我正在使用FragmentStatePagerAdapter
。 getItem(int position)
返回错误的位置。我有5个片段。这是我改变片段时的位置:
Fragment0 -> Fragment1: position = 2
Fragment1 -> Fragment2: position = 3
Fragment2 -> Fragment3: position = 4
Fragment3 -> Fragment4: getItem is not called!
Fragment4 -> Fragment3: position = 2
Fragment3 -> Fragment2: position = 1
Fragment2 -> Fragment1: position = 0
Fragment1 -> Fragment0: getItem is not called!
以下是我的适配器的代码:
import android.content.Context;
import android.support.v4.app.Fragment;
import android.support.v4.app.FragmentManager;
import android.support.v4.app.FragmentStatePagerAdapter;
public class AppFragmentPageAdapter extends FragmentStatePagerAdapter {
final int PAGE_COUNT = 5;
private String tabTitles[] = new String[] { "اخبار", "حقیقتسنج", "ویدیوها", "زندگینامه", "برنامهها" };
private Context context;
public AppFragmentPageAdapter(FragmentManager fm, Context context) {
super(fm);
this.context = context;
}
@Override
public int getCount() {
return PAGE_COUNT;
}
@Override
public Fragment getItem(int position) {
switch (position) {
case (0):
return NewsFragment.newInstance(position);
default:
return VideosFragment.newInstance(position);
}
}
@Override
public CharSequence getPageTitle(int position) {
return tabTitles[position];
}
}
答案 0 :(得分:4)
这是正常行为。默认情况下,FragmentStatePagerAdapter保留当前显示的片段及其邻居的链接。首先,适配器创建Fragment0和Fragment1。当您滑动到Fragment1时,他将创建Fragment2并为此调用getItem(2)。滑动到Fragment2后,适配器将破坏Fragment0并创建Fragment3。