如何使用 ViewPager 和嵌套片段正确处理屏幕旋转

Question

我有这个活动，它包含一个片段。此片段布局由一个具有多个片段（实际上是两个）的视图寻呼机组成。

创建视图寻呼机时，将创建其适配器，调用 getItem并创建我的子片段。伟大。

现在，当我旋转屏幕时，框架会处理片段重新创建，适配器在我的onCreate中从主片段再次创建，但getItem永远不会被调用，所以我的适配器持有错误的引用（实际上是 null）而不是两个片段。

我发现片段管理器（即子片段管理器）包含一个名为 mActive 的片段数组，当然无法从代码访问。但是，有以下getFragment方法：

@Override
public Fragment getFragment(Bundle bundle, String key) {
    int index = bundle.getInt(key, -1);
    if (index == -1) {
        return null;
    }
    if (index >= mActive.size()) {
        throwException(new IllegalStateException("Fragement no longer exists for key "
                + key + ": index " + index));
    }
    Fragment f = mActive.get(index);
    if (f == null) {
        throwException(new IllegalStateException("Fragement no longer exists for key "
                + key + ": index " + index));
    }
    return f;
}

我不会评论错别字:)
这是我为了在我的适配器构造函数中更新对我的片段的引用而实现的黑客：

// fm holds a reference to a FragmentManager
Bundle hack = new Bundle();
try {
    for (int i = 0; i < mFragments.length; i++) {
        hack.putInt("hack", i);
        mFragments[i] = fm.getFragment(hack, "hack");
    }
} catch (Exception e) {
    // No need to fail here, likely because it's the first creation and mActive is empty
}

我并不骄傲。这行得通，但它很丑陋。屏幕旋转后拥有有效适配器的实际方法是什么？

PS：这是完整的代码

Answer 1

我遇到了同样的问题 - 我假设您正在为寻呼机适配器FragmentPagerAdapter子类化（因为getItem()特定于FragmentPagerAdapter）。

我的解决方案是PagerAdapter子类化并自己处理片段创建/删除（重新实现一些FragmentPagerAdapter代码）：

public class ListPagerAdapter extends PagerAdapter {
    FragmentManager fragmentManager;
    Fragment[] fragments;
    public ListPagerAdapter(FragmentManager fm){
        fragmentManager = fm;
        fragments = new Fragment[5];
    }
    @Override
    public void destroyItem(ViewGroup container, int position, Object object) {
        assert(0 <= position && position < fragments.length);
        FragmentTransaction trans = fragmentManager.beginTransaction();
        trans.remove(fragments[position]);
        trans.commit();
        fragments[position] = null;
}
    @Override
    public Fragment instantiateItem(ViewGroup container, int position){
        Fragment fragment = getItem(position);
        FragmentTransaction trans = fragmentManager.beginTransaction();
        trans.add(container.getId(),fragment,"fragment:"+position);
        trans.commit();
        return fragment;
    }
    @Override
    public int getCount() {
        return fragments.length;
    }
    @Override
    public boolean isViewFromObject(View view, Object fragment) {
        return ((Fragment) fragment).getView() == view;
    }
    public Fragment getItem(int position){
        assert(0 <= position && position < fragments.length);
        if(fragments[position] == null){
            fragments[position] = ; //make your fragment here
        }
        return fragments[position];
    }
}

希望这有帮助。

Answer 2

为什么上面的解决方案如此复杂？看起来矫枉过正。我只是通过将旧的引用替换为从 FragmentPagerAdapter 扩展的类中的新引用来解决它

 @Override
public Object instantiateItem(ViewGroup container, int position) {
    frags[position] = (Fragment) super.instantiateItem(container, position);
    return frags[position];
}

所有适配器的代码如下所示

public class RelationsFragmentsAdapter extends FragmentPagerAdapter {
private final String titles[] = new String[3];
private final Fragment frags[] = new Fragment[titles.length];
public RelationsFragmentsAdapter(FragmentManager fm) {
    super(fm);
    frags[0] = new FriendsFragment();
    frags[1] = new FriendsRequestFragment();
    frags[2] = new FriendsDeclinedFragment();
    Resources resources = AppController.getAppContext().getResources();
    titles[0] = resources.getString(R.string.my_friends);
    titles[1] = resources.getString(R.string.my_new_friends);
    titles[2] = resources.getString(R.string.followers);
}
@Override
public CharSequence getPageTitle(int position) {
    return titles[position];
}
@Override
public Fragment getItem(int position) {
    return frags[position];
}
@Override
public int getCount() {
    return frags.length;
}
@Override
public Object instantiateItem(ViewGroup container, int position) {
    frags[position] = (Fragment) super.instantiateItem(container, position);
    return frags[position];
}

