我的活动中有一些碎片

[1], [2], [3], [4], [5], [6]

如果当前活动片段是[2],那么在返回按钮上按下我必须从[2]返回到[1],否则什么也不做。

最好的做法是什么?

编辑:应用程序不能从[3]…[6]返回[2]


当前回答

使用addToBackStack方法替换一个片段:

getFragmentManager().beginTransaction().replace(R.id.content_frame, fragment).addToBackStack("my_fragment").commit();

然后在您的活动中,使用以下代码从一个片段返回到另一个片段(前一个片段)。

@Override
public void onBackPressed() {
    if (getParentFragmentManager().getBackStackEntryCount() > 0) {
        getParentFragmentManager().popBackStack();
    } else {
        super.onBackPressed();
    }
}

其他回答

如果你重写了片段视图的onKey方法,你需要:

    view.setFocusableInTouchMode(true);
    view.requestFocus();
    view.setOnKeyListener(new View.OnKeyListener() {
            @Override
            public boolean onKey(View v, int keyCode, KeyEvent event) {
                Log.i(tag, "keyCode: " + keyCode);
                if( keyCode == KeyEvent.KEYCODE_BACK && event.getAction() == KeyEvent.ACTION_UP) {
                    Log.i(tag, "onKey Back listener is working!!!");
                    getFragmentManager().popBackStack(null, FragmentManager.POP_BACK_STACK_INCLUSIVE);
                    return true;
                } 
                return false;
            }
        });

我正在与SlidingMenu和Fragment一起工作,在这里展示我的案例,希望能帮助到别人。

按[后退]键时的逻辑:

When SlidingMenu shows, close it, no more things to do. Or when 2nd(or more) Fragment showing, slide back to previous Fragment, and no more things to do. SlidingMenu not shows, current Fragment is #0, do the original [Back] key does. public class Main extends SherlockFragmentActivity { private SlidingMenu menu=null; Constants.VP=new ViewPager(this); //Some stuff... @Override public void onBackPressed() { if(menu.isMenuShowing()) { menu.showContent(true); //Close SlidingMenu when menu showing return; } else { int page=Constants.VP.getCurrentItem(); if(page>0) { Constants.VP.setCurrentItem(page-1, true); //Show previous fragment until Fragment#0 return; } else {super.onBackPressed();} //If SlidingMenu is not showing and current Fragment is #0, do the original [Back] key does. In my case is exit from APP } } }

创建接口:

后退按钮处理程序接口

public interface BackButtonHandlerInterface {
    void addBackClickListener (OnBackClickListener onBackClickListener);
    void removeBackClickListener (OnBackClickListener onBackClickListener);
}

OnBackClickListener

public interface OnBackClickListener {
     boolean onBackClick();
}

在活动:

public class MainActivity extends AppCompatActivity implements BackButtonHandlerInterface {

    private ArrayList<WeakReference<OnBackClickListener>> backClickListenersList = new ArrayList<>();

    @Override
    public void addBackClickListener(OnBackClickListener onBackClickListener) {
        backClickListenersList.add(new WeakReference<>(onBackClickListener));
    }

    @Override
    public void removeBackClickListener(OnBackClickListener onBackClickListener) {
        for (Iterator<WeakReference<OnBackClickListener>> iterator = backClickListenersList.iterator();
         iterator.hasNext();){
            WeakReference<OnBackClickListener> weakRef = iterator.next();
            if (weakRef.get() == onBackClickListener){
                iterator.remove();
            }
        }
    }

    @Override
    public void onBackPressed() {
        if(!fragmentsBackKeyIntercept()){
            super.onBackPressed();
        }
    }

    private boolean fragmentsBackKeyIntercept() {
        boolean isIntercept = false;
        for (WeakReference<OnBackClickListener> weakRef : backClickListenersList) {
            OnBackClickListener onBackClickListener = weakRef.get();
            if (onBackClickListener != null) {
                boolean isFragmIntercept = onBackClickListener.onBackClick();
                if (!isIntercept) isIntercept = isFragmIntercept;
            }
        }
        return isIntercept;
    }
}

在片段:

public class MyFragment extends Fragment implements OnBackClickListener{

    private BackButtonHandlerInterface backButtonHandler;

    @Override
    public void onAttach(Activity activity) {
        super.onAttach(activity);
        backButtonHandler = (BackButtonHandlerInterface) activity;
        backButtonHandler.addBackClickListener(this);
    }

    @Override
    public void onDetach() {
        super.onDetach();
        backButtonHandler.removeBackClickListener(this);
        backButtonHandler = null;
    }

    @Override
    public boolean onBackClick() {
        //This method handle onBackPressed()! return true or false
        return false;
    }

}

更新

提供自定义向后导航

class MyFragment : Fragment() {

    override fun onCreate(savedInstanceState: Bundle?) {
        super.onCreate(savedInstanceState)

        // This callback will only be called when MyFragment is at least Started.
        val callback = requireActivity().onBackPressedDispatcher.addCallback(this) {
            // Handle the back button event
        }

        // The callback can be enabled or disabled here or in the lambda
    }

}

如果你想处理硬件返回键事件,那么你必须在Fragment的onActivityCreated()方法中执行以下代码。

你还需要检查Action_Down或Action_UP事件。如果你不检查,那么onKey()方法将调用2次。

同样,如果你的rootview(getView())将不包含焦点,那么它将无法工作。如果你点击了任何控件,那么你需要再次使用getView().requestFocus()给rootview的焦点;在此之后,只有onKeydown()将调用。

getView().setFocusableInTouchMode(true);
getView().requestFocus();

getView().setOnKeyListener(new OnKeyListener() {
        @Override
        public boolean onKey(View v, int keyCode, KeyEvent event) {
                if (event.getAction() == KeyEvent.ACTION_DOWN) {
                    if (keyCode == KeyEvent.KEYCODE_BACK) {
                        Toast.makeText(getActivity(), "Back Pressed", Toast.LENGTH_SHORT).show();
                    return true;
                    }
                }
                return false;
            }
        });

对我来说很好。

检查后台工作完美


@Override
public boolean onKeyDown(int keyCode, KeyEvent event)
{
    if (keyCode == KeyEvent.KEYCODE_BACK)
    {
        if (getFragmentManager().getBackStackEntryCount() == 1)
        {
            // DO something here since there is only one fragment left
            // Popping a dialog asking to quit the application
            return false;
        }
    }
    return super.onKeyDown(keyCode, event);
}