Question

我有一个ListView，我正在使用支持库中的Palette。当使用BitmapFactory.decodeStream从URL生成位图时，这会引发异常（网络在UI线程上）或者可能非常昂贵。我如何使这个异步？我想不出任何有效的方法来做到这一点。什么是最好的方法？

    @Override
    public View getView(int position, View convertView, ViewGroup viewGroup) {
        final ViewHolder holder;
        if (convertView == null) {
            convertView = mInflater.inflate(R.layout.item_grid, null);
            holder = new ViewHolder();
            holder.image = (ImageView) convertView.findViewById(R.id.img_photo);
            holder.bg = (LinearLayout) convertView.findViewById(R.id.bg_title);
            holder.text = (TextView) convertView.findViewById(R.id.txt_title);
            convertView.setTag(holder);
        } else {
            holder = (ViewHolder) convertView.getTag();
        }

        Picasso.with(mContext)
                .load(mShows.get(position).poster)
                .into(holder.image);

        try {
            URL url = new URL(mShows.get(position).poster);
            Bitmap bitmap = BitmapFactory.decodeStream(
                    url.openConnection().getInputStream()); // Too expensive!!!
            Palette palette = Palette.generate(bitmap);

            holder.text.setText(mShows.get(position).title);
            holder.text.setTextColor(palette.getVibrantColor().getRgb());

            holder.bg.setAlpha(0.4f);
            holder.bg.setBackgroundColor(palette.getDarkMutedColor().getRgb());

        } catch (IOException e) {
            e.printStackTrace();
        }

        return convertView;
    }

Answer 1

你可以使用Picasso的into(...)和Callback参数来表示图片成功加载的时间：

Picasso.with(mContext)
            .load(mShows.get(position).poster)
            .into(holder.image, new Callback() {

    @Override public void onSuccess() {
        Bitmap bitmap = ((BitmapDrawable)holder.image.getDrawable()).getBitmap();
        // do your processing here....
    }

    @Override public void onError() {
        // reset your views to default colors, etc.
    }

});

Answer 2

尝试在工作线程中进行网络和位图解码，并在获取位图时进行异步通知，并且还需要位图缓存以避免重复解码工作。一个简单的实现是这样的：

Interface IBitmapCache {
    Bitmap get(String key);
    put(String key, Bitmap map);
}

Class YourAdapter {

    Class ViewHolder {
        //your other views
        //use url as key.
        String url;
    }

    IBitmapCache mCache;

    //you need to hold an instance of your listview here.
    WeakReference<ListView> mAttachedListView;

    View getView() {
        //... handle other things
        Bitmap bitmap = mCache.get(url);
        if (bitmap == null) {
            //retrieve bitmap asynchronous
        }
    }

    //callback when bitmap is retrieved
    void onBitmapRetrived(String url, Bitmap bitmap) {
        if (mAttachedListView.get() != null) {
            final ListView list = mAttachedListView.get();
            final int count = list.getLastVisiblePosition() - list.getFirstVisiblePosition + 1;
            for (int i = 0; i < count; i++) {
                View v = lv.getChildAt(i);
                if (v == null) {
                    continue;
                }
                ViewHolder holder = (ViewHolder) v.getTag();
                if (url.equals(holder.url)) {
                    //do your Palette related things here.
                    break;
                }
            }
        }
    }
}

你还应该做两件事，把解码后的Bitmap放到你的位图缓存中，设置回调你的工作线程并确保在UI线程上调用它，这可以通过Handler轻松实现。 / p>

使用Picasso在ListView Android中调色板

2 个答案: