Works Great with AndroidTeam's solution, however I found that I needed the ability to go back much like FrgmentTransaction.addToBackStack(null) But merely adding this will only cause the Fragment to be replaced without notifying the ViewPager. Combining the provided solution with this minor enhancement will allow you to return to the previous state by merely overriding the activity's onBackPressed() method. The biggest drawback is that it will only go back one at a time which may result in multiple back clicks

private ArrayList<Fragment> bFragments = new ArrayList<Fragment>();
private ArrayList<Integer> bPosition = new ArrayList<Integer>();

public void replaceFragmentsWithBackOut(ViewPager container, Fragment oldFragment, Fragment newFragment) {
    startUpdate(container);

    // remove old fragment

    if (mCurTransaction == null) {
         mCurTransaction = mFragmentManager.beginTransaction();
     }
    int position = getFragmentPosition(oldFragment);
     while (mSavedState.size() <= position) {
         mSavedState.add(null);
     }

     //Add Fragment to Back List
     bFragments.add(oldFragment);

     //Add Pager Position to Back List
     bPosition.add(position);

     mSavedState.set(position, null);
     mFragments.set(position, null);

     mCurTransaction.remove(oldFragment);

     // add new fragment

     while (mFragments.size() <= position) {
         mFragments.add(null);
     }
     mFragments.set(position, newFragment);
     mCurTransaction.add(container.getId(), newFragment);

    finishUpdate(container);

    // ensure getItem returns newFragemtn after calling handleGetItemInbalidated()
    handleGetItemInvalidated(container, oldFragment, newFragment);

    container.notifyItemChanged(oldFragment, newFragment);
 }


public boolean popBackImmediate(ViewPager container){
    int bFragSize = bFragments.size();
    int bPosSize = bPosition.size();

    if(bFragSize>0 && bPosSize>0){
        if(bFragSize==bPosSize){
            int last = bFragSize-1;
            int position = bPosition.get(last);

            //Returns Fragment Currently at this position
            Fragment replacedFragment = mFragments.get(position);               
            Fragment originalFragment = bFragments.get(last);

            this.replaceFragments(container, replacedFragment, originalFragment);

            bPosition.remove(last);
            bFragments.remove(last);

            return true;
        }
    }

    return false;       
}

希望这能帮助到一些人。

同样，就getFragmentPosition()而言，它几乎是反向的getItem()。你知道哪些片段去哪里，只是确保你返回正确的位置它将在。这里有一个例子:

    @Override
    protected int getFragmentPosition(Fragment fragment) {
            if(fragment.equals(originalFragment1)){
                return 0;
            }
            if(fragment.equals(replacementFragment1)){
                return 0;
            }
            if(fragment.equals(Fragment2)){
                return 1;
            }
        return -1;
    }

我按照@wize和@mdelolmo的答案，我得到了解决方案。由于吨。但是，我稍微调整了这些解决方案，以提高内存消耗。

我观察到的问题:

它们保存被替换的Fragment实例。在我的情况下，它是一个持有MapView的片段，我认为它的成本很高。所以，我维护FragmentPagerPositionChanged (POSITION_NONE或POSITION_UNCHANGED)而不是Fragment本身。

这是我的实现。

  public static class DemoCollectionPagerAdapter extends FragmentStatePagerAdapter {

    private SwitchFragListener mSwitchFragListener;
    private Switch mToggle;
    private int pagerAdapterPosChanged = POSITION_UNCHANGED;
    private static final int TOGGLE_ENABLE_POS = 2;


    public DemoCollectionPagerAdapter(FragmentManager fm, Switch toggle) {
        super(fm);
        mToggle = toggle;

        mSwitchFragListener = new SwitchFragListener();
        mToggle.setOnCheckedChangeListener(new CompoundButton.OnCheckedChangeListener() {
            @Override
            public void onCheckedChanged(CompoundButton buttonView, boolean isChecked) {
                mSwitchFragListener.onSwitchToNextFragment();
            }
        });
    }

