问题
i use the ViewPager for switching views with left/right swipe.
The ViewPager needs an Adapter, so I\'ve built this one:
public class ListViewPagerAdapter extends PagerAdapter {
protected static final String TAG = \"ListViewPagerAdapter\";
protected static final int NUM_VIEWS = 3;
protected final Activity mActivity;
public ListViewPagerAdapter(Activity activity) {
mActivity = activity;
}
@Override
public int getCount() {
return NUM_VIEWS;
}
@Override
public void startUpdate(View container) {}
@Override
public Object instantiateItem(View container, int position) {
// ViewPager
ViewPager viewPager = (ViewPager) container;
// Wird verwendet, um die Views aufzurufen
LayoutInflater layoutInflater = mActivity.getLayoutInflater();
// Standardmäßig ist news eingeblendet
View view = layoutInflater.inflate(R.layout.news_fragment, null);
// Falls sich die Position verändert, so verändert sich auch die View
if (position == 0) {
view = layoutInflater.inflate(R.layout.favorite_fragment, null);
} else if (position == 2) {
view = layoutInflater.inflate(R.layout.videos_fragment, null);
}
// View einblenden
viewPager.addView(view, 0);
return view;
}
@Override
public void destroyItem(View container, int position, Object object) {
// ViewPager
ViewPager viewPager = (ViewPager) container;
// View
View view = (View) object;
// View löschen
viewPager.removeView(view);
}
@Override
public void finishUpdate(View container) {}
@Override
public boolean isViewFromObject(View view, Object object) {
View _view = (View) object;
return view == _view;
}
@Override
public Parcelable saveState() {
return null;
}
@Override
public void restoreState(Parcelable state, ClassLoader loader) {}
}
Now, i want to get the current focused view by the viewpager. I tried getChildAt(x), but it does not work.
Are there some example, or do you have any idea how to get the current view?
Thanks
回答1:
It is possible to save the currently active object (View, Fragment, ...) by overriding PagerAdapter.setPrimaryItem method. For example:
private View mCurrentView;
@Override
public void setPrimaryItem(ViewGroup container, int position, Object object) {
mCurrentView = (View)object;
}
回答2:
You have to register a listener to your ViewPager :
pager.setOnPageChangeListener(new MyPageChangeListener());
You have to customize the listener by extending a stub listener:
private int focusedPage = 0;
private class MyPageChangeListener extends ViewPager.SimpleOnPageChangeListener {
@Override
public void onPageSelected(int position) {
focusedPage = position;
}
}
I found this by looking at the ViewPager.java source code in the compatibility library. I read that we can do more, for example catch onPageScrollStateChanged.
To build the adapter, I used the code on this blog post. You might want to have a look.
回答3:
You can add a tag to the created view in the instantiateItem method:
view.setTag(position);
Later you can access the current selected view by:
mPager.findViewWithTag(mPager.getCurrentItem());
回答4:
I recently needed to implement this exact solution. Here's the way I did it:
Map<Integer, Object> views = Maps.newHashMap();
@Override
public Object instantiateItem(View container, int position) {
/* Create and add your View here */
Object result = ???
views.put(position, result);
return result;
}
@Override
public void destroyItem(View container, int position, Object object) {
/** Remove your View here */
views.remove(position);
}
protected View findViewForPosition(int position) {
Object object = views.get(position);
if (object != null) {
for (int i = 0; i < getChildCount(); i++) {
View view = getChildAt(i);
if (isViewFromObject(view, object)) {
return view;
}
}
}
return null;
}
回答5:
Try this:
public View getCurrentView(ViewPager pager) {
for (int i = 0; i < pager.getChildCount(); i++) {
View child = pager.getChildAt(i);
if (child.getX() <= pager.getScrollX() + pager.getWidth() &&
child.getX() + child.getWidth() >= pager.getScrollX() + pager.getWidth()) {
return child;
}
}
return getChildAt(0);
}
回答6:
To get focused view I use this way:
When I inflate my view before add it on ViewPager I set tag to it. Then I check this tag.
private View getCurrentView()
{
for (int i = 0; i < pager.getChildCount(); i++)
{
View v = pager.getChildAt(i);
if (v != null)
{
if (v.getTag().equals(pageList.get(pager.getCurrentItem()).getTag())) return v;
// pageList is a list of pages that I pass to page adapter
}
}
return null;
}
回答7:
the given answers were not really suitable for me, they all have unwanted side effects. Either I have to add unnecessary variables (I like clean code) or I have to work around the Android framework itself.
My solution is based on reflection, which accesses the array list of all objects within the ViewPager and returns the current selected object in the ViewPager. It has almost no side effects (reflection is slow, subclassing of an existing class) and it keeps the code clean.
public class ViewPagerEx extends ViewPager {
private static final String TAG = ViewPagerEx.class.getSimpleName();
public ViewPagerEx(Context context, AttributeSet attrs) {
super(context, attrs);
}
public ViewPagerEx(Context context) {
super(context);
}
public Object getCurrentObject() {
try {
final Field itemsField = ViewPager.class.getDeclaredField("mItems");
itemsField.setAccessible(true);
final ArrayList<Object> items = (ArrayList<Object>) itemsField.get(this);
final int currentItemIndex = getCurrentItem();
if (currentItemIndex < 0 || currentItemIndex >= items.size()) {
return null;
}
final Object infoItem = items.get(getCurrentItem());
final Class<?> itemInfoClass = findItemInfoClass(ViewPager.class.getDeclaredClasses());
final Field objectField = itemInfoClass.getDeclaredField("object");
objectField.setAccessible(true);
return objectField.get(infoItem);
} catch (NoSuchFieldException e) {
Log.e(TAG, e.toString());
} catch (IllegalArgumentException e) {
Log.e(TAG, e.toString());
} catch (IllegalAccessException e) {
Log.e(TAG, e.toString());
}
return null;
}
private Class<?> findItemInfoClass(final Class<?>[] classes) throws IllegalArgumentException {
for (int i = 0; i < classes.length; i++) {
if (classes[i].getSimpleName().equals("ItemInfo")) {
return classes[i];
}
}
throw new IllegalArgumentException("cannot find class ItemInfo");
}
}
回答8:
Inside your FragmentStatePagerAdapter:
private View mCurrentView;
@Override
public void setPrimaryItem(ViewGroup container, int position, Object object) {
Fragment f = (Fragment) object;
mCurrentView = f.getView();
}
回答9:
You Can Use This Code
viewPager.addOnPageChangeListener(new ViewPager.OnPageChangeListener() {
@Override
public void onPageScrolled(int position, float positionOffset, int positionOffsetPixels) {
}
@Override
public void onPageSelected(int position) {
Toast.makeText(getContext(), String.valueOf(position), Toast.LENGTH_SHORT).show();
}
@Override
public void onPageScrollStateChanged(int state) {
}
});
回答10:
If you examine carefully, there are at most 3 views saved by ViewPager. You can easily get the current view by
view = MyActivity.mViewPager.getChildAt(1);
回答11:
Am I missing something? There's only ever 3 children inside the ViewGroup, so it boils down to:
int current = viewPager.getCurrentItem();
int childCount = viewPager.getChildCount();
// If there's a single page or we're at the beginning, return the first view
if (childCount == 1 || current == 0) {
return viewPager.getChildAt(0);
} else { //For any other case, we want the second child. This is either the last page or the page in the middle for any other case.
return viewPager.getChildAt(1);
}
来源：https://stackoverflow.com/questions/6807262/get-focused-view-from-viewpager