Here's my relatively simple solution to this problem. The keys to this solution are to use FragmentStatePagerAdapter instead of FragmentPagerAdapter as the former will remove unused fragments for you while the later still retains their instances. The second is the use of POSITION_NONE in getItem(). I've used a simple List to keep track of my fragments. My requirement was to replace the entire list of fragments at once with a new list, but the below could be easily modified to replace individual fragments:

public class MyFragmentAdapter extends FragmentStatePagerAdapter {
    private List<Fragment> fragmentList = new ArrayList<Fragment>();
    private List<String> tabTitleList = new ArrayList<String>();

    public MyFragmentAdapter(FragmentManager fm) {
        super(fm);
    }

    public void addFragments(List<Fragment> fragments, List<String> titles) {
        fragmentList.clear();
        tabTitleList.clear();
        fragmentList.addAll(fragments);
        tabTitleList.addAll(titles);
        notifyDataSetChanged();
    }

    @Override
    public int getItemPosition(Object object) {
        if (fragmentList.contains(object)) {
            return POSITION_UNCHANGED;
        }
        return POSITION_NONE;
    }

    @Override
    public Fragment getItem(int item) {
        if (item >= fragmentList.size()) {
            return null;
        }
        return fragmentList.get(item);
    }

    @Override
    public int getCount() {
        return fragmentList.size();
    }

    @Override
    public CharSequence getPageTitle(int position) {
        return tabTitleList.get(position);
    }
}

从2012年11月13日起，在ViewPager中重新调整片段似乎变得容易多了。谷歌发布的Android 4.2支持嵌套片段，在新的Android支持库v11中也支持它，所以这将一直工作到1.6

It's very similiar to the normal way of replacing a fragment except you use getChildFragmentManager. It seems to work except the nested fragment backstack isn't popped when the user clicks the back button. As per the solution in that linked question, you need to manually call the popBackStackImmediate() on the child manager of the fragment. So you need to override onBackPressed() of the ViewPager activity where you'll get the current fragment of the ViewPager and call getChildFragmentManager().popBackStackImmediate() on it.

获取当前正在显示的片段也有点棘手，我使用了这个肮脏的“android:switcher:VIEWPAGER_ID:INDEX”解决方案，但你也可以跟踪ViewPager的所有片段，正如本页上第二个解决方案所解释的那样。

这是我的代码，当用户单击一行时，ViewPager会显示一个详细的视图，后退按钮工作。为了简洁起见，我尝试只包括相关代码，所以如果你想要完整的应用程序上传到GitHub，请留下评论。

HomeActivity.java

 public class HomeActivity extends SherlockFragmentActivity {
FragmentAdapter mAdapter;
ViewPager mPager;
TabPageIndicator mIndicator;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    mAdapter = new FragmentAdapter(getSupportFragmentManager());
    mPager = (ViewPager)findViewById(R.id.pager);
    mPager.setAdapter(mAdapter);
    mIndicator = (TabPageIndicator)findViewById(R.id.indicator);
    mIndicator.setViewPager(mPager);
}

// This the important bit to make sure the back button works when you're nesting fragments. Very hacky, all it takes is some Google engineer to change that ViewPager view tag to break this in a future Android update.
@Override
public void onBackPressed() {
    Fragment fragment = (Fragment) getSupportFragmentManager().findFragmentByTag("android:switcher:" + R.id.pager + ":"+mPager.getCurrentItem());
    if (fragment != null) // could be null if not instantiated yet
    {
        if (fragment.getView() != null) {
            // Pop the backstack on the ChildManager if there is any. If not, close this activity as normal.
            if (!fragment.getChildFragmentManager().popBackStackImmediate()) {
                finish();
            }
        }
    }
}

class FragmentAdapter extends FragmentPagerAdapter {        
    public FragmentAdapter(FragmentManager fm) {
        super(fm);
    }

    @Override
    public Fragment getItem(int position) {
        switch (position) {
        case 0:
            return ListProductsFragment.newInstance();
        case 1:
            return ListActiveSubstancesFragment.newInstance();
        case 2:
            return ListProductFunctionsFragment.newInstance();
        case 3:
            return ListCropsFragment.newInstance();
        default:
            return null;
        }
    }

    @Override
    public int getCount() {
        return 4;
    }

 }
}

ListProductsFragment.java

public class ListProductsFragment extends SherlockFragment {
private ListView list;

public static ListProductsFragment newInstance() {
    ListProductsFragment f = new ListProductsFragment();
    return f;
}

@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
    View V = inflater.inflate(R.layout.list, container, false);
    list = (ListView)V.findViewById(android.R.id.list);
    list.setOnItemClickListener(new OnItemClickListener() {
        public void onItemClick(AdapterView<?> parent, View view,
            int position, long id) {
          // This is important bit
          Fragment productDetailFragment = FragmentProductDetail.newInstance();
          FragmentTransaction transaction = getChildFragmentManager().beginTransaction();
          transaction.addToBackStack(null);
          transaction.replace(R.id.products_list_linear, productDetailFragment).commit();
        }
      });       
    return V;
}
}

