Question

我正在尝试使用为其定义的按钮（操作）创建一个简单的通知。我已设法正确显示它并为我的操作创建一个PendingIntent。我还创建了一个BroadcastReceiver，它应该在单击我的动作时被调用。但它的onReceive（）方法不会被调用。我不知道为什么。我还在AndroidManifest.xml中注册了BroadcastReceiver

MainActivity.java

public class MainActivity extends AppCompatActivity {

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    Button notify = (Button) findViewById(R.id.notify);

    notify.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {

            Intent dismissIntent = new Intent("action1");
            dismissIntent.setAction("action1");
            PendingIntent action1intent = PendingIntent.getBroadcast(MainActivity.this, (int) System.currentTimeMillis(), dismissIntent, PendingIntent.FLAG_CANCEL_CURRENT);

            NotificationCompat.Builder mBuilder =
                    new NotificationCompat.Builder(MainActivity.this)
                            .setSmallIcon(R.drawable.ic_menu)
                            .setContentTitle("My notification")
                            .setContentText("Hello World!")
                            .addAction(R.drawable.ic_menu, "action 1", action1intent);

            Intent resultIntent = new Intent(MainActivity.this, MainActivity.class);
            TaskStackBuilder stackBuilder = TaskStackBuilder.create(MainActivity.this);
            stackBuilder.addParentStack(MainActivity.class);
            stackBuilder.addNextIntent(resultIntent);

            PendingIntent resultPendingIntent = stackBuilder.getPendingIntent(0, PendingIntent.FLAG_UPDATE_CURRENT);

            mBuilder.setContentIntent(resultPendingIntent);
            NotificationManager mNotificationManager =
                    (NotificationManager) getSystemService(Context.NOTIFICATION_SERVICE);

            mNotificationManager.notify(0, mBuilder.build());
        }
    });

}

public class Receiver extends BroadcastReceiver {

    @Override
    public void onReceive(Context context, Intent intent) {

        Log.d("log", "Receiving somthn");

        String whichAction = intent.getAction();

        if (whichAction.equals("action1")) {
            MainActivity.makeToast("This is action 1", context);
        }
    }
}


public static void makeToast(String string, Context context) {
    Toast.makeText(context, string, Toast.LENGTH_SHORT).show();
}
}

的AndroidManifest.xml

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="com.example.orglce.notification">

<application
    android:allowBackup="true"
    android:icon="@mipmap/ic_launcher"
    android:label="@string/app_name"
    android:supportsRtl="true"
    android:theme="@style/AppTheme">

    <receiver android:name=".MainActivity$Receiver">
        <intent-filter>
            <action android:name="com.example.orglce.notification.BROADCAST" />
        </intent-filter>
    </receiver>

    <activity android:name=".MainActivity">
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />
            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
    </activity>
</application>

Answer 1

首先，如果您要使用.MainActivity$Receiver，则Receiver必须是static类。

其次，dismissIntent（冗余地）使用action1作为操作字符串，但<intent-filter>使用com.example.orglce.notification.BROADCAST。这些不匹配。我建议删除<intent-filter>，删除action1，然后使用new Intent(this, Receiver.class)创建一个明确的Intent来标识您的BroadcastReceiver。

BroadcastReceiver未触发通知操作单击

1 个答案: