0

这不起作用，我从来没有在接收器中收到消息。我已经工作了好几个小时了。我想要做的是使用 LocalBroadcastManager 没有 Activity 使用另一个 AppContext 类。

1级

import android.content.Context;

public class AppContext extends Application {

    private static AppContext instance;

    public AppContext() {
        instance = this;
    }

    public static Context getContext() {
        return instance;
    }

}

2 级

@Override
public void onResume() {

    receiver = new BroadcastReceiver() {

        @Override
        public void onReceive(Context context, Intent intent) {
            //receive your message here

            String message = intent.getStringExtra("message");
            Log.d("receiver", "Got message: " + message);

            Toast.makeText(getApplicationContext(), message, Toast.LENGTH_LONG).show();

        }

    };

    LocalBroadcastManager.getInstance(AppContext.getContext()).registerReceiver(receiver,
            new IntentFilter("custom-event-name"));

3级

                Log.d("sender", "Broadcasting message");
                Intent intent = new Intent("custom-event-name");
                // You can also include some extra data.
                intent.putExtra("message", "99999");
                LocalBroadcastManager.getInstance(AppContext.getContext()).sendBroadcast(intent);
4

1 回答 1

1

我也有同样的问题。不要更改 class3 中的任何内容。我认为它完美。在 class2 中，创建一个扩展 BroadcastReceiver 的内部类。

private class Class2Receiver extends BroadcastReceiver 
{
        @Override
        public void onReceive(Context context, Intent intent) 
{
            Class2.this.class2Refresh();
         }
    }

//class2Refresh(): 在此方法中接收您的消息

在 class2 的 onResume() 方法中，实例化如下：

Class2 class2Rec=new Class2.Class2Receiver();
        LocalBroadcastManager.getInstance(getContext()).registerReceiver(class2Rec,new IntentFilter("custom-event-name"));
        class2Refresh();

让我知道这是否有帮助！

于 2017-02-16T03:42:18.240 回答