作为ViewPager的一部分更新ListFragment中的数据

时间:2011-09-11 15:17:08

标签: android android-fragments android-viewpager android-adapter android-listfragment

我在Android中使用v4兼容性ViewPager。我的FragmentActivity有一堆数据,可以在我的ViewPager中的不同页面上以不同的方式显示。到目前为止,我只有3个相同ListFragment的实例,但将来我将有3个不同的ListFragments实例。 ViewPager在垂直电话屏幕上,列表不是并排的。

现在,ListFragment上的一个按钮启动一个单独的整页活动(通过FragmentActivity),它返回并且FragmentActivity修改数据,保存数据,然后尝试更新其ViewPager中的所有视图。就在这里,我被卡住了。

public class ProgressMainActivity extends FragmentActivity
{
    MyAdapter mAdapter;
    ViewPager mPager;

    @Override
    public void onCreate(Bundle savedInstanceState)
    {
    ...
        mAdapter = new MyAdapter(getSupportFragmentManager());

        mPager = (ViewPager) findViewById(R.id.viewpager);
        mPager.setAdapter(mAdapter);
    }

    @Override
    protected void onActivityResult(int requestCode, int resultCode, Intent data)
    {
        ...
        updateFragments();
        ...
    }
    public void updateFragments()
    {
        //Attempt 1:
        //mAdapter.notifyDataSetChanged();
        //mPager.setAdapter(mAdapter);

        //Attempt 2:
        //HomeListFragment fragment = (HomeListFragment) getSupportFragmentManager().findFragmentById(mAdapter.fragId[0]);
        //fragment.updateDisplay();
    }

    public static class MyAdapter extends FragmentPagerAdapter implements
         TitleProvider
    {
      int[] fragId = {0,0,0,0,0};
      public MyAdapter(FragmentManager fm)
      {
         super(fm);
      }
      @Override
      public String getTitle(int position){
         return titles[position];
      }
      @Override
      public int getCount(){
         return titles.length;
      }

      @Override
      public Fragment getItem(int position)
      {

         Fragment frag = HomeListFragment.newInstance(position);
         //Attempt 2:
         //fragId[position] = frag.getId();
         return frag;
      }

      @Override
      public int getItemPosition(Object object) {
         return POSITION_NONE; //To make notifyDataSetChanged() do something
     }
   }

    public class HomeListFragment extends ListFragment
    {
    ...
        public static HomeListFragment newInstance(int num)
        {
            HomeListFragment f = new HomeListFragment();
            ...
            return f;
        }
   ...

现在你可以看到,我的第一次尝试是在整个FragmentPagerAdapter上发送notifyDataSetChanged,这表明有时会更新数据,但是其他的我得到了IllegalStateException:在onSaveInstanceState之后无法执行此操作。

我的第二次尝试是在我的ListFragment中尝试调用更新函数,但getItem中的getId返回0.根据我尝试过的文档

  

使用从FragmentManager获取对片段的引用   findFragmentById()或findFragmentByTag()

但我不知道我的片段的标签或ID!我有一个用于ViewPager的android:id =“@ + id / viewpager”,以及一个用于ListFragment布局中的ListView的android:id =“@ android:id / list”,但我不认为这些是有用的。< / p>

那么,我怎么能: a)一次性安全地更新整个ViewPager(理想情况下将用户返回到之前所在的页面) - 用户可以看到视图更改。 或者最好是, b)在每个受影响的ListFragment中调用一个函数来手动更新ListView。

任何帮助都将被感激地接受!

10 个答案:

答案 0 :(得分:252)

Barkside的答案适用于FragmentPagerAdapter,但不适用于FragmentStatePagerAdapter,因为它没有在传递给FragmentManager的片段上设置标记。

使用FragmentStatePagerAdapter似乎我们可以使用instantiateItem(ViewGroup container, int position)调用。它返回对位置position处的片段的引用。如果FragmentStatePagerAdapter已经存在对相关片段的引用,instantiateItem只返回对该片段的引用,并且不会调用getItem()再次实例化它。

因此,假设我正在查看片段#50,并希望访问片段#49。由于它们很接近,#49很可能已经被实例化。所以,

ViewPager pager = findViewById(R.id.viewpager);
FragmentStatePagerAdapter a = (FragmentStatePagerAdapter) pager.getAdapter();
MyFragment f49 = (MyFragment) a.instantiateItem(pager, 49)

