32

我有这个活动，其中包含一个片段。这个片段布局由一个带有多个片段（实际上是两个）的视图分页器组成。

创建视图寻呼机时，会创建它的适配器，getItem调用它并创建我的子片段。伟大的。

现在，当我旋转屏幕时，框架处理片段的重新创建，适配器在我onCreate的主片段中再次创建，但从getItem未被调用，因此我的适配器持有错误的引用（实际上是空值）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 的片段数组mActive，这当然不能从代码中访问。不过有这个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;
}

我不会评论错字:)
这是我为了更新对我的片段的引用而在我的适配器构造函数中实现的 hack：

// 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。这有效，但它很丑。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

4

6 回答 6

29

我遇到了同样的问题 - 我假设您正在FragmentPagerAdapter为您的寻呼机适配器进行子类化（getItem()具体到FragmentPagerAdapter）。

我的解决方案是改为子类PagerAdapter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];
    }
}

希望这可以帮助。

于 2014-02-02T22:40:10.827 回答
7

我的回答有点类似于 Joshua Hunt 的回答，但是通过在finishUpdate方法中提交事务，您可以获得更好的性能。每次更新一次交易而不是两次。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;
    }

}
于 2015-02-12T19:44:49.163 回答
7

为什么上面的解决方案如此复杂？看起来有点矫枉过正。我只是通过替换从 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];
}

}

于 2017-07-19T09:01:33.447 回答
0

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());
        }
    }
}
于 2013-11-07T14:46:53.597 回答
0

参考simekadam 的解决方案，mFragments 没有被填充到 instantiateItem 中并且需要 mFragments.put(position, fragment);在里面，否则你最终会遇到这个错误：Trying to remove fragment from view 在 mNextAnim 上给我 NullPointerException。

于 2015-11-03T18:44:39.437 回答
0

问题是getItem()inFragmentPageAdapter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()是创建片段，调用它来查询/查找片段是不安全的。

我的建议是制作当前 FragmentPagerAdapter 的副本并以这种方式进行更改：

添加：

    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instanceiteItem：

    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，并且与谷歌工程师的原始适配器具有相同的性能。

于 2016-10-18T11:07:43.020 回答