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我确实有一个有很多静态方便方法的类。其中一个应该动态启动一个 BroadcastReceiver - 但它总是返回一个 InstantiationException。BroadcastReceiver 有一个无参数的构造函数,所以它应该可以工作——但它没有。这是我到目前为止所做的:

这是其类中的便捷方法:

// Static convenience methods in a tools class
public class MyTools {

    // Start BroadcastReceiver dynamically
    public static BroadcastReceiver startBroadcastReceiver(Context context,
            Class<? extends BroadcastReceiver> receiverClass, String receiverTag) {

        BroadcastReceiver receiver = null;

        try {
            receiver = (BroadcastReceiver) receiverClass.newInstance();
            if (receiver != null) {
                IntentFilter intentFilter = new IntentFilter(receiverTag);
                if (intentFilter != null) {
                    context.registerReceiver(receiver, intentFilter);
                }
            }
        } catch (Exception exception) {
            // --> InstantiationException
        }

        return receiver;
    }

    // ...
}

这是一个带有 InnerClass BroadcastReceiver 的活动,它尝试使用这种便捷方法启动 BroadcastReceiver:

// An activity with an InnerClass BroadcastReceiver
public class MyActivity extends Activity {

    public class MyBroadcastReceiver extends BroadcastReceiver {

        public static final String TAG = "aa.bb.cc.MyActivity.MyBroadcastReceiver";

        public static final long ACTION_UNDEFINED = 0;
        public static final long ACTION_DOSOMETHING = 1;

        @Override
        public void onReceive(Context context, Intent intent) {
            if (intent != null) {
                Bundle bundleExtras = intent.getExtras();
                if (bundleExtras != null) {
                    long action = bundleExtras.getLong("ACTION");
                    if (action == ACTION_DOSOMETHING) {
                        doSomething();
                    }
                }
            }
        }
    }

    private MyBroadcastReceiver receiver;

    @Override
    protected void onResume() {
        super.onResume();

        // Start BroadcastReceiver
        receiver = (MyBroadcastReceiver) MyTools.startBroadcastReceiver(this,
                MyBroadcastReceiver.class, MyBroadcastReceiver.TAG);
    }


    public void doSomething() {
        // ...
    }
}

这种方法有什么问题?

非常感谢任何帮助。

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1 回答 1

1

那就是问题所在

使您广播接收器成为静态内部类 public static class MyBroadcastReceiver extends BroadcastReceiver

或在自己的文件中声明

来自实例化异常的无空构造函数消息可能会令人困惑

于 2012-09-08T08:03:16.277 回答