Lru Cache listView 图片问题

Question

我的 ListView 在我的活动中包含图像，在我添加 LRU 缓存之前效果很好。当我第一次加载时（如果我不滚动它），而图像正在加载，第一个元素会在我正在加载的所有图像之间更改其图像。

我的 LRU 缓存看起来像

public class Image {
    private static LruCache<String, Bitmap>     mMemoryCache    = null;
    private static int                          cacheSize       = 1024 * 1024 * 10;

    private static class AsyncLoader extends AsyncTask<String, Void, Bitmap> {
        private ImageView   mTarget;

        public AsyncLoader(ImageView target) {
            this.mTarget = target;
        }

        @Override
        protected void onPreExecute() {
            mTarget.setTag(this);
        }

        @Override
        protected Bitmap doInBackground(String...urls) {
            String url = urls[0];

            Bitmap result = null;

            if (url != null) {
                result = load(url);

                if (result != null) {
                    mMemoryCache.put(url, result);
                }
            }

            return result;
        }

        @Override
        protected void onPostExecute(Bitmap result) {
            if (mTarget.getTag() == this) {
                mTarget.setTag(null);
                if (result != null) mTarget.setImageBitmap(result);
            }
        }
    }

    public static Bitmap load(String urlString) {
        if (urlString == null || urlString.length() == 0) return null;

        Bitmap bitmap = null;
        URL url = null;

        try {
            url = new URL(urlString);
        } catch (MalformedURLException e) {
            e.printStackTrace();
        }

        try {
            HttpURLConnection conn = (HttpURLConnection) url.openConnection();
            conn.setDoInput(true);
            conn.connect();
            InputStream is = conn.getInputStream();
            bitmap = BitmapFactory.decodeStream(is);
            is.close();
        } catch (IOException e) {
            e.printStackTrace();
        }

        return bitmap;
    }

    public static void loadToView(String url, ImageView view) {
        if (url == null || url.length() == 0) return;
        if (mMemoryCache == null) {
            mMemoryCache = new LruCache<String, Bitmap>(cacheSize) {
                @Override
                protected int sizeOf(String key, Bitmap bitmap) {
                    return (bitmap.getRowBytes() * bitmap.getHeight());
                }
            };
        }

        Bitmap bitmap = getBitmapFromMemCache(url);
        if (bitmap == null) {
            final AsyncLoader task = (AsyncLoader) new AsyncLoader(view);
            view.setTag(task);
            task.execute(url);
        } else {
            view.setImageBitmap(bitmap);
        }
    }
}

用法：

Image.loadToView("http://image_url", (ImageView) view.findViewById(R.id.img_thumbnail));

这就是我的 BaseAdapter getView 的样子

@Override
public View getView(final int position, View view, ViewGroup parent) {
    if (view == null) view = inflater.inflate(R.layout.list_item, null);

    ((TextView) view.findViewById(R.id.txt_name)).setText(filtered_data.get(position).name);

    Image.loadToView("http://graph.facebook.com/" + filtered_data.get(position).id + "/picture", (ImageView) view.findViewById(R.id.img_thumbnail));

    return view;
}

Answer 1

我发现阅读了很多关于 listview 问题的图片，它与 listview 回收有关，一些相关的帖子：

• Android Listview AsyncTask image problem
• Android listview image loading problem

我添加的是 onPostExecute 方法中的 else 语句，如果标签不是设置在标签上的那个我取消它。

它的固定代码

public class Image {
    private static LruCache<String, Bitmap>     mMemoryCache    = null;
    private static int                          cacheSize       = 1024 * 1024 * 10;

    private static class AsyncLoader extends AsyncTask<String, Void, Bitmap> {
        private ImageView   mTarget;

        public AsyncLoader(ImageView target) {
            this.mTarget = target;
        }

        @Override
        protected void onPreExecute() {
            mTarget.setTag(this);
        }

        @Override
        protected Bitmap doInBackground(String...urls) {
            String url = urls[0];

            Bitmap result = null;

            if (url != null) {
                result = load(url);

                if (result != null) {
                    mMemoryCache.put(url, result);
                }
            }

            return result;
        }

        @Override
        protected void onPostExecute(Bitmap result) {
            if (mTarget.getTag() == this) {
                mTarget.setTag(null);
                if (result != null) mTarget.setImageBitmap(result);
            } else if (mTarget.getTag() != null) {
                ((AsyncLoader) mTarget.getTag()).cancel(true);
                mTarget.setTag(null);
            }
        }
    }

    public static Bitmap load(String urlString) {
        if (urlString == null || urlString.length() == 0) return null;

        Bitmap bitmap = null;
        URL url = null;

        try {
            url = new URL(urlString);
        } catch (MalformedURLException e) {
            e.printStackTrace();
        }

        try {
            HttpURLConnection conn = (HttpURLConnection) url.openConnection();
            conn.setDoInput(true);
            conn.connect();
            InputStream is = conn.getInputStream();
            bitmap = BitmapFactory.decodeStream(is);
            is.close();
        } catch (IOException e) {
            e.printStackTrace();
        }

        return bitmap;
    }

    public static void loadToView(String url, ImageView view) {
        if (url == null || url.length() == 0) return;
        if (mMemoryCache == null) {
            mMemoryCache = new LruCache<String, Bitmap>(cacheSize) {
                @Override
                protected int sizeOf(String key, Bitmap bitmap) {
                    return (bitmap.getRowBytes() * bitmap.getHeight());
                }
            };
        }

        Bitmap bitmap = getBitmapFromMemCache(url);
        if (bitmap == null) {
            final AsyncLoader task = (AsyncLoader) new AsyncLoader(view);
            view.setTag(task);
            task.execute(url);
        } else {
            view.setImageBitmap(bitmap);
        }
    }
}

我认为 google 需要解决此问题或提供更简洁的解决方法，因为（在我看来）google 在构建某些东西时失败，开发人员只需要关心构建自己的应用程序而不是浪费时间寻找方法解决sdk问题