    @Override
    public Fragment getItem(int i) {
        switch (i)
        {
            case TOGGLE_ENABLE_POS:
                if(mToggle.isChecked())
                {
                    return TabReplaceFragment.getInstance();
                }else
                {
                    return DemoTab2Fragment.getInstance(i);
                }

            default:
                return DemoTabFragment.getInstance(i);
        }
    }

    @Override
    public int getCount() {
        return 5;
    }

    @Override
    public CharSequence getPageTitle(int position) {
        return "Tab " + (position + 1);
    }

    @Override
    public int getItemPosition(Object object) {

        //  This check make sures getItem() is called only for the required Fragment
        if (object instanceof TabReplaceFragment
                ||  object instanceof DemoTab2Fragment)
            return pagerAdapterPosChanged;

        return POSITION_UNCHANGED;
    }

    /**
     * Switch fragments Interface implementation
     */
    private final class SwitchFragListener implements
            SwitchFragInterface {

        SwitchFragListener() {}

        public void onSwitchToNextFragment() {

            pagerAdapterPosChanged = POSITION_NONE;
            notifyDataSetChanged();
        }
    }

    /**
     * Interface to switch frags
     */
    private interface SwitchFragInterface{
        void onSwitchToNextFragment();
    }
}

演示链接在这里..https://youtu.be/l_62uhKkLyM

出于演示目的，在位置2处使用了两个片段TabReplaceFragment和DemoTab2Fragment。在所有其他情况下，我使用DemoTabFragment实例。

解释:

我将Switch从Activity传递到DemoCollectionPagerAdapter。基于这个开关的状态，我们将显示正确的片段。当开关检查被更改时，我调用SwitchFragListener的onSwitchToNextFragment方法，其中我将pagerAdapterPosChanged变量的值更改为POSITION_NONE。查看更多关于POSITION_NONE的信息。这将使getItem无效，我有逻辑实例化正确的片段在那里。抱歉，如果解释得有点乱的话。

再次感谢@wize和@mdelolmo的原创想法。

希望这对你有帮助。：）

如果这个实现有任何缺陷，请告诉我。那将对我的项目大有帮助。

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;
}
}

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!

要替换ViewPager中的片段，你可以将ViewPager、PagerAdapter和FragmentStatePagerAdapter类的源代码移动到你的项目中，并添加以下代码。

分为:viewpage:

public void notifyItemChanged(Object oldItem, Object newItem) {
    if (mItems != null) {
            for (ItemInfo itemInfo : mItems) {
                        if (itemInfo.object.equals(oldItem)) {
                                itemInfo.object = newItem;
                        }
                    }
       }
       invalidate();
    }

FragmentStatePagerAdapter:

public void replaceFragmetns(ViewPager container, Fragment oldFragment, Fragment newFragment) {
       startUpdate(container);

       // remove old fragment

       if (mCurTransaction == null) {
            mCurTransaction = mFragmentManager.beginTransaction();
        }
       int position = getFragmentPosition(oldFragment);
        while (mSavedState.size() <= position) {
            mSavedState.add(null);
        }
        mSavedState.set(position, null);
        mFragments.set(position, null);

        mCurTransaction.remove(oldFragment);

        // add new fragment

        while (mFragments.size() <= position) {
            mFragments.add(null);
        }
        mFragments.set(position, newFragment);
        mCurTransaction.add(container.getId(), newFragment);

       finishUpdate(container);

       // ensure getItem returns newFragemtn after calling handleGetItemInbalidated()
       handleGetItemInbalidated(container, oldFragment, newFragment);

       container.notifyItemChanged(oldFragment, newFragment);
    }

protected abstract void handleGetItemInbalidated(View container, Fragment oldFragment, Fragment newFragment);
protected abstract int  getFragmentPosition(Fragment fragment);

handleGetItemInvalidated()确保下一次调用getItem()后返回newFragment getFragmentPosition()返回片段在适配器中的位置。

现在，替换片段调用

mAdapter.replaceFragmetns(mViewPager, oldFragment, newFragment);

如果你对一个示例项目感兴趣，请向我索取资源。