基于@wize的回答，我发现它很有用，也很优雅，我可以部分实现我想要的，因为我想要电缆在被替换后回到第一个片段。我稍微修改了一下他的代码就实现了。

这将是FragmentPagerAdapter:

public static class MyAdapter extends FragmentPagerAdapter {
    private final class CalendarPageListener implements
            CalendarPageFragmentListener {
        public void onSwitchToNextFragment() {
            mFragmentManager.beginTransaction().remove(mFragmentAtPos0)
                    .commit();
            if (mFragmentAtPos0 instanceof FirstFragment){
                mFragmentAtPos0 = NextFragment.newInstance(listener);
            }else{ // Instance of NextFragment
                mFragmentAtPos0 = FirstFragment.newInstance(listener);
            }
            notifyDataSetChanged();
        }
    }

    CalendarPageListener listener = new CalendarPageListener();;
    private Fragment mFragmentAtPos0;
    private FragmentManager mFragmentManager;

    public MyAdapter(FragmentManager fm) {
        super(fm);
        mFragmentManager = fm;
    }

    @Override
    public int getCount() {
        return NUM_ITEMS;
    }

    @Override
    public int getItemPosition(Object object) {
        if (object instanceof FirstFragment && mFragmentAtPos0 instanceof NextFragment)
            return POSITION_NONE;
        if (object instanceof NextFragment && mFragmentAtPos0 instanceof FirstFragment)
            return POSITION_NONE;
        return POSITION_UNCHANGED;
    }

    @Override
    public Fragment getItem(int position) {
        if (position == 0)
            return Portada.newInstance();
        if (position == 1) { // Position where you want to replace fragments
            if (mFragmentAtPos0 == null) {
                mFragmentAtPos0 = FirstFragment.newInstance(listener);
            }
            return mFragmentAtPos0;
        }
        if (position == 2)
            return Clasificacion.newInstance();
        if (position == 3)
            return Informacion.newInstance();

        return null;
    }
}

public interface CalendarPageFragmentListener {
    void onSwitchToNextFragment();
}

要执行替换，只需定义一个静态字段，类型为CalendarPageFragmentListener，并通过相应片段的newInstance方法初始化，并分别调用firstfragment . pagelister . onswitchtonextfragment()或nextfragment . pagelister . onswitchtonextfragment()。

替换viewpager中的片段是相当复杂的，但这是非常可能的，可以看起来超级光滑。首先，您需要让viewpager本身处理片段的删除和添加。发生的事情是，当你替换SearchFragment内部的片段时，你的viewpager保留它的片段视图。因此，您最终会得到一个空白页，因为当您试图替换它时，SearchFragment会被删除。

解决方案是在viewpager内部创建一个侦听器，它将处理在viewpager外部所做的更改，因此首先将这段代码添加到适配器的底部。

public interface nextFragmentListener {
    public void fragment0Changed(String newFragmentIdentification);
}

然后你需要在你的viewpager中创建一个私有类，当你想要改变你的片段时，它会成为一个监听器。例如，你可以加上这样的东西。注意，它实现了刚刚创建的接口。因此，无论何时调用此方法，它都将在下面的类中运行代码。

private final class fragmentChangeListener implements nextFragmentListener {


    @Override
    public void fragment0Changed(String fragment) {
        //I will explain the purpose of fragment0 in a moment
        fragment0 = fragment;
        manager.beginTransaction().remove(fragAt0).commit();

        switch (fragment){
            case "searchFragment":
                fragAt0 = SearchFragment.newInstance(listener);
                break;
            case "searchResultFragment":
                fragAt0 = Fragment_Table.newInstance(listener);
                break;
        }

        notifyDataSetChanged();
    }

这里主要有两点需要指出:

fragAt0 is a "flexible" fragment. It can take on whatever fragment type you give it. This allows it to become your best friend in changing the fragment at position 0 to the fragment you desire. Notice the listeners that are placed in the 'newInstance(listener)constructor. These are how you will callfragment0Changed(String newFragmentIdentification)`. The following code shows how you create the listener inside of your fragment. static nextFragmentListener listenerSearch; public static Fragment_Journals newInstance(nextFragmentListener listener){ listenerSearch = listener; return new Fragment_Journals(); }

您可以在onPostExecute内部调用更改

private class SearchAsyncTask extends AsyncTask<Void, Void, Void>{

    protected Void doInBackground(Void... params){
        .
        .//some more operation
        .
    }
    protected void onPostExecute(Void param){

        listenerSearch.fragment0Changed("searchResultFragment");
    }

}

这将触发viewpager内部的代码来切换位置为零位fragAt0的片段，成为一个新的searchResultFragment。在viewpager发挥作用之前，还需要添加另外两个小部件。

一个是在viewpager的getItem重写方法中。

@Override
public Fragment getItem(int index) {

    switch (index) {
    case 0:
        //this is where it will "remember" which fragment you have just selected. the key is to set a static String fragment at the top of your page that will hold the position that you had just selected.  

