Question

我创建了一个自定义视图，可以放在应用程序的不同位置。我无法避免在视图中使用BroadcastReceiver从应用程序的其余部分获取消息。

我读过它不推荐（Where should I unregisterReceiver in my own view?），但是如果我选择使用它，是否有一个从BroadcastManager取消注册视图的地方？

Answer 1

我建议你使用LocalBroadcastManager。它就像一个BroadcastReceiver，其Intent只能在你的应用程序中看到。

private BroadcastReceiver receiver = new BroadcastReceiver() {
    @Override
    public void onReceive(Context context, Intent intent) {
        // your code here
    }
};

@Override
protected void onPause() {
    LocalBroadcastManager.getInstance(this).unregisterReceiver(receiver);
    super.onPause();
}

@Override
protected void onResume() {
    LocalBroadcastManager lbm = LocalBroadcastManager.getInstance(this);
    IntentFilter filter = new IntentFilter();
    filter.addAction(MyClass.MY_ACTION);
    lbm.registerReceiver(receiver, filter);
    super.onResume();
}

Answer 2

如@vggonz所说，视图没有像活动那样清晰的生命周期，但是，如果您在onAttachedToWindow()中注册广播接收器并取消注册onDetachedFromWindow()，则它会按预期工作。

public class MyView extends View {

    private BroadcastReceiver receiver;

    public MyView(Context context) {
        super(context);
    }

    @Override
    protected void onAttachedToWindow() {
        super.onAttachedToWindow();
        IntentFilter filter = new IntentFilter();
        filter.addAction("SOME_ACTION");
        filter.addAction("SOME_OTHER_ACTION");
        receiver = new BroadcastReceiver() {
            @Override
            public void onReceive(Context context, Intent intent) {
                //do something based on the intent's action
            }
        };
        getContext().registerReceiver(receiver, filter);
    }

    @Override
    protected void onDetachedFromWindow() {
        super.onDetachedFromWindow();
        if (receiver != null) {
            getContext().unregisterReceiver(receiver);
            receiver = null;
        }
    }
}

在自定义视图中注册本地广播

2 个答案: