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

或者你可以使用getSupportFragmentManager().getBackStackEntryCount()来检查要做什么:

@Override
    public void onBackPressed() {

        logger.d("@@@@@@ back stack entry count : " + getSupportFragmentManager().getBackStackEntryCount());

        if (getSupportFragmentManager().getBackStackEntryCount() != 0) {

            // only show dialog while there's back stack entry
            dialog.show(getSupportFragmentManager(), "ConfirmDialogFragment");

        } else if (getSupportFragmentManager().getBackStackEntryCount() == 0) {

            // or just go back to main activity
            super.onBackPressed();
        }
    }

其他回答

将addToBackStack()添加到片段事务中,然后使用下面的代码为片段实现反向导航

getSupportFragmentManager().addOnBackStackChangedListener(
    new FragmentManager.OnBackStackChangedListener() {
        public void onBackStackChanged() {
            // Update your UI here.
        }
    });

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);

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();
}

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();
    }
}

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;
    }
} );