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。这个片段布局由一个包含多个片段（实际上是两个）的视图寻呼机组成。

当创建视图分页器时，它的适配器被创建，getItem被调用，我的子片段被创建。太棒了。

现在，当我旋转屏幕时，框架处理片段的重新创建，适配器从主片段在我的onCreate中再次创建，但是getItem从未被调用，因此我的适配器持有错误的引用（实际上为空），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，这当然是代码无法访问的。但是有一种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;
}

我不会对输入错误进行评论：）
这是我在适配器构造函数中为了更新对片段的引用而实施的一种攻击：

// 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。这很管用，但很难看。屏幕旋转后，使用有效适配器的实际方式是什么？

PS：这是完整的代码

Answer 1

为什么上面的解决方案如此复杂？看起来太过分了。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这个问题

 @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 2

我的答案有点类似于Joshua Hunt的答案，但是通过在finishUpdate方法中提交事务，您可以获得更好的性能。每次更新一个事务而不是两个事务。下面是代码：

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 3

我也有同样的问题-我假设您正在为您的寻呼机适配器子类化FragmentPagerAdapter（因为getItem（）特定于FragmentPagerAdapter）。

我的解决方案是将PagerAdapter子类化，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。