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

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;
    }
} );

对于那些谁使用静态片段

在这种情况下,如果你有一个静态片段,那么它会更可取。 为片段创建一个实例对象

private static MyFragment instance=null;

在MyFragment的onCreate()中初始化该实例

  instance=this;

也可以创建一个函数来获取Instance

 public static MyFragment getInstance(){
   return instance;
}

也可以创建函数

public boolean allowBackPressed(){
    if(allowBack==true){
        return true;
    }
    return false;
}


 //allowBack is a boolean variable that will be set to true at the action 
 //where you want that your backButton should not close activity. In my case I open 
 //Navigation Drawer then I set it to true. so when I press backbutton my 
 //drawer should be get closed

public void performSomeAction(){
    //.. Your code
    ///Here I have closed my drawer
}

在你能做的活动中

@Override
public void onBackPressed() {

    if (MyFragment.getInstance().allowBackPressed()) { 
        MyFragment.getInstance().performSomeAction();
    }
    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();
    }
}

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);
}