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();
    }
}

其他回答

如果你正在使用FragmentActivity。然后像这样做

首先在你的碎片中调用这个。

public void callParentMethod(){
    getActivity().onBackPressed();
}

然后调用你的父类FragmentActivity中的onBackPressed方法。

@Override
public void onBackPressed() {
  //super.onBackPressed();
  //create a dialog to ask yes no question whether or not the user wants to exit
  ...
}

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;
    }
} );

在fragment类中,为back event放以下代码:

 rootView.setFocusableInTouchMode(true);
        rootView.requestFocus();
        rootView.setOnKeyListener( new OnKeyListener()
        {
            @Override
            public boolean onKey( View v, int keyCode, KeyEvent event )
            {
                if( keyCode == KeyEvent.KEYCODE_BACK )
                {
                    FragmentManager fragmentManager = getFragmentManager();
                    fragmentManager.beginTransaction()
                            .replace(R.id.frame_container, new Book_service_provider()).commit();

                    return true;
                }
                return false;
            }
        } );

如果您管理将每个事务添加到后退堆栈的流程,那么您可以这样做,以便在用户按后退按钮时显示上一个片段(您也可以映射home按钮)。

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

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();
    }
}