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我正在開發由3個List碎片組成的Android應用程序。我正在使用ViewPager
對象來讓用戶在片段之間滑動。我用碎片使用了相同的代碼。但當我改變它使用列表片段我得到一個錯誤GetItem(int)返回FragmentPagerAdapter中的Listfragment
the return type is incompatible with FragmentPagerAdapter.getItem(int)
我在這裏呆了幾個小時。請幫助..
我的片段活動
public class PageViewActivity extends FragmentActivity {
MyPageAdapter pageAdapter;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_page_view);
List<ListFragment> fragments = getFragments();
pageAdapter = new MyPageAdapter(getSupportFragmentManager(), fragments);
ViewPager pager = (ViewPager)findViewById(R.id.viewpager);
pager.setAdapter(pageAdapter);
}
private List<ListFragment> getFragments(){
List<ListFragment> fList = new ArrayList<ListFragment>();
fList.add(AllMsgFragment.newInstance("Fragment 1"));
fList.add(ErrorMsgFragment.newInstance("Fragment 2"));
fList.add(SuccessMsgFragment.newInstance("Fragment 3"));
return fList;
}
和我FragmentPageAdapter
class MyPageAdapter extends FragmentPagerAdapter {
private List<ListFragment> fragments;
public MyPageAdapter(FragmentManager fm, List<ListFragment> fragments) {
super(fm);
this.fragments = fragments;
}
@Override
public Fragment getItem(int position) {//The return type is incompatible with FragmentPagerAdapter.getItem(int)
return this.fragments.get(position);
}
@Override
public int getCount() {
return this.fragments.size();
}
@Override
public CharSequence getPageTitle(int position) {
return "Page #" + (position + 1);
}
}
在此先感謝
確保導入正確的類 「進口android.support.v4.app.ListFragment;」不要使用「import android.app.ListFragment;」或其他 – hieuxit