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

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;
    }
} );

其他回答

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 } } }

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);
}

在你的oncreateView()方法中,你需要写这些代码,在KEYCODE_BACk条件下,你可以写任何你想要的功能

View v = inflater.inflate(R.layout.xyz, container, false);
//Back pressed Logic for fragment 
v.setFocusableInTouchMode(true); 
v.requestFocus(); 
v.setOnKeyListener(new View.OnKeyListener() { 
    @Override 
    public boolean onKey(View v, int keyCode, KeyEvent event) {
        if (event.getAction() == KeyEvent.ACTION_DOWN) {
            if (keyCode == KeyEvent.KEYCODE_BACK) {
                getActivity().finish(); 
                Intent intent = new Intent(getActivity(), MainActivity.class);
                startActivity(intent);

                return true; 
            } 
        } 
        return false; 
    } 
}); 

你可以使用from getActionBar().setDisplayHomeAsUpEnabled():

@Override
public void onBackStackChanged() {
    int backStackEntryCount = getFragmentManager().getBackStackEntryCount();

    if(backStackEntryCount > 0){
        getActionBar().setDisplayHomeAsUpEnabled(true);
    }else{
        getActionBar().setDisplayHomeAsUpEnabled(false);
    }
}

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;
    }
} );