答案 1 :(得分:152)

好吧,我想我已经找到了一种方法来执行请求b)在我自己的问题中,所以我会分享其他人的利益。 ViewPager中的片段标记采用"android:switcher:VIEWPAGER_ID:INDEX"形式,其中VIEWPAGER_ID是XML布局中的R.id.viewpager,INDEX是viewpager中的位置。因此,如果位置已知(例如0),我可以在updateFragments()中执行:

      HomeListFragment fragment = 
          (HomeListFragment) getSupportFragmentManager().findFragmentByTag(
                       "android:switcher:"+R.id.viewpager+":0");
      if(fragment != null)  // could be null if not instantiated yet
      {
         if(fragment.getView() != null) 
         {
            // no need to call if fragment's onDestroyView() 
            //has since been called.
            fragment.updateDisplay(); // do what updates are required
         }
      }

我不知道这是否是一种有效的方式,但它会做到更好的建议。

答案 2 :(得分:57)

每次实例化片段时尝试记录标记。

public class MPagerAdapter extends FragmentPagerAdapter {
    private Map<Integer, String> mFragmentTags;
    private FragmentManager mFragmentManager;

    public MPagerAdapter(FragmentManager fm) {
        super(fm);
        mFragmentManager = fm;
        mFragmentTags = new HashMap<Integer, String>();
    }

    @Override
    public int getCount() {
        return 10;
    }

    @Override
    public Fragment getItem(int position) {
        return Fragment.instantiate(mContext, AFragment.class.getName(), null);
    }

    @Override
    public Object instantiateItem(ViewGroup container, int position) {
        Object obj = super.instantiateItem(container, position);
        if (obj instanceof Fragment) {
            // record the fragment tag here.
            Fragment f = (Fragment) obj;
            String tag = f.getTag();
            mFragmentTags.put(position, tag);
        }
        return obj;
    }

    public Fragment getFragment(int position) {
        String tag = mFragmentTags.get(position);
        if (tag == null)
            return null;
        return mFragmentManager.findFragmentByTag(tag);
    }
}

答案 3 :(得分:35)

如果您问我,下一页的第二个解决方案,跟踪所有“活动”片段页面更好:http://tamsler.blogspot.nl/2011/11/android-viewpager-and-fragments-part-ii.html

来自比德赛德的回答对我来说太难了。

您可以跟踪所有“活动”片段页面。在这种情况下,您将跟踪ViewPager使用的FragmentStatePagerAdapter中的片段页面。

private final SparseArray<Fragment> mPageReferences = new SparseArray<Fragment>();

public Fragment getItem(int index) {
    Fragment myFragment = MyFragment.newInstance();
    mPageReferences.put(index, myFragment);
    return myFragment;
}

为了避免保留对“非活动”片段页面的引用,我们需要实现FragmentStatePagerAdapter的destroyItem(...)方法:

public void destroyItem(View container, int position, Object object) {
    super.destroyItem(container, position, object);
    mPageReferences.remove(position);
}

...当您需要访问当前可见的页面时,请调用:

int index = mViewPager.getCurrentItem();
MyAdapter adapter = ((MyAdapter)mViewPager.getAdapter());
MyFragment fragment = adapter.getFragment(index);

... MyAdapter的getFragment(int)方法如下所示:

public MyFragment getFragment(int key) {
    return mPageReferences.get(key);
}

答案 4 :(得分:13)

好的,在上面的@barkside测试方法后,我无法使用我的应用程序。然后我记得IOSched2012 app也使用viewpager,这就是我找到解决方案的地方。它不使用任何片段ID或标签,因为这些不是viewpager以易于访问的方式存储的。

以下是IOSched应​​用HomeActivity的important parts。要特别注意评论,因为其中有关键。:

@Override
protected void onSaveInstanceState(Bundle outState) {
    super.onSaveInstanceState(outState);

