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2024-08-21 07:00:03

在ViewPager中替换Fragment

我试图使用片段与一个ViewPager使用FragmentPagerAdapter。 我想要实现的是替换一个片段，位于ViewPager的第一页，与另一个。

寻呼机由两个页面组成。第一个是FirstPagerFragment，第二个是SecondPagerFragment。点击第一页的一个按钮。我想用NextFragment替换FirstPagerFragment。

下面是我的代码。

public class FragmentPagerActivity extends FragmentActivity {

    static final int NUM_ITEMS = 2;

    MyAdapter mAdapter;
    ViewPager mPager;

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

        mAdapter = new MyAdapter(getSupportFragmentManager());

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

    }


    /**
     * Pager Adapter
     */
    public static class MyAdapter extends FragmentPagerAdapter {
        public MyAdapter(FragmentManager fm) {
            super(fm);
        }

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

        @Override
        public Fragment getItem(int position) {

            if(position == 0) {
                return FirstPageFragment.newInstance();
            } else {
                return SecondPageFragment.newInstance();
            }

        }
    }


    /**
     * Second Page FRAGMENT
     */
    public static class SecondPageFragment extends Fragment {

        public static SecondPageFragment newInstance() {
            SecondPageFragment f = new SecondPageFragment();
            return f;
        }

        @Override
        public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
            //Log.d("DEBUG", "onCreateView");
            return inflater.inflate(R.layout.second, container, false);

        }
    }

    /**
     * FIRST PAGE FRAGMENT
     */
    public static class FirstPageFragment extends Fragment {

        Button button;

        public static FirstPageFragment newInstance() {
            FirstPageFragment f = new FirstPageFragment();
            return f;
        }

        @Override
        public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
            //Log.d("DEBUG", "onCreateView");
            View root = inflater.inflate(R.layout.first, container, false);
            button = (Button) root.findViewById(R.id.button);
            button.setOnClickListener(new OnClickListener() {

                @Override
                public void onClick(View v) {
                    FragmentTransaction trans = getFragmentManager().beginTransaction();
                                    trans.replace(R.id.first_fragment_root_id, NextFragment.newInstance());
                    trans.setTransition(FragmentTransaction.TRANSIT_FRAGMENT_OPEN);
                    trans.addToBackStack(null);
                    trans.commit();

                }

            });

            return root;
        }

        /**
     * Next Page FRAGMENT in the First Page
     */
    public static class NextFragment extends Fragment {

        public static NextFragment newInstance() {
            NextFragment f = new NextFragment();
            return f;
        }

        @Override
        public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
            //Log.d("DEBUG", "onCreateView");
            return inflater.inflate(R.layout.next, container, false);

        }
    }
}

.．.这里是XML文件

fragment_pager.xml

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
        android:orientation="vertical" android:padding="4dip"
        android:gravity="center_horizontal"
        android:layout_width="match_parent" android:layout_height="match_parent">

    <android.support.v4.view.ViewPager
            android:id="@+id/pager"
            android:layout_width="match_parent"
            android:layout_height="match_parent"
            android:layout_weight="1">
    </android.support.v4.view.ViewPager>

</LinearLayout>

first.xml

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
  android:id="@+id/first_fragment_root_id"
  android:orientation="vertical"
  android:layout_width="match_parent"
  android:layout_height="match_parent">

  <Button android:id="@+id/button"
     android:layout_width="wrap_content" android:layout_height="wrap_content"
     android:text="to next"/>

</LinearLayout>

现在的问题是……我应该使用哪个ID

trans.replace(R.id.first_fragment_root_id, NextFragment.newInstance());

?

如果我用r。id。first_fragment_root_id，替换工作，但Hierarchy Viewer显示一个奇怪的行为，如下所示。

一开始的情况是

更换后的情况是

正如你所看到的，有一些错误，我希望在替换片段后找到与第一张图片中相同的状态。

当前回答

我找到了一个简单的解决方案，即使你想在中间添加新的片段或替换当前片段，它也能很好地工作。在我的解决方案中，您应该重写getItemId()，它应该为每个片段返回唯一的id。不是默认的位置。

就是这样:

public class DynamicPagerAdapter extends FragmentPagerAdapter {

private ArrayList<Page> mPages = new ArrayList<Page>();
private ArrayList<Fragment> mFragments = new ArrayList<Fragment>();

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

public void replacePage(int position, Page page) {
    mPages.set(position, page);
    notifyDataSetChanged();
}

public void setPages(ArrayList<Page> pages) {
    mPages = pages;
    notifyDataSetChanged();
}

