31

我有寻呼机的片段活动：

List<Fragment> fragments = new Vector<Fragment>();
    fragments.add(Fragment.instantiate(this, PastEventListFragment.class.getName(),bundle));
    fragments.add(Fragment.instantiate(this, EventListFragment.class.getName(),bundle));

    this.mPagerAdapter  = new EventPagerAdapter(super.getSupportFragmentManager(), fragments);
    //
    ViewPager pager = (ViewPager)super.findViewById(R.id.viewpager1);

    pager.setAdapter(this.mPagerAdapter);
    pager.setCurrentItem(1);

我赶上 onKeyDown 事件：

@Override
public boolean onKeyDown(int keyCode, KeyEvent event) {
    if (keyCode == KeyEvent.KEYCODE_MENU) {

    }
    return super.onKeyDown(keyCode, event);
}

问题是：如何在我在此活动中实例化的所有片段中使用事件。谢谢

4

6 回答 6

39

您可以做的是在您的片段类中定义一个自定义方法。例如：

public void myOnKeyDown(int key_code){
   //do whatever you want here
}

并在 Activity 类中引发按键事件时调用此方法。例如：

@Override
public boolean onKeyDown(int keyCode, KeyEvent event) {
    if (keyCode == KeyEvent.KEYCODE_MENU) {
        ((PastEventListFragment)fragments.get(0)).myOnKeyDown(keyCode);
        ((EventListFragment)fragments.get(1)).myOnKeyDown(keyCode);

        //and so on...
    }
    return super.onKeyDown(keyCode, event);
}
于 2012-08-31T08:17:42.050 回答
9

如果有人对如何使用 Boradcast 感兴趣：

在 onViewCreated 的片段中

@Override
public void onViewCreated(View view, Bundle savedInstanceState) {
    super.onViewCreated(view, savedInstanceState);


// Register to receive messages.
// We are registering an observer (mMessageReceiver) to receive Intents
// with actions named "custom-event-name".
 LocalBroadcastManager.getInstance(this).registerReceiver(mMessageReceiver,
 new IntentFilter("activity-says-hi"));

...}

 // Our handler for received Intents. This will be called whenever an Intent
 // with an action named "custom-event-name" is broadcasted.
 private BroadcastReceiver mMessageReceiver = new BroadcastReceiver() {
 @Override
 public void onReceive(Context context, Intent intent) {
 // Get extra data included in the Intent

 doSomethingCauseVolumeKeyPressed();

 }
};

你的 keyevent - 投入活动的代码

@Override
public boolean dispatchKeyEvent(KeyEvent event) {
    int action = event.getAction();
    int keyCode = event.getKeyCode();
    switch (keyCode) {
        case KeyEvent.KEYCODE_VOLUME_UP:
            if (action == KeyEvent.ACTION_DOWN) {
                sendBroadcast();
            }
            return true;
        case KeyEvent.KEYCODE_VOLUME_DOWN:
            if (action == KeyEvent.ACTION_DOWN) {
                sendBroadcast();
            }
            return true;
        default:
            return super.dispatchKeyEvent(event);
    }
}

您的广播发件人：

private void  sendVolumeBroadcast(){
    Intent intent = new Intent("activity-says-hi");
    LocalBroadcastManager.getInstance(this).sendBroadcast(intent);
}
于 2016-02-25T14:43:07.123 回答
5

正如其他人所提到的，接受的答案导致活动与其片段之间的紧密耦合。

我建议改用某种基于事件的实现。这更可重用，并产生更好的软件架构。在以前的项目中，我使用了以下解决方案之一（Kotlin）：

广播

使用 Android 的 LocalBroadcastManager：文档

创建一个广播接收器：

class SomeBroadcastReceiver : BroadcastReceiver() {

    override fun onReceive(context: Context?, intent: Intent?) {
        val keyCode = intent?.getIntExtra("KEY_CODE", 0)
        // Do something with the event
    }

}

在您的活动中：

class SomeActivity : AppCompatActivity() {

    override fun onKeyDown(keyCode: Int, event: KeyEvent?): Boolean {
        val intent = Intent("SOME_TAG").apply { putExtra("KEY_CODE", keyCode) }
        LocalBroadcastManager.getInstance(this).sendBroadcast(intent)
        return super.onKeyDown(keyCode, event)
    }

