在AndroidManifest.xml中注册本地BroadcastReceiver？

Question

无论如何在AndroidManifest.xml中注册BroadcastReceiver并接收由LocalBroadcastManager发送的广播？

目前我必须打电话

registerReceiver(BroadcastReceiver receiver, IntentFilter filter) 

要注册一个Receiver,在AndroidManifest.xml中声明将无法正常工作.但这意味着我必须确切地知道接收者的包名和类名,而不仅仅是意图过滤器.是否可以在清单文件中声明接收器？

以下是我目前的代码.

AndroidManifest.xml中:

...
<receiver 
    android:name="com.example.test.MessageReceiver"
    android:enabled="true" >
    <intent-filter>
        <action android:name="com.m2x.test.intent.MESSAGE_RECEIVED" />
    </intent-filter>
</receiver>
...

MainActivity.java:

Intent intent = new Intent();
intent.setAction("com.m2x.test.intent.MESSAGE_RECEIVED");

LocalBroadcastManager manager = LocalBroadcastManager.getInstance(mContext.get());
manager.sendBroadcast(intent);

MessageReceiver.java

public class MessageReceiver extends BroadcastReceiver {

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

        if (intent.getAction().equals("com.m2x.test.intent.MESSAGE_RECEIVED")) {
            Toast.makeText(context, "user message received", Toast.LENGTH_SHORT).show();
        }

    }
}

Answer 1

不,你不能.

本地BroadcastReceiver不是真实的BroadcastReceiver,基本上是它的callbacks功能列表.
您可以检查LocalBroadcastManager.java的源代码.