@Override
public Fragment getItem(int position) {
    if (mPages.get(position).mPageType == PageType.FIRST) {
        return FirstFragment.newInstance(mPages.get(position));
    } else {
        return SecondFragment.newInstance(mPages.get(position));
    }
}

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

@Override
public long getItemId(int position) {
    // return unique id
    return mPages.get(position).getId();
}

@Override
public Object instantiateItem(ViewGroup container, int position) {
    Fragment fragment = (Fragment) super.instantiateItem(container, position);
    while (mFragments.size() <= position) {
        mFragments.add(null);
    }
    mFragments.set(position, fragment);
    return fragment;
}

@Override
public void destroyItem(ViewGroup container, int position, Object object) {
    super.destroyItem(container, position, object);
    mFragments.set(position, null);
}

@Override
public int getItemPosition(Object object) {
    PagerFragment pagerFragment = (PagerFragment) object;
    Page page = pagerFragment.getPage();
    int position = mFragments.indexOf(pagerFragment);
    if (page.equals(mPages.get(position))) {
        return POSITION_UNCHANGED;
    } else {
        return POSITION_NONE;
    }
}
}

注意:在这个例子中，FirstFragment和SecondFragment扩展了抽象类PageFragment，它有方法getPage()。

2014-04-22 11:39:01

其他回答

基于@wize的回答，我发现它很有用，也很优雅，我可以部分实现我想要的，因为我想要电缆在被替换后回到第一个片段。我稍微修改了一下他的代码就实现了。

这将是FragmentPagerAdapter:

public static class MyAdapter extends FragmentPagerAdapter {
    private final class CalendarPageListener implements
            CalendarPageFragmentListener {
        public void onSwitchToNextFragment() {
            mFragmentManager.beginTransaction().remove(mFragmentAtPos0)
                    .commit();
            if (mFragmentAtPos0 instanceof FirstFragment){
                mFragmentAtPos0 = NextFragment.newInstance(listener);
            }else{ // Instance of NextFragment
                mFragmentAtPos0 = FirstFragment.newInstance(listener);
            }
            notifyDataSetChanged();
        }
    }

    CalendarPageListener listener = new CalendarPageListener();;
    private Fragment mFragmentAtPos0;
    private FragmentManager mFragmentManager;

    public MyAdapter(FragmentManager fm) {
        super(fm);
        mFragmentManager = fm;
    }

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

    @Override
    public int getItemPosition(Object object) {
        if (object instanceof FirstFragment && mFragmentAtPos0 instanceof NextFragment)
            return POSITION_NONE;
        if (object instanceof NextFragment && mFragmentAtPos0 instanceof FirstFragment)
            return POSITION_NONE;
        return POSITION_UNCHANGED;
    }

    @Override
    public Fragment getItem(int position) {
        if (position == 0)
            return Portada.newInstance();
        if (position == 1) { // Position where you want to replace fragments
            if (mFragmentAtPos0 == null) {
                mFragmentAtPos0 = FirstFragment.newInstance(listener);
            }
            return mFragmentAtPos0;
        }
        if (position == 2)
            return Clasificacion.newInstance();
        if (position == 3)
            return Informacion.newInstance();

        return null;
    }
}

public interface CalendarPageFragmentListener {
    void onSwitchToNextFragment();
}

要执行替换，只需定义一个静态字段，类型为CalendarPageFragmentListener，并通过相应片段的newInstance方法初始化，并分别调用firstfragment . pagelister . onswitchtonextfragment()或nextfragment . pagelister . onswitchtonextfragment()。

2012-08-28 08:39:34

tl;dr:使用一个主机片段，负责替换其托管内容，并跟踪回溯导航历史(如在浏览器中)。

由于您的用例由固定数量的选项卡组成，我的解决方案工作得很好:想法是用自定义类HostFragment的实例填充ViewPager，这能够替换其托管内容并保持自己的回溯导航历史。要替换托管的片段，你需要调用hostfragment.replaceFragment()方法:

public void replaceFragment(Fragment fragment, boolean addToBackstack) {
    if (addToBackstack) {
        getChildFragmentManager().beginTransaction().replace(R.id.hosted_fragment, fragment).addToBackStack(null).commit();
    } else {
        getChildFragmentManager().beginTransaction().replace(R.id.hosted_fragment, fragment).commit();
    }
}

该方法所做的就是用id R.id替换框架布局。Hosted_fragment和提供给方法的片段。

查看我关于这个主题的教程，了解更多细节和GitHub上完整的工作示例!