}

然后，在任何片段（或服务等）中：

class SomeFragment : Fragment() {

    val receiver = SomeBroadcastReceiver()

    override fun onCreateView(inflater: LayoutInflater, container: ViewGroup?, savedInstanceState: Bundle?): View? {

        val filter = IntentFilter().apply { addAction("SOME_TAG") }
        LocalBroadcastManager.getInstance(context!!).registerReceiver(receiver, filter)

        return super.onCreateView(inflater, container, savedInstanceState)
    }

}

事件总线

使用事件总线

创建一个事件类：

data class Event(val keyCode: Int)

在您的活动中：

class SomeActivity : AppCompatActivity() {

    override fun onKeyDown(keyCode: Int, event: KeyEvent?): Boolean {
        EventBus.getDefault().post(Event(keyCode))
        return super.onKeyDown(keyCode, event)
    }

}

然后，在您的片段中：

class SomeFragment : Fragment() {

    override fun onCreateView(inflater: LayoutInflater, container: ViewGroup?, savedInstanceState: Bundle?): View? {

        // Register for events
        EventBus.getDefault().register(this)

        return super.onCreateView(inflater, container, savedInstanceState)
    }

    @Subscribe
    public fun onKeyEvent(event : Event) {
        // Called by eventBus when an event occurs
    }

    override fun onDestroyView() {
        super.onDestroyView()
        EventBus.getDefault().unregister(this)
    }

}
于 2018-12-06T09:44:09.777 回答
4

按照@hsu.tw 的回答以避免紧密耦合，我发现了这个要点。

避免紧密耦合是有代价的：您需要一个可聚焦的视图（幸运的是，这是我的情况，因为我已经在前台有一个可以监听其他触摸事件的视图，所以我只是添加了View.OnKeyListener它）。

View.OnKeyListener独立于 Activity将 a 附加到 Fragment 中的视图所需的步骤是（检查要点）：

 view.setFocusableInTouchMode(true);
 view.requestFocus();
 view.setOnKeyListener(pressKeyListener);

onViewCreated我在Fragment的回调中实现了这一点

于 2019-08-05T15:24:51.450 回答
1

我对 Activity 和 Fragment 类进行了子类化以执行 KeyEvents 传递。对我来说，它看起来比发送本地广播更清晰。但是这个解决方案可能不是那么灵活。自行选择首选方式。

这是活动：

public abstract class KeyEventPassingActivity extends Activity {

    public interface KeyEventListener extends View.OnKeyListener {
        boolean isVisible();
        View getView();
    }

    private final List<KeyEventListener> keyEventHandlerList = new ArrayList<>();

    @Override
    public boolean dispatchKeyEvent(KeyEvent event) {
        for (KeyEventListener handler : keyEventHandlerList) {
            if (handleKeyEvent(handler, event)) {
                return true;
            }
        }
        return super.dispatchKeyEvent(event);
    }

    void addKeyEventHandler(@NonNull KeyEventListener handler) {
        keyEventHandlerList.add(handler);
    }

    void removeKeyEventHandler(@NonNull KeyEventListener handler) {
        keyEventHandlerList.remove(handler);
    }

    /**
     * @return <tt>true</tt> if the event was handled, <tt>false</tt> otherwise
     */
    private boolean handleKeyEvent(@Nullable KeyEventListener listener, KeyEvent event) {
        return listener != null
                && listener.isVisible()
                && listener.onKey(listener.getView(), event.getKeyCode(), event);
    }
}

和片段：

public abstract class KeyEventHandlingFragment extends Fragment
        implements KeyEventPassingActivity.KeyEventListener {

    @SuppressWarnings("deprecation")
    @Override
    public void onAttach(Activity activity) {
        super.onAttach(activity);
        if (activity instanceof KeyEventPassingActivity) {
            ((KeyEventPassingActivity) activity).addKeyEventHandler(this);
        }
    }

    @Override
    public void onDetach() {
        Activity activity = getActivity();
        if (activity instanceof KeyEventPassingActivity) {
            ((KeyEventPassingActivity) activity).removeKeyEventHandler(this);
        }
        super.onDetach();
    }
}

要点：https ://gist.github.com/0neel/7d1ed5d26f2148b4168b6616337159ed

于 2017-05-04T09:22:41.330 回答
1

我在开发 Android TV 应用程序时遇到了同样的问题。

我这样解决这个问题：

在 onCreateView 方法中，我通过某个 View 调用“requestFocus”。（我将其标记为 ViewA。）然后我将 KeyEventListener 设置为 ViewA。

在您的情况下，您应该在 Adapter 和 PagerChangeListener 中执行此操作（set-KeyEventListener）。

于 2019-06-14T08:02:03.577 回答