我正在使用带有视图V1、V2、V3等的ViewPager,我尝试通过点击按钮来设置每个视图中使用的LinearLayout的可见性。通过这段代码,它将应用更改到下一个视图而不是当前视图。例如,当我在V5上时,当我点击它时,它会隐藏/显示V6上的对象。如果我从V6向后移动到V5,则它将在V4上应用更改。
这是代码:
这是代码:
public class FragmentStatePagerSupport extends FragmentActivity {
static final int NUM_ITEMS = 10;
MyAdapter mAdapter;
ViewPager mPager;
static int mNum;
private Button btn_zoom;
static LinearLayout LL_Head;
@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);
mPager.setCurrentItem(5);
btn_zoom = (Button)findViewById(R.id.btn_zoom);
btn_zoom.setOnClickListener(new OnClickListener(){
@Override
public void onClick(View v) {
if (LL_Head.getVisibility() == View.VISIBLE) {
LL_Head.setVisibility(View.GONE);
}else{
LL_Head.setVisibility(View.VISIBLE);
}
}
});
.
.
.
}
public static class MyAdapter extends FragmentStatePagerAdapter {
public MyAdapter(FragmentManager fm) {
super(fm);
}
@Override
public int getCount() {
return NUM_ITEMS;
}
@Override
public Fragment getItem(int position) {
return ArrayListFragment.newInstance(position);
}
}
public static class ArrayListFragment extends ListFragment {
static ArrayListFragment newInstance(int num) {
ArrayListFragment f = new ArrayListFragment();
Bundle args = new Bundle();
args.putInt("num", num);
f.setArguments(args);
return f;
}
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
mNum = getArguments() != null ? getArguments().getInt("num") : 1;
}
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
Bundle savedInstanceState) {
View v = inflater.inflate(R.layout.sura_vpager, container, false);
TextView tv1=(TextView) v.findViewById(R.id.txtHead);
tv1.setText("Fragment #" + mNum);
LL_Head = (LinearLayout)v.findViewById(R.id.LL_Head);
return v;
}
请提供建议,感谢。