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

在您的活动中添加此代码

@Override

public void onBackPressed() {
    if (getFragmentManager().getBackStackEntryCount() == 0) {
        super.onBackPressed();
    } else {
        getFragmentManager().popBackStack();
    }
}

并在commit()之前在Fragment中添加这一行

ft.addToBackStack(任何名称);

其他回答

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
    }
}

我宁愿这样做:

private final static String TAG_FRAGMENT = "TAG_FRAGMENT";

private void showFragment() {
    final Myfragment fragment = new MyFragment();
    final FragmentTransaction transaction = getSupportFragmentManager().beginTransaction();
    transaction.replace(R.id.fragment, fragment, TAG_FRAGMENT);
    transaction.addToBackStack(null);
    transaction.commit();
}

@Override
public void onBackPressed() {
    final Myfragment fragment = (Myfragment) getSupportFragmentManager().findFragmentByTag(TAG_FRAGMENT);

    if (fragment.allowBackPressed()) { // and then you define a method allowBackPressed with the logic to allow back pressed or not
        super.onBackPressed();
    }
}

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;
            }
        });

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;
    }
} );

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 } } }