    // Since the pager fragments don't have known tags or IDs, the only way to persist the
    // reference is to use putFragment/getFragment. Remember, we're not persisting the exact
    // Fragment instance. This mechanism simply gives us a way to persist access to the
    // 'current' fragment instance for the given fragment (which changes across orientation
    // changes).
    //
    // The outcome of all this is that the "Refresh" menu button refreshes the stream across
    // orientation changes.
    if (mSocialStreamFragment != null) {
        getSupportFragmentManager().putFragment(outState, "stream_fragment",
                mSocialStreamFragment);
    }
}

@Override
protected void onRestoreInstanceState(Bundle savedInstanceState) {
    super.onRestoreInstanceState(savedInstanceState);
    if (mSocialStreamFragment == null) {
        mSocialStreamFragment = (SocialStreamFragment) getSupportFragmentManager()
                .getFragment(savedInstanceState, "stream_fragment");
    }
}

并将Fragments的实例存储在FragmentPagerAdapter中,如下所示:

    private class HomePagerAdapter extends FragmentPagerAdapter {
    public HomePagerAdapter(FragmentManager fm) {
        super(fm);
    }

    @Override
    public Fragment getItem(int position) {
        switch (position) {
            case 0:
                return (mMyScheduleFragment = new MyScheduleFragment());

            case 1:
                return (mExploreFragment = new ExploreFragment());

            case 2:
                return (mSocialStreamFragment = new SocialStreamFragment());
        }
        return null;
    }

另外,请记住保护Fragment这样的电话:

    if (mSocialStreamFragment != null) {
        mSocialStreamFragment.refresh();
    }

答案 5 :(得分:2)

您可以复制FragmentPagerAdapter并修改一些源代码,添加getTag()方法

例如

public abstract class AppFragmentPagerAdapter extends PagerAdapter {
private static final String TAG = "FragmentPagerAdapter";
private static final boolean DEBUG = false;

private final FragmentManager mFragmentManager;
private FragmentTransaction mCurTransaction = null;
private Fragment mCurrentPrimaryItem = null;

public AppFragmentPagerAdapter(FragmentManager fm) {
    mFragmentManager = fm;
}


public abstract Fragment getItem(int position);

@Override
public void startUpdate(ViewGroup container) {
}

@Override
public Object instantiateItem(ViewGroup container, int position) {
    if (mCurTransaction == null) {
        mCurTransaction = mFragmentManager.beginTransaction();
    }

    final long itemId = getItemId(position);


    String name = getTag(position);
    Fragment fragment = mFragmentManager.findFragmentByTag(name);
    if (fragment != null) {
        if (DEBUG) Log.v(TAG, "Attaching item #" + itemId + ": f=" + fragment);
        mCurTransaction.attach(fragment);
    } else {
        fragment = getItem(position);
        if (DEBUG) Log.v(TAG, "Adding item #" + itemId + ": f=" + fragment);

        mCurTransaction.add(container.getId(), fragment,
                getTag(position));
    }
    if (fragment != mCurrentPrimaryItem) {
        fragment.setMenuVisibility(false);
        fragment.setUserVisibleHint(false);
    }

    return fragment;
}

@Override
public void destroyItem(ViewGroup container, int position, Object object) {
    if (mCurTransaction == null) {
        mCurTransaction = mFragmentManager.beginTransaction();
    }
    if (DEBUG) Log.v(TAG, "Detaching item #" + getItemId(position) + ": f=" + object
            + " v=" + ((Fragment) object).getView());
    mCurTransaction.detach((Fragment) object);
}

@Override
public void setPrimaryItem(ViewGroup container, int position, Object object) {
    Fragment fragment = (Fragment) object;
    if (fragment != mCurrentPrimaryItem) {
        if (mCurrentPrimaryItem != null) {
            mCurrentPrimaryItem.setMenuVisibility(false);
            mCurrentPrimaryItem.setUserVisibleHint(false);
        }
        if (fragment != null) {
            fragment.setMenuVisibility(true);
            fragment.setUserVisibleHint(true);
        }
        mCurrentPrimaryItem = fragment;
    }
}

@Override
public void finishUpdate(ViewGroup container) {
    if (mCurTransaction != null) {
        mCurTransaction.commitAllowingStateLoss();
        mCurTransaction = null;
        mFragmentManager.executePendingTransactions();
    }
}

@Override
public boolean isViewFromObject(View view, Object object) {
    return ((Fragment) object).getView() == view;
}

