5
我有這個getView適配器:getView被稱爲多次爲第一位置的GridView
public View getView(int position, View convertView, ViewGroup parent) {
Log.d("getView gv", position+"");
NewsLine holder = null;
if (convertView == null) {
LayoutInflater inflater = ((Activity) context).getLayoutInflater();
convertView = inflater.inflate(R.layout.grid_entry, parent, false);
holder = new NewsLine();
holder.iv= (ImageView) convertView.findViewById(R.id.photo);
convertView.setTag(holder);
} else {
holder = (NewsLine) convertView.getTag();
}
NewsItem item=news.ITEMS.get(position);
//---------
if(item.imgurl!=null && item.imgurl.compareToIgnoreCase("null")!=0)
{
holder.iv.setVisibility(View.VISIBLE);
mMemoryCache.loadBitmap(item.imgurl, holder.iv,position);
}
else
holder.iv.setVisibility(View.INVISIBLE);
//-------------
return convertView;
}
我有兩個問題:
的getView被稱爲幾次位置0 (如果在LruCache中遺漏了位圖,則使用AsyncTask下載位圖)。我有一個動畫(從0-1的alpha)重新啓動幾次該位置。
因爲我正在回收視圖,有時您可以在幾分之一秒內看到舊的imageView內容。
// ----
這裏是緩存類(僅限堆):
public class SetImgAT extends LruCache<String, Bitmap> {
private static SetImgAT instance;
private Animation FadeInAnimation;
private SetImgAT(int size, Context context) {
super(size);
FadeInAnimation = AnimationUtils.loadAnimation(context, R.anim.fadein);
}
public static synchronized SetImgAT getInstance(int size, Context context) {
if (instance == null) {
instance = new SetImgAT(size, context);
}
return instance;
}
@Override
protected int sizeOf(String key, Bitmap value) {
return (value.getRowBytes() * value.getHeight());
}
public void loadBitmap(String url, ImageView imageView,int pos) {
Bitmap bitmap = instance.get(url.hashCode() + "");
if (bitmap != null) {
Log.d("ImageCache", "hit - "+url.hashCode()+"pos:"+pos);
imageView.setImageBitmap(bitmap);
imageView.invalidate();
} else {
Log.d("ImageCache", "miss");
BitmapWorkerTask task = new BitmapWorkerTask(imageView);
task.execute(url);
}
}
class BitmapWorkerTask extends AsyncTask<String, Void, Bitmap> {
ImageView mImageView;
public BitmapWorkerTask(ImageView imageView) {
mImageView = imageView;
}
@Override
protected Bitmap doInBackground(String... url) {
Bitmap Picture = null;
if (url[0] != null && url[0].compareToIgnoreCase("null") != 0) {
Log.d("GetBMP from", url[0]);
URL img_value = null;
try {
img_value = new URL(url[0]);
} catch (MalformedURLException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
try {
Picture = BitmapFactory.decodeStream(img_value
.openConnection().getInputStream());
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
if (Picture == null) {
Log.d("deb", "no bitmap");
} else {
Log.d("got deb", "got bitmap to "+url[0].hashCode());
instance.put(url[0].hashCode()+"", Picture);
}
}
return Picture;
}
@Override
protected void onPostExecute(Bitmap result) {
// super.onPostExecute(result);
if (result != null) {
Log.d("deb", "set bitmap");
mImageView.setImageBitmap(result);
//mImageView.startAnimation(FadeInAnimation);
}
}
}
// ------------- ---
}
謝謝! :)
對於問題2,嘗試調用'mMemoryCache.loadBitmap()''之前setVisibility()'防止舊圖像出現。 – Sam
「getView被調用多次位置0」 - 這是完全正常的行爲。沒有規定多少次getView()會被調用。總的來說,StackOverflow是針對編程問題的,並且您沒有提出任何問題。 – CommonsWare
山姆它沒有工作,似乎有更多的照顧,我要通過http://developer.android.com/training/displaying-bitmaps/cache-bitmap緩存位圖的樣本。 html。在CommonsWare謝謝你,我不認爲我需要把這個改爲一個問題,也許你可以試試。 – Misca