对于GridView中的第一个位置，多次调用getView

Question

我有一个带有此getView:的适配器

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;
}

我有两个问题：

1. 对于位置0，会多次调用getView(如果位图在LruCache中丢失，则会使用AsyncTask下载位图)。我有一个动画(从0-1开始的alpha)，它为那个位置重新启动了几次。

2. 因为我在循环使用视图，有时你只需几秒钟就可以看到旧的imageView内容。

//----

这里是缓存类(只有堆)：

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);
}
}
}

//----------------

}

谢谢！：)

Answer 1

我在来回滚动或随意调用notifyDataSetChanged()时看到过类似的行为。

作为对你现在正在做的事情的增强，我建议使用毕加索，因为除了动画中的渐变之外，它处理这种情况非常好。

getView()中的一行：

Picasso.with(context).load(urlToLoad).into(imageView);

请参阅：http://square.github.io/picasso/