2016-01-03 17:41:01

从2012年11月13日起，在ViewPager中重新调整片段似乎变得容易多了。谷歌发布的Android 4.2支持嵌套片段，在新的Android支持库v11中也支持它，所以这将一直工作到1.6

It's very similiar to the normal way of replacing a fragment except you use getChildFragmentManager. It seems to work except the nested fragment backstack isn't popped when the user clicks the back button. As per the solution in that linked question, you need to manually call the popBackStackImmediate() on the child manager of the fragment. So you need to override onBackPressed() of the ViewPager activity where you'll get the current fragment of the ViewPager and call getChildFragmentManager().popBackStackImmediate() on it.

获取当前正在显示的片段也有点棘手，我使用了这个肮脏的“android:switcher:VIEWPAGER_ID:INDEX”解决方案，但你也可以跟踪ViewPager的所有片段，正如本页上第二个解决方案所解释的那样。

这是我的代码，当用户单击一行时，ViewPager会显示一个详细的视图，后退按钮工作。为了简洁起见，我尝试只包括相关代码，所以如果你想要完整的应用程序上传到GitHub，请留下评论。

HomeActivity.java

 public class HomeActivity extends SherlockFragmentActivity {
FragmentAdapter mAdapter;
ViewPager mPager;
TabPageIndicator mIndicator;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    mAdapter = new FragmentAdapter(getSupportFragmentManager());
    mPager = (ViewPager)findViewById(R.id.pager);
    mPager.setAdapter(mAdapter);
    mIndicator = (TabPageIndicator)findViewById(R.id.indicator);
    mIndicator.setViewPager(mPager);
}

// This the important bit to make sure the back button works when you're nesting fragments. Very hacky, all it takes is some Google engineer to change that ViewPager view tag to break this in a future Android update.
@Override
public void onBackPressed() {
    Fragment fragment = (Fragment) getSupportFragmentManager().findFragmentByTag("android:switcher:" + R.id.pager + ":"+mPager.getCurrentItem());
    if (fragment != null) // could be null if not instantiated yet
    {
        if (fragment.getView() != null) {
            // Pop the backstack on the ChildManager if there is any. If not, close this activity as normal.
            if (!fragment.getChildFragmentManager().popBackStackImmediate()) {
                finish();
            }
        }
    }
}

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

    @Override
    public Fragment getItem(int position) {
        switch (position) {
        case 0:
            return ListProductsFragment.newInstance();
        case 1:
            return ListActiveSubstancesFragment.newInstance();
        case 2:
            return ListProductFunctionsFragment.newInstance();
        case 3:
            return ListCropsFragment.newInstance();
        default:
            return null;
        }
    }

    @Override
    public int getCount() {
        return 4;
    }

 }
}

ListProductsFragment.java

public class ListProductsFragment extends SherlockFragment {
private ListView list;

public static ListProductsFragment newInstance() {
    ListProductsFragment f = new ListProductsFragment();
    return f;
}

@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
    View V = inflater.inflate(R.layout.list, container, false);
    list = (ListView)V.findViewById(android.R.id.list);
    list.setOnItemClickListener(new OnItemClickListener() {
        public void onItemClick(AdapterView<?> parent, View view,
            int position, long id) {
          // This is important bit
          Fragment productDetailFragment = FragmentProductDetail.newInstance();
          FragmentTransaction transaction = getChildFragmentManager().beginTransaction();
          transaction.addToBackStack(null);
          transaction.replace(R.id.products_list_linear, productDetailFragment).commit();
        }
      });       
    return V;
}
}
2012-12-18 01:47:49

我按照@wize和@mdelolmo的答案，我得到了解决方案。由于吨。但是，我稍微调整了这些解决方案，以提高内存消耗。

我观察到的问题:

它们保存被替换的Fragment实例。在我的情况下，它是一个持有MapView的片段，我认为它的成本很高。所以，我维护FragmentPagerPositionChanged (POSITION_NONE或POSITION_UNCHANGED)而不是Fragment本身。

这是我的实现。

  public static class DemoCollectionPagerAdapter extends FragmentStatePagerAdapter {

    private SwitchFragListener mSwitchFragListener;
    private Switch mToggle;
    private int pagerAdapterPosChanged = POSITION_UNCHANGED;
    private static final int TOGGLE_ENABLE_POS = 2;


    public DemoCollectionPagerAdapter(FragmentManager fm, Switch toggle) {
        super(fm);
        mToggle = toggle;

        mSwitchFragListener = new SwitchFragListener();
        mToggle.setOnCheckedChangeListener(new CompoundButton.OnCheckedChangeListener() {
            @Override
            public void onCheckedChanged(CompoundButton buttonView, boolean isChecked) {
                mSwitchFragListener.onSwitchToNextFragment();
            }
        });
    }

