这是基于上面Steven的回答。这将返回已经附加到父活动的片段的实际实例。

FragmentPagerAdapter fragmentPagerAdapter = (FragmentPagerAdapter) mViewPager.getAdapter();
    for(int i = 0; i < fragmentPagerAdapter.getCount(); i++) {

        Fragment viewPagerFragment = (Fragment) mViewPager.getAdapter().instantiateItem(mViewPager, i);
        if(viewPagerFragment != null && viewPagerFragment.isAdded()) {

            if (viewPagerFragment instanceof FragmentOne){
                FragmentOne oneFragment = (FragmentOne) viewPagerFragment;
                if (oneFragment != null){
                    oneFragment.update(); // your custom method
                }
            } else if (viewPagerFragment instanceof FragmentTwo){
                FragmentTwo twoFragment = (FragmentTwo) viewPagerFragment;

                if (twoFragment != null){
                    twoFragment.update(); // your custom method
                }
            }
        }
    }

另一个简单的解决方案:

    public class MyPagerAdapter extends FragmentPagerAdapter {
        private Fragment mCurrentFragment;

        public Fragment getCurrentFragment() {
            return mCurrentFragment;
        }
//...    
        @Override
        public void setPrimaryItem(ViewGroup container, int position, Object object) {
            if (getCurrentFragment() != object) {
                mCurrentFragment = ((Fragment) object);
            }
            super.setPrimaryItem(container, position, object);
        }
    }

为了从ViewPager中获取当前可见片段。我用了这个简单的语句，效果很好。

      public Fragment getFragmentFromViewpager()  
      {
         return ((Fragment) (mAdapter.instantiateItem(mViewPager, mViewPager.getCurrentItem())));
      }

在/values/integers.xml中创建整数资源id

<integer name="page1">1</integer>
<integer name="page2">2</integer>
<integer name="page3">3</integer>

然后在PagerAdapter中使用getItem函数:

public Fragment getItem(int position) {

        Fragment fragment = null;

        if (position == 0) {
            fragment = FragmentOne.newInstance();
            mViewPager.setTag(R.integer.page1,fragment);

        }
        else if (position == 1) {

            fragment = FragmentTwo.newInstance();
            mViewPager.setTag(R.integer.page2,fragment);

        } else if (position == 2) {

            fragment = FragmentThree.newInstance();
            mViewPager.setTag(R.integer.page3,fragment);

        }

        return fragment;
        }

然后在activity中写这个函数来获取片段引用:

private Fragment getFragmentByPosition(int position) {
    Fragment fragment = null;

    switch (position) {
        case 0:
            fragment = (Fragment) mViewPager.getTag(R.integer.page1);
            break;

        case 1:

            fragment = (Fragment) mViewPager.getTag(R.integer.page2);
            break;

        case 2:
            fragment = (Fragment) mViewPager.getTag(R.integer.page3);
            break;
            }

            return fragment;
    }

通过调用上面的函数来获取片段引用，然后将其转换为自定义片段:

Fragment fragment = getFragmentByPosition(position);

        if (fragment != null) {
                    FragmentOne fragmentOne = (FragmentOne) fragment;
                    }

Fragment yourFragment = yourviewpageradapter.getItem(int index);

Index是fragment在适配器中的位置，比如你先添加fragment1，所以检索fragment1时将Index传递为0，以此类推

主要答案依赖于框架生成的名称。如果这种情况发生变化，那么它将不再起作用。

这个解决方案怎么样，重写你的Fragment(State)PagerAdapter的instantiateItem()和destroyItem():

public class MyPagerAdapter extends FragmentStatePagerAdapter {
    SparseArray<Fragment> registeredFragments = new SparseArray<Fragment>();

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

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

    @Override
    public Fragment getItem(int position) {
        return MyFragment.newInstance(...); 
    }

    @Override
    public Object instantiateItem(ViewGroup container, int position) {
        Fragment fragment = (Fragment) super.instantiateItem(container, position);
        registeredFragments.put(position, fragment);
        return fragment;
    }

    @Override
    public void destroyItem(ViewGroup container, int position, Object object) {
        registeredFragments.remove(position);
        super.destroyItem(container, position, object);
    }

    public Fragment getRegisteredFragment(int position) {
        return registeredFragments.get(position);
    }
}

在处理可用的Fragments时，这似乎对我有用。尚未实例化的片段在调用getRegisteredFragment时将返回null。但我一直在使用这个主要是为了获得当前片段的ViewPager: adapter . getregisteredfragment (viewpage . getcurrentitem())，这不会返回null。

我不知道这个解决方案还有什么其他缺点。如果有的话，我想知道。