我有一个寻呼机适配器，可以假设代表日历的复杂视图。
每年的日历需要大约350毫秒的时间。
为了提高性能，我想实现回收视图的ListView数组适配器中存在的相同机制（convertView中的getView()参数）。
以下是来自适配器的当前getView()。
@Override
protected View getView(VerticalViewPager pager, final DateTileGrid currentDataItem, int position)
{
mInflater = LayoutInflater.from(pager.getContext());
// This is were i would like to understand weather is should use a recycled view or create a new one.
View datesGridView = mInflater.inflate(R.layout.fragment_dates_grid_page, pager, false);
DateTileGridView datesGrid = (DateTileGridView) datesGridView.findViewById(R.id.datesGridMainGrid);
TextView yearTitle = (TextView) datesGridView.findViewById(R.id.datesGridYearTextView);
yearTitle.setText(currentDataItem.getCurrentYear() + "");
DateTileView[] tiles = datesGrid.getTiles();
for (int i = 0; i < 12; i++)
{
String pictureCount = currentDataItem.getTile(i).getPictureCount().toString();
tiles[i].setCenterLabel(pictureCount);
final int finalI = i;
tiles[i].setOnCheckedChangeListener(new DateTileView.OnCheckedChangeListener() {
@Override
public void onCheckedChanged(DateTileView tileChecked, boolean isChecked)
{
DateTile tile = currentDataItem.getTile(finalI);
tile.isSelected(isChecked);
}
});
}
return datesGridView;
}
实现此类行为的任何指针或方向？
特别是如何在适配器中知道其中一个DateTileGridViews被刷过屏幕，以便我可以将其保存在内存中以便下次重复使用。
答案 0 :(得分：32)
所以我已经弄明白了。
destroyItem(ViewGroup container, int position, Object view)并保存您缓存的视图这是代码..我使用Stack of view来缓存我的寻呼机中所有已移除的视图
private View inflateOrRecycleView(Context context)
{
View viewToReturn;
mInflater = LayoutInflater.from(context);
if (mRecycledViewsList.isEmpty())
{
viewToReturn = mInflater.inflate(R.layout.fragment_dates_grid_page, null, false);
}
else
{
viewToReturn = mRecycledViewsList.pop();
Log.i(TAG,"Restored recycled view from cache "+ viewToReturn.hashCode());
}
return viewToReturn;
}
@Override
public void destroyItem(ViewGroup container, int position, Object view)
{
VerticalViewPager pager = (VerticalViewPager) container;
View recycledView = (View) view;
pager.removeView(recycledView);
mRecycledViewsList.push(recycledView);
Log.i(TAG,"Stored view in cache "+ recycledView.hashCode());
}
不要忘记在适配器构造函数中实例化堆栈。
答案 1 :(得分：3)
我这样做..首先创建抽象的softCache类：
public abstract class SoftCache<T> {
private Stack<Reference<T>> mRecyclingStack;
final Class<T> classType;
public SoftCache(Class<T> typeParameterClass) {
this.classType = typeParameterClass;
mRecyclingStack = new Stack<Reference<T>>();
}
/* implement this to create new object of type T if cache is empty */
public abstract T runWhenCacheEmpty();
/*
* retrieves last item from cache or creates a new T object if cache is
* empty
*/
public T get() {
T itemCached = null;
if (mRecyclingStack.isEmpty()) {
itemCached = runWhenCacheEmpty();
} else {
SoftReference<T> softRef = (SoftReference<T>) mRecyclingStack
.pop();
Object obj = softRef.get();
/*
* if referent object is empty(due to GC) then create a new
* object
*/
if (obj == null) {
itemCached = runWhenCacheEmpty();
}
/*
* otherwise restore from cache by casting the referent as the
* class Type that was passed to constructor
*/
else {
itemCached = (classType.cast(softRef.get()));
}
}
return itemCached;
}
现在从SoftCache继承，因此我们可以实现runWhenCacheEmpty方法：
public class ViewCache extends SoftCache<View>{
public ViewCache(Class<View> typeParameterClass) {
super(typeParameterClass);
}
@Override
public View runWhenCacheEmpty() {
return mFragment.getActivity().getLayoutInflater()
.inflate(R.layout.mypagelayout, null);
}
}
然后在你的构造函数中实例化它，如果你希望它是一个View类（但它可以适用于任何类型的类）：
SoftCache<View> myViewCache = new ViewCache(View.class);
现在在destroyItem中将视图保存到缓存中：
@Override
public void destroyItem(final ViewGroup container, final int position, final Object object) {
final View v = (View) object;
if(v.getId() == R.id.mypagelayout)
myViewCache.put(v); //this saves it
}
现在方法instantiateItem就像这样使用它：
@Override
public Object instantiateItem(final ViewGroup container, final int position) {
View MyPageView=myViewCache.get();
}
更新：如果您想将缓存用于不同的布局或者不想扩展它，我想出了一个解决方案，您可以在多个布局中使用相同的缓存，您可以使用布局ID检索放入的布局：
public class SoftViewCache {
private HashMap<Integer,ArrayList<SoftReference<View>>> multiMap;
public SoftViewCache() {
multiMap= new HashMap<Integer, ArrayList<SoftReference<View>>>();
}
/*
* retrieves cached item or return null if cache is
* empty
*/
public View get(int id) {
View itemCached = null;
if (!multiMap.containsKey(id)) {
return null;
}
else {
/*get the referent object and check if its already been GC if not we re-use*/
SoftReference<View> softRef =multiMap.get(id).get(0);
Object obj = softRef.get();
/*
* if referent object is empty(due to GC) then caller must create a new
* object
*/
if (null == obj) {
return null;
}
/*
* otherwise restore from cache
*/
else {
itemCached = (softRef.get());
}
}
return itemCached;
}
/* saves a view object to the cache by reference, we use a multiMap to allow
* duplicate IDs*/
public void put(View item) {
SoftReference<View> ref = new SoftReference<View>(item);
int key = item.getId();
/*check if we already have a reuseable layouts saved if so just add to the list
* of reusable layouts*/
if (multiMap.containsKey(key)) {
multiMap.get(key).add(ref);
} else {
/*otherwise we have no reusable layouts lets create a list of reusable layouts
* and add it to the multiMap*/
ArrayList<SoftReference<View>> list = new ArrayList<SoftReference<View>>();
list.add(ref);
multiMap.put(key, list);
}
}
}
答案 2 :(得分：3)
我通过定义RecycleCache来解决这个问题，就像这个
protected static class RecycleCache {
private final RecyclerPagerAdapter mAdapter;
private final ViewGroup mParent;
private final int mViewType;
private List<ViewHolder> mCaches;
public RecycleCache(RecyclerPagerAdapter adapter, ViewGroup parent, int viewType) {
mAdapter = adapter;
mParent = parent;
mViewType = viewType;
mCaches = new ArrayList<>();
}
public ViewHolder getFreeViewHolder() {
int i = 0;
ViewHolder viewHolder;
for (int n = mCaches.size(); i < n; i++) {
viewHolder = mCaches.get(i);
if (!viewHolder.mIsAttached) {
return viewHolder;
}
}
viewHolder = mAdapter.onCreateViewHolder(mParent, mViewType);
mCaches.add(viewHolder);
return viewHolder;
}
}
在此处查看我的示例代码RecyclerPagerAdapter