这就是我实现目标的方法。

首先，在viewPager标签中添加Root_fragment，你想在其中实现按钮点击片段事件。例子;

@Override
public Fragment getItem(int position) {
  if(position==0)
      return RootTabFragment.newInstance();
  else
      return SecondPagerFragment.newInstance();
}

首先，RootTabFragment应该包含FragmentLayout来改变片段。

<FrameLayout xmlns:android="http://schemas.android.com/apk/res/android"
   xmlns:tools="http://schemas.android.com/tools"
   android:id="@+id/root_frame"
   android:layout_width="match_parent"
   android:layout_height="match_parent">
</FrameLayout>

然后，在RootTabFragment onCreateView内部，实现FirstPagerFragment的fragmentChange

getChildFragmentManager().beginTransaction().replace(R.id.root_frame, FirstPagerFragment.newInstance()).commit();

在那之后，实现onClick事件为您的按钮在FirstPagerFragment和使片段改变像那样再次。

getChildFragmentManager().beginTransaction().replace(R.id.root_frame, NextFragment.newInstance()).commit();

希望这对你们有帮助。

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()。

我已经创建了一个ViewPager 3个元素和2个子元素的索引2和3和这里我想做的。

我已经在StackOverFlow之前的问题和答案的帮助下实现了这一点，这里是链接。

ViewPagerChildFragments

在你的onCreateView方法中，容器实际上是一个ViewPager实例。

所以，只是打电话

ViewPager vpViewPager = (ViewPager) container;
vpViewPager.setCurrentItem(1);

将改变当前片段在你的ViewPager。

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()。

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;
    }