public void onReceive (Context context, Intent intent) {
            if (intent.getAction().equals(ACTION_PING)) {
                LocalBroadcastManager manager = LocalBroadcastManager.getInstance(getApplicationContext());
                manager.sendBroadcast(new Intent(ACTION_PONG));
            }
        }
    };
}

public class MyActivity extends Activity {
    private boolean isSvcRunning = false;

    @Override
    protected void onStart() {
        LocalBroadcastManager manager = LocalBroadcastManager.getInstance(getApplicationContext());
        manager.registerReceiver(mReceiver, new IntentFilter(PingableService.ACTION_PONG));
        // the service will respond to this broadcast only if it's running
        manager.sendBroadcast(new Intent(PingableService.ACTION_PING));
        super.onStart();
    }

    @Override
    protected void onStop() {
        LocalBroadcastManager.getInstance(this).unregisterReceiver(mReceiver);
        super.onStop();
    }

    protected BroadcastReceiver mReceiver = new BroadcastReceiver() {
        @Override
        public void onReceive (Context context, Intent intent) {
            // here you receive the response from the service
            if (intent.getAction().equals(PingableService.ACTION_PONG)) {
                isSvcRunning = true;
            }
        }
    };
}

得到它!

你必须调用startService()为你的服务正确注册和传递BIND_AUTO_CREATE是不够的。

Intent bindIntent = new Intent(this,ServiceTask.class);
startService(bindIntent);
bindService(bindIntent,mConnection,0);

现在是ServiceTools类:

public class ServiceTools {
    private static String LOG_TAG = ServiceTools.class.getName();

    public static boolean isServiceRunning(String serviceClassName){
        final ActivityManager activityManager = (ActivityManager)Application.getContext().getSystemService(Context.ACTIVITY_SERVICE);
        final List<RunningServiceInfo> services = activityManager.getRunningServices(Integer.MAX_VALUE);

        for (RunningServiceInfo runningServiceInfo : services) {
            if (runningServiceInfo.service.getClassName().equals(serviceClassName)){
                return true;
            }
        }
        return false;
     }
}

可以有多个具有相同类名的服务。

我刚刚创建了两个应用程序。第一个应用程序的包名是com.example.mock。我在应用程序中创建了一个名为lorem的子包和一个名为Mock2Service的服务。因此它的完全限定名是com.example.mock.lorem.Mock2Service。

然后我创建了第二个应用程序和一个名为Mock2Service的服务。第二个应用程序的包名是com.example.mock.lorem。服务的完全限定名是com.example.mock.lorem。Mock2Service。

这是我的logcat输出。

03-27 12:02:19.985: D/TAG(32155): Mock-01: com.example.mock.lorem.Mock2Service
03-27 12:02:33.755: D/TAG(32277): Mock-02: com.example.mock.lorem.Mock2Service

更好的方法是比较ComponentName实例，因为ComponentName的equals()比较包名和类名。一个设备上不能安装两个包名相同的应用程序。

ComponentName的equals()方法。

@Override
public boolean equals(Object obj) {
    try {
        if (obj != null) {
            ComponentName other = (ComponentName)obj;
            // Note: no null checks, because mPackage and mClass can
            // never be null.
            return mPackage.equals(other.mPackage)
                    && mClass.equals(other.mClass);
        }
    } catch (ClassCastException e) {
    }
    return false;
}

ComponentName

在Kotlin课上，geekQ的反应。由于geekQ

fun isMyServiceRunning(serviceClass : Class<*> ) : Boolean{
    var manager = getSystemService(Context.ACTIVITY_SERVICE) as ActivityManager
    for (service in manager.getRunningServices(Integer.MAX_VALUE)) {
        if (serviceClass.name.equals(service.service.className)) {
            return true
        }
    }
    return false
}

调用

isMyServiceRunning(NewService::class.java)

我的kotlin转换ActivityManager::getRunningServices基于答案。把这个函数放在一个活动中-

private fun isMyServiceRunning(serviceClass: Class<out Service>) =
    (getSystemService(ACTIVITY_SERVICE) as ActivityManager)
        .getRunningServices(Int.MAX_VALUE)
        ?.map { it.service.className }
        ?.contains(serviceClass.name) ?: false

这更适用于Intent Service调试，因为它们会生成一个线程，但也适用于常规服务。我找到这个线程感谢宾

在我的例子中，我摆弄了调试器并找到了线程视图。它看起来有点像微软Word里的项目符号图标。无论如何，您不必在调试器模式下使用它。单击该流程并单击该按钮。任何Intent服务都会在运行时显示出来，至少在模拟器上是这样。