@Override
public Parcelable saveState() {
    return null;
}

@Override
public void restoreState(Parcelable state, ClassLoader loader) {
}


public long getItemId(int position) {
    return position;
}

private static String makeFragmentName(int viewId, long id) {
    return "android:switcher:" + viewId + ":" + id;
}

protected abstract String getTag(int position);
}

然后扩展它,覆盖这些抽象方法,不需要害怕Android组改变

FragmentPageAdapter未来的源代码

 class TimeLinePagerAdapter extends AppFragmentPagerAdapter {


    List<Fragment> list = new ArrayList<Fragment>();


    public TimeLinePagerAdapter(FragmentManager fm) {
        super(fm);
        list.add(new FriendsTimeLineFragment());
        list.add(new MentionsTimeLineFragment());
        list.add(new CommentsTimeLineFragment());
    }


    public Fragment getItem(int position) {
        return list.get(position);
    }

    @Override
    protected String getTag(int position) {
        List<String> tagList = new ArrayList<String>();
        tagList.add(FriendsTimeLineFragment.class.getName());
        tagList.add(MentionsTimeLineFragment.class.getName());
        tagList.add(CommentsTimeLineFragment.class.getName());
        return tagList.get(position);
    }


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


}

答案 6 :(得分:1)

或者,您可以像setPrimaryItem一样覆盖FragmentPagerAdapter方法:

public void setPrimaryItem(ViewGroup container, int position, Object object) { 
    if (mCurrentFragment != object) {
        mCurrentFragment = (Fragment) object; //Keep reference to object
        ((MyInterface)mCurrentFragment).viewDidAppear();//Or call a method on the fragment
    }

    super.setPrimaryItem(container, position, object);
}

public Fragment getCurrentFragment(){
    return mCurrentFragment;
}

答案 7 :(得分:1)

也没有问题:

页面片段布局中的某处:

<FrameLayout android:layout_width="0dp" android:layout_height="0dp" android:visibility="gone" android:id="@+id/fragment_reference">
     <View android:layout_width="0dp" android:layout_height="0dp" android:visibility="gone"/>
</FrameLayout>

在片段的onCreateView()中:

...
View root = inflater.inflate(R.layout.fragment_page, container, false);
ViewGroup ref = (ViewGroup)root.findViewById(R.id.fragment_reference);
ref.setTag(this);
ref.getChildAt(0).setTag("fragment:" + pageIndex);
return root;

以及从ViewPager返回Fragment的方法(如果存在):

public Fragment getFragment(int pageIndex) {        
        View w = mViewPager.findViewWithTag("fragment:" + pageIndex);
        if (w == null) return null;
        View r = (View) w.getParent();
        return (Fragment) r.getTag();
}

答案 8 :(得分:0)

我想提出我的方法,以防它可以帮助其他人:

这是我的寻呼机适配器:

 public class CustomPagerAdapter extends FragmentStatePagerAdapter{
    private Fragment[] fragments;

    public CustomPagerAdapter(FragmentManager fm) {
        super(fm);
        fragments = new Fragment[]{
                new FragmentA(),
                new FragmentB()
        };
    }

    @Override
    public Fragment getItem(int arg0) {
        return fragments[arg0];
    }

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

}

在我的活动中,我有:

public class MainActivity {
        private ViewPager view_pager;
        private CustomPagerAdapter adapter;


        @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        adapter = new CustomPagerAdapter(getSupportFragmentManager());
        view_pager = (ViewPager) findViewById(R.id.pager);
        view_pager.setAdapter(adapter);
        view_pager.setOnPageChangeListener(this);
          ...
         }

}

然后获取当前片段我所做的是:

int index = view_pager.getCurrentItem();
Fragment currentFragment = adapter.getItem(index);

答案 9 :(得分:0)

这是我的解决方案,因为我不需要跟踪我的标签,无论如何都需要刷新它们。

    Bundle b = new Bundle();
    b.putInt(Constants.SharedPreferenceKeys.NUM_QUERY_DAYS,numQueryDays);
    for(android.support.v4.app.Fragment f:getSupportFragmentManager().getFragments()){
        if(f instanceof HomeTermFragment){
            ((HomeTermFragment) f).restartLoader(b);
        }
    }