}

Answer 3

我的答案有点类似于Joshua Hunt的答案，但是通过以finishUpdate方法提交事务，您可以获得更好的性能。一个事务，而不是每次更新两个事务。这是代码：

private class SuchPagerAdapter extends PagerAdapter{
    private final FragmentManager mFragmentManager;
    private SparseArray<Fragment> mFragments;
    private FragmentTransaction mCurTransaction;
    private SuchPagerAdapter(FragmentManager fragmentManager) {
        mFragmentManager = fragmentManager;
        mFragments = new SparseArray<>();
    }
    @Override
    public Object instantiateItem(ViewGroup container, int position) {
        Fragment fragment = getItem(position);
        if (mCurTransaction == null) {
            mCurTransaction = mFragmentManager.beginTransaction();
        }
        mCurTransaction.add(container.getId(),fragment,"fragment:"+position);
        return fragment;
    }
    @Override
    public void destroyItem(ViewGroup container, int position, Object object) {
        if (mCurTransaction == null) {
            mCurTransaction = mFragmentManager.beginTransaction();
        }
        mCurTransaction.detach(mFragments.get(position));
        mFragments.remove(position);
    }
    @Override
    public boolean isViewFromObject(View view, Object fragment) {
        return ((Fragment) fragment).getView() == view;
    }
    public Fragment getItem(int position) {         
        return YoursVeryFragment.instantiate();
    }
    @Override
    public void finishUpdate(ViewGroup container) {
        if (mCurTransaction != null) {
            mCurTransaction.commitAllowingStateLoss();
            mCurTransaction = null;
            mFragmentManager.executePendingTransactions();
        }
    }

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

Answer 4

问题是FragmentPageAdapter中的getItem()有一个错误的名称。它应该被命名为createItem().因为它的工作方式getItem()用于创建片段，并且调用它来查询/查找片段是不安全的。

我的建议是复制当前的FragmentPagerAdapter并以这种方式进行更改：

加：

    public abstract Fragment createFragment(int position);

并将 getItem 更改为：

public Fragment getItem(int position) {
    if(containerId!=null) {
        final long itemId = getItemId(position);
        String name = makeFragmentName(containerId, itemId);
        return mFragmentManager.findFragmentByTag(name);
    } else {
        return null;
    }
}

最后将其添加到实例化项：

    if(containerId==null)
        containerId = container.getId();
    else if(containerId!=container.getId())
        throw new RuntimeException("Container id not expected to change");

此要点的完整代码

我认为这种实现更安全，更易于使用，并且与Google工程师的原始适配器具有相同的性能。

Answer 5

我的代码：

    public class SampleAdapter extends FragmentStatePagerAdapter {
    private Fragment mFragmentAtPos2;
    private FragmentManager mFragmentManager;
    private Fragment[] mFragments = new Fragment[3];

    public SampleAdapter(FragmentManager mgr) {
        super(mgr);
        mFragmentManager = mgr;
        Bundle hack = new Bundle();
        try {
            for (int i = 0; i < mFragments.length; i++) {
                hack.putInt("hack", i);
                mFragments[i] = mFragmentManager.getFragment(hack, "hack");
            }
        } catch (Exception e) {
            // No need to fail here, likely because it's the first creation and mActive is empty
        }
    }
    public void switchFrag(Fragment frag) {
        if (frag == null) {
            Dbg.e(TAG, "- switch(frag) frag is NULL");
            return;
        } else Dbg.v(TAG, "- switch(frag) - frag is " + frag.getClass());
        // We have to check for mFragmentAtPos2 null in case of first time (only Mytrips fragment being instatiante).
        if (mFragmentAtPos2!= null) 
            mFragmentManager.beginTransaction()  
                .remove(mFragmentAtPos2)
                .commit();
        mFragmentAtPos2 = frag;
        notifyDataSetChanged();
    }
    @Override
    public int getCount() {
        return(3);
    }
    @Override
    public int getItemPosition(Object object) {
        Dbg.v(TAG,"getItemPosition : "+object.getClass());
        if (object instanceof MyTripsFragment
                || object instanceof FindingDriverFragment
                || object instanceof BookingAcceptedFragment
                || object instanceof RideStartedFragment
                || object instanceof RideEndedFragment
                || object instanceof ContactUsFragment
                )
            return POSITION_NONE;
        else return POSITION_UNCHANGED;
    }
    @Override
    public Fragment getItem(int position) {
        Dbg.v("SampleAdapter", "getItem called on: "+position);
        switch (position) {
        case 0: 
            if (snapbookFrag==null) {
                snapbookFrag = new SnapBookingFragment();
                Dbg.e(TAG, "snapbookFrag created");
            }
            return snapbookFrag;
        case 1: 
            if(bookingFormFrag==null) {
                bookingFormFrag = new BookingFormFragment();
                Dbg.e(TAG, "bookingFormFrag created");
            }
            return bookingFormFrag;
        case 2: 
            if (mFragmentAtPos2 == null) {
                myTripsFrag = new MyTripsFragment();
                mFragmentAtPos2 = myTripsFrag;
                return mFragmentAtPos2;
            }
            return mFragmentAtPos2;
        default:
            return(new SnapBookingFragment());
        }
    }
}

Answer 6

参考 simekadam 的解决方案，mFragments 没有填充在实例化项和需求中 mFragments.put(position, fragment);里面，否则你最终会遇到这个错误：试图从视图中删除片段会给我 mNextAnim 上的 NullPointerException。