    @Override
    public Fragment getItem(int i) {
        switch (i)
        {
            case TOGGLE_ENABLE_POS:
                if(mToggle.isChecked())
                {
                    return TabReplaceFragment.getInstance();
                }else
                {
                    return DemoTab2Fragment.getInstance(i);
                }

            default:
                return DemoTabFragment.getInstance(i);
        }
    }

    @Override
    public int getCount() {
        return 5;
    }

    @Override
    public CharSequence getPageTitle(int position) {
        return "Tab " + (position + 1);
    }

    @Override
    public int getItemPosition(Object object) {

        //  This check make sures getItem() is called only for the required Fragment
        if (object instanceof TabReplaceFragment
                ||  object instanceof DemoTab2Fragment)
            return pagerAdapterPosChanged;

        return POSITION_UNCHANGED;
    }

    /**
     * Switch fragments Interface implementation
     */
    private final class SwitchFragListener implements
            SwitchFragInterface {

        SwitchFragListener() {}

        public void onSwitchToNextFragment() {

            pagerAdapterPosChanged = POSITION_NONE;
            notifyDataSetChanged();
        }
    }

    /**
     * Interface to switch frags
     */
    private interface SwitchFragInterface{
        void onSwitchToNextFragment();
    }
}

演示链接在这里..https://youtu.be/l_62uhKkLyM

出于演示目的，在位置2处使用了两个片段TabReplaceFragment和DemoTab2Fragment。在所有其他情况下，我使用DemoTabFragment实例。

解释:

我将Switch从Activity传递到DemoCollectionPagerAdapter。基于这个开关的状态，我们将显示正确的片段。当开关检查被更改时，我调用SwitchFragListener的onSwitchToNextFragment方法，其中我将pagerAdapterPosChanged变量的值更改为POSITION_NONE。查看更多关于POSITION_NONE的信息。这将使getItem无效，我有逻辑实例化正确的片段在那里。抱歉，如果解释得有点乱的话。

再次感谢@wize和@mdelolmo的原创想法。

希望这对你有帮助。：）

如果这个实现有任何缺陷，请告诉我。那将对我的项目大有帮助。

2016-06-22 05:16:54

我找到了一个简单的解决方案，即使你想在中间添加新的片段或替换当前片段，它也能很好地工作。在我的解决方案中，您应该重写getItemId()，它应该为每个片段返回唯一的id。不是默认的位置。

就是这样:

public class DynamicPagerAdapter extends FragmentPagerAdapter {

private ArrayList<Page> mPages = new ArrayList<Page>();
private ArrayList<Fragment> mFragments = new ArrayList<Fragment>();

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

public void replacePage(int position, Page page) {
    mPages.set(position, page);
    notifyDataSetChanged();
}

public void setPages(ArrayList<Page> pages) {
    mPages = pages;
    notifyDataSetChanged();
}

@Override
public Fragment getItem(int position) {
    if (mPages.get(position).mPageType == PageType.FIRST) {
        return FirstFragment.newInstance(mPages.get(position));
    } else {
        return SecondFragment.newInstance(mPages.get(position));
    }
}

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

@Override
public long getItemId(int position) {
    // return unique id
    return mPages.get(position).getId();
}

@Override
public Object instantiateItem(ViewGroup container, int position) {
    Fragment fragment = (Fragment) super.instantiateItem(container, position);
    while (mFragments.size() <= position) {
        mFragments.add(null);
    }
    mFragments.set(position, fragment);
    return fragment;
}

@Override
public void destroyItem(ViewGroup container, int position, Object object) {
    super.destroyItem(container, position, object);
    mFragments.set(position, null);
}

@Override
public int getItemPosition(Object object) {
    PagerFragment pagerFragment = (PagerFragment) object;
    Page page = pagerFragment.getPage();
    int position = mFragments.indexOf(pagerFragment);
    if (page.equals(mPages.get(position))) {
        return POSITION_UNCHANGED;
    } else {
        return POSITION_NONE;
    }
}
}

注意:在这个例子中，FirstFragment和SecondFragment扩展了抽象类PageFragment，它有方法getPage()。

2014-04-22 11:39:01

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