        if(fragAt0 == null){

            switch(fragment0){
            case "searchFragment":
                fragAt0 = FragmentSearch.newInstance(listener);
                break;
            case "searchResultsFragment":
                fragAt0 = FragmentSearchResults.newInstance(listener);
                break;
            }
        }
        return fragAt0;
    case 1:
        // Games fragment activity
        return new CreateFragment();

    }

现在，如果没有这最后一块，你仍然会得到一张白纸。有点蹩脚，但它是viewPager的必要部分。你必须重写viewpager的getItemPosition方法。通常这个方法将返回POSITION_UNCHANGED，它告诉viewpager保持所有内容相同，因此永远不会调用getItem来将新的片段放在页面上。这里有一个你可以做的例子

public int getItemPosition(Object object)
{
    //object is the current fragment displayed at position 0.  
    if(object instanceof SearchFragment && fragAt0 instanceof SearchResultFragment){
        return POSITION_NONE;
    //this condition is for when you press back
    }else if{(object instanceof SearchResultFragment && fragAt0 instanceof SearchFragment){
        return POSITION_NONE;
    }
        return POSITION_UNCHANGED
}

如我所说，代码非常复杂，但您基本上必须为您的情况创建一个自定义适配器。我所提到的内容将使更改片段成为可能。这可能需要很长时间来消化一切，所以我会有耐心，但一切都会有意义的。这是完全值得花时间的，因为它可以做出一个非常漂亮的应用程序。

这里是处理后退按钮的块。你把这个放到MainActivity中

 public void onBackPressed() {
    if(mViewPager.getCurrentItem() == 0) {
        if(pagerAdapter.getItem(0) instanceof FragmentSearchResults){
            ((Fragment_Table) pagerAdapter.getItem(0)).backPressed();
        }else if (pagerAdapter.getItem(0) instanceof FragmentSearch) {
            finish();
        }
    }

您需要在FragmentSearchResults中创建一个名为backPressed()的方法，该方法调用fragment0changed。这与我之前展示的代码一起处理按下后退按钮。祝您在更改viewpager的代码中好运。这需要大量的工作，据我所知，没有任何快速的适应。就像我说的，你基本上是在创建一个自定义viewpager适配器，并让它使用侦听器处理所有必要的更改

tl;dr:使用一个主机片段，负责替换其托管内容，并跟踪回溯导航历史(如在浏览器中)。

由于您的用例由固定数量的选项卡组成，我的解决方案工作得很好:想法是用自定义类HostFragment的实例填充ViewPager，这能够替换其托管内容并保持自己的回溯导航历史。要替换托管的片段，你需要调用hostfragment.replaceFragment()方法:

public void replaceFragment(Fragment fragment, boolean addToBackstack) {
    if (addToBackstack) {
        getChildFragmentManager().beginTransaction().replace(R.id.hosted_fragment, fragment).addToBackStack(null).commit();
    } else {
        getChildFragmentManager().beginTransaction().replace(R.id.hosted_fragment, fragment).commit();
    }
}

该方法所做的就是用id R.id替换框架布局。Hosted_fragment和提供给方法的片段。

查看我关于这个主题的教程，了解更多细节和GitHub上完整的工作示例!

我找到了一个简单的解决方案，即使你想在中间添加新的片段或替换当前片段，它也能很好地工作。在我的解决方案中，您应该重写getItemId()，它应该为每个片段返回唯一的id。不是默认的位置。

就是这样:

public class DynamicPagerAdapter extends FragmentPagerAdapter {

private ArrayList<Page> mPages = new ArrayList<Page>();
private ArrayList<Fragment> mFragments = new ArrayList<Fragment>();

public DynamicPagerAdapter(FragmentManager fm) {
    super(fm);
}

public void replacePage(int position, Page page) {
    mPages.set(position, page);
    notifyDataSetChanged();
}

public void setPages(ArrayList<Page> pages) {
    mPages = pages;
    notifyDataSetChanged();
}

@Override
public Fragment getItem(int position) {
    if (mPages.get(position).mPageType == PageType.FIRST) {
        return FirstFragment.newInstance(mPages.get(position));
    } else {
        return SecondFragment.newInstance(mPages.get(position));
    }
}

@Override
public int getCount() {
    return mPages.size();
}

@Override
public long getItemId(int position) {
    // return unique id
    return mPages.get(position).getId();
}

@Override
public Object instantiateItem(ViewGroup container, int position) {
    Fragment fragment = (Fragment) super.instantiateItem(container, position);
    while (mFragments.size() <= position) {
        mFragments.add(null);
    }
    mFragments.set(position, fragment);
    return fragment;
}

@Override
public void destroyItem(ViewGroup container, int position, Object object) {
    super.destroyItem(container, position, object);
    mFragments.set(position, null);
}

@Override
public int getItemPosition(Object object) {
    PagerFragment pagerFragment = (PagerFragment) object;
    Page page = pagerFragment.getPage();
    int position = mFragments.indexOf(pagerFragment);
    if (page.equals(mPages.get(position))) {
        return POSITION_UNCHANGED;
    } else {
        return POSITION_NONE;
    }
}
}

注意:在这个例子中，FirstFragment和SecondFragment扩展了抽象类PageFragment，它有方法getPage()。