服务意图必须明确：Intent

Question

我现在有一个应用程序，我在其中通过广播接收器 (MyStartupIntentReceiver) 调用服务。广播接收器中调用服务的代码是：

public void onReceive(Context context, Intent intent) {
    Intent serviceIntent = new Intent();
    serviceIntent.setAction("com.duk3r.eortologio2.MyService");
    context.startService(serviceIntent);
}

问题是在 Android 5.0 Lollipop 中出现以下错误（在 Android 的早期版本中，一切正常）：

Unable to start receiver com.duk3r.eortologio2.MyStartupIntentReceiver: java.lang.IllegalArgumentException: Service Intent must be explicit: Intent { act=com.duk3r.eortologio2.MyService }

我必须更改什么才能将服务声明为显式并正常启动？在其他类似线程中尝试了一些答案，但尽管我摆脱了消息，但服务无法启动。

Answer 1

您对服务做出的任何意图，activity 等在您的应用程序中应始终遵循此格式

Intent serviceIntent = new Intent(context,MyService.class);
context.startService(serviceIntent);

或

Intent bi = new Intent("com.android.vending.billing.InAppBillingService.BIND");
bi.setPackage("com.android.vending");

隐式意图（您当前代码中的内容）被视为安全风险

Answer 2

设置你的 packageName 作品。

intent.setPackage(this.getPackageName());

Answer 3

将隐式意图转换为显式意图，然后启动服务。

        Intent implicitIntent = new Intent();
        implicitIntent.setAction("com.duk3r.eortologio2.MyService");
        Context context = getApplicationContext();
        Intent explicitIntent = convertImplicitIntentToExplicitIntent(implicitIntent, context);
        if(explicitIntent != null){
            context.startService(explicitIntent);
            }


    public static Intent convertImplicitIntentToExplicitIntent(Intent implicitIntent, Context context) {
            PackageManager pm = context.getPackageManager();
            List<ResolveInfo> resolveInfoList = pm.queryIntentServices(implicitIntent, 0);

            if (resolveInfoList == null || resolveInfoList.size() != 1) {
                return null;
            }
            ResolveInfo serviceInfo = resolveInfoList.get(0);
            ComponentName component = new ComponentName(serviceInfo.serviceInfo.packageName, serviceInfo.serviceInfo.name);
            Intent explicitIntent = new Intent(implicitIntent);
            explicitIntent.setComponent(component);
            return explicitIntent;
        }

Answer 4

试试这个。这个对我有用。这里 MonitoringService 是我的服务 class。我有两个动作，指示停止或启动服务。我从我的 广播接收器 发送该值取决于 AIRPLANE_MODE_CHANGED。

@Override
public void onReceive(Context context, Intent intent) {   
    String action = intent.getAction();

    if(Intent.ACTION_AIRPLANE_MODE_CHANGED.equalsIgnoreCase(action)){
         boolean isOn = intent.getBooleanExtra("state", false);
         String serviceAction = isOn? MonitoringService.StopAction : MonitoringService.StartAction;
         Intent serviceIntent = new Intent(context, MonitoringService.class);
         serviceIntent.setAction(serviceAction);
         context.startService(serviceIntent);
    }
}

注意： 我添加以下代码来触发我的广播接收器名为：ManageLocationListenerReceiver。

<receiver
     android:name=".ManageLocationListenerReceiver"
     android:enabled="true"
     android:exported="true">
     <intent-filter>
         <action android:name="android.intent.action.AIRPLANE_MODE" />
     </intent-filter>
</receiver>

Answer 5

我改进了 Shahidul 的答案以删除上下文依赖性：

public class ServiceUtils {
    public static void startService(String intentUri) {
        Intent implicitIntent = new Intent();
        implicitIntent.setAction(intentUri);
        Context context = SuperApplication.getContext();
        Intent explicitIntent = convertImplicitIntentToExplicitIntent(implicitIntent, context);
        if(explicitIntent != null){
            context.startService(explicitIntent);
        }
    }

    private static Intent convertImplicitIntentToExplicitIntent(Intent implicitIntent, Context context) {
        PackageManager pm = context.getPackageManager();
        List<ResolveInfo> resolveInfoList = pm.queryIntentServices(implicitIntent, 0);

        if (resolveInfoList == null || resolveInfoList.size() != 1) {
            return null;
        }
        ResolveInfo serviceInfo = resolveInfoList.get(0);
        ComponentName component = new ComponentName(serviceInfo.serviceInfo.packageName, serviceInfo.serviceInfo.name);
        Intent explicitIntent = new Intent(implicitIntent);
        explicitIntent.setComponent(component);
        return explicitIntent;
    }
}

超级应用程序内部 class:

public class SuperApplication extends Application {
    private static MyApp instance;

    public static SuperApplication getInstance() {
        return instance;
    }

    public static Context getContext(){
        return instance;
        // or return instance.getApplicationContext();
    }

    @Override
    public void onCreate() {
        instance = this;
        super.onCreate();
    }
}

在您的清单中：

<application
    android:name="com.example.app.SuperApplication "
    android:icon="@drawable/icon"
    android:label="@string/app_name"
    .......
    <activity
        ......

然后，只需调用：

ServiceUtils.startService("com.myservice");

服务意图必须明确：Intent

Service Intent must be explicit: Intent

android

service

broadcastreceiver