java - android广播接收器和服务来获取通知

Question

嗨，我在接收器类中使用了以下代码

public class AlarmReceiver extends BroadcastReceiver {

    @Override
    public void onReceive(Context context, Intent intent) {
        Log.e("the time is right","yay!");
        Intent i = new Intent(context, AlarmServie.class);
        context.startService(i);
    }
 }

这里是我在服务类中使用的代码

public class AlarmServie extends Service {
    @Nullable
    @Override
    public IBinder onBind(Intent intent) {
        return null;
    }

    @Override
    public void onStart(Intent intent, int startId) {
        super.onStart(intent, startId);
        Log.e("onStart:","came" );
     /*  NotificationManager notifyman = (NotificationManager) getSystemService(NOTIFICATION_SERVICE);
        Intent main_activity = new Intent(this.getApplicationContext(), MainActivity.class);
        PendingIntent o = PendingIntent.getActivity(this, 0, main_activity, 0);

        /*Notification noti = new Notification.Builder(this)
                .setContentTitle("Reminder to Pill")
                .setContentText("Click for info")
                .setAutoCancel(true)
                .setContentIntent(o)
                .build();*/
        NotificationManager nm = (NotificationManager) this.getApplicationContext().getSystemService(NOTIFICATION_SERVICE);
        Intent in = new Intent(this.getApplicationContext(), MainActivity.class);
        PendingIntent pending = PendingIntent.getactivity(this, 0, in, 0);
        NotificationCompat.Builder mBuilder =new NotificationCompat.Builder(AlarmServie.this);
        mBuilder.setContentTitle("Pill Reminder");
        mBuilder.setContentText("CLick here to View");
        //mBuilder.setSound(sound);

        TaskStackBuilder ts=TaskStackBuilder.create(this);
        ts.addParentStack(MainActivity.class);
        nm.notify(9999,mBuilder.build());

    }}

我创建了这段代码来获取通知。当我运行应用程序时，接收器类被触发，但它没有移动到服务类来调用通知。任何人都可以告诉我代码中有什么问题，或者在详细教程中告诉我如何使用广播接收器和服务获取通知

这是 list 文件，请告诉我将代码准确放置在哪里

<uses-permission android:name="android.permission.VIBRATE"/>
    <uses-permission android:name="android.permission.WAKE_LOCK"/>
    <uses-permission android:name="com.android.alarm.permission.SET_ALARM"/>


    <application
        android:allowBackup="true"
        android:icon="@mipmap/ic_launcher"
        android:label="@string/app_name"
        android:supportsRtl="true"
        android:theme="@style/AppTheme">
        <activity
            android:name=".MainActivity"
            android:label="@string/app_name"
            android:theme="@style/AppTheme.NoActionBar">
            <intent-filter>
                <action android:name="android.intent.action.MAIN" />

                <category android:name="android.intent.category.LAUNCHER" />



            </intent-filter>
        </activity>
        <receiver android:name=".AlarmReceiver"></receiver>
        <activity android:name=".Set"></activity><!-- ATTENTION: This was auto-generated to add Google Play services to your project for
     App Indexing.  See https://g.co/AppIndexing/AndroidStudio for more information. -->
        <meta-data
            android:name="com.google.android.gms.version"
            android:value="@integer/google_play_services_version" />
    </application>

</manifest>

请详细解释代码中的问题。请准确说明服务标签的放置位置

Answer 1

您必须declare service于显也。启动服务的 Intent 被发送到系统，然后系统尝试使用 list 文件中传递的信息找到正确的应用程序组件来处理此特定 Intent 。

应该足够了，但我不确定你是否将服务放在默认包中。

<application>
    ....
    <service android:name=".AlarmServie"/>
    ....
</application>