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

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 } } }

其他回答

当你在Fragments之间转换时,调用addToBackStack()作为FragmentTransaction的一部分:

FragmentTransaction tx = fragmentManager.beginTransation();
tx.replace( R.id.fragment, new MyFragment() ).addToBackStack( "tag" ).commit();

如果你需要更详细的控制(例如,当一些片段可见时,你想要抑制返回键),你可以在你的片段的父视图上设置一个OnKeyListener:

//You need to add the following line for this solution to work; thanks skayred
fragment.getView().setFocusableInTouchMode(true);
fragment.getView().requestFocus();
fragment.getView().setOnKeyListener( new OnKeyListener()
{
    @Override
    public boolean onKey( View v, int keyCode, KeyEvent event )
    {
        if( keyCode == KeyEvent.KEYCODE_BACK )
        {
            return true;
        }
        return false;
    }
} );

这是一个非常好的和可靠的解决方案:http://vinsol.com/blog/2014/10/01/handling-back-button-press-inside-fragments/

这家伙已经制作了一个抽象片段,处理backPress行为,并使用策略模式在活动片段之间切换。

对于你们中的一些人来说,抽象课程可能会有一些缺陷。

简单地说,链接中的解决方案是这样的:

// Abstract Fragment handling the back presses

public abstract class BackHandledFragment extends Fragment {
    protected BackHandlerInterface backHandlerInterface;
    public abstract String getTagText();
    public abstract boolean onBackPressed();

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        if(!(getActivity()  instanceof BackHandlerInterface)) {
            throw new ClassCastException("Hosting activity must implement BackHandlerInterface");
        } else {
            backHandlerInterface = (BackHandlerInterface) getActivity();
        }
    }

    @Override
    public void onStart() {
        super.onStart();

        // Mark this fragment as the selected Fragment.
        backHandlerInterface.setSelectedFragment(this);
    }

    public interface BackHandlerInterface {
        public void setSelectedFragment(BackHandledFragment backHandledFragment);
    }
}   

和在活动中的用法:

// BASIC ACTIVITY CODE THAT LETS ITS FRAGMENT UTILIZE onBackPress EVENTS 
// IN AN ADAPTIVE AND ORGANIZED PATTERN USING BackHandledFragment

public class TheActivity extends FragmentActivity implements BackHandlerInterface {
    private BackHandledFragment selectedFragment;

    @Override
    public void onBackPressed() {
        if(selectedFragment == null || !selectedFragment.onBackPressed()) {
            // Selected fragment did not consume the back press event.
            super.onBackPressed();
        }
    }

    @Override
    public void setSelectedFragment(BackHandledFragment selectedFragment) {
        this.selectedFragment = selectedFragment;
    }
}

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。

最理想的方法如下: 片段:当按下后退按钮并自定义时调用的回调

public class MyActivity extends Activity
{
    //...
    //Defined in Activity class, so override
    @Override
    public void onBackPressed()
    {
        super.onBackPressed();
        myFragment.onBackPressed();
    }
}

public class MyFragment extends Fragment
{
    //Your created method
    public static void onBackPressed()
    {
        //Pop Fragments off backstack and do your other checks
    }
}

我认为最简单的方法是创建一个接口,并在Activity中检查片段是否属于接口类型,如果是,则调用它的方法来处理弹出。下面是要在片段中实现的接口。

public interface BackPressedFragment {

    // Note for this to work, name AND tag must be set anytime the fragment is added to back stack, e.g.
    // getActivity().getSupportFragmentManager().beginTransaction()
    //                .replace(R.id.fragment_container, MyFragment.newInstance(), "MY_FRAG_TAG")
    //                .addToBackStack("MY_FRAG_TAG")
    //                .commit();
    // This is really an override. Should call popBackStack itself.
    void onPopBackStack();
}

下面是如何实现它。

public class MyFragment extends Fragment implements BackPressedFragment
    @Override
    public void onPopBackStack() {
        /* Your code goes here, do anything you want. */
        getActivity().getSupportFragmentManager().popBackStack();
}

在你的Activity中,当你处理弹出时(可能在onBackPressed和onOptionsItemSelected中),使用这个方法弹出backstack:

public void popBackStack() {
    FragmentManager fm = getSupportFragmentManager();
    // Call current fragment's onPopBackStack if it has one.
    String fragmentTag = fm.getBackStackEntryAt(fm.getBackStackEntryCount() - 1).getName();
    Fragment currentFragment = getSupportFragmentManager().findFragmentByTag(fragmentTag);
    if (currentFragment instanceof BackPressedFragment)
        ((BackPressedFragment)currentFragment).onPopBackStack();
    else
        fm.popBackStack();
}