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我的 ListView 在我的活动中有图像，在我添加 LRU 缓存之前效果很好。当我第一次加载（如果我不滚动它）同时加载图像时，第一个元素会在我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;
}
4

1 回答 1

3

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

我添加的是 onPostExecute 方法中的 else 语句，如果标签不是在标签上设置的标签，我将取消它。

其固定代码

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);
        }
    }
}

我认为谷歌需要解决这个问题或提供更清洁的解决方法，因为（在我看来）谷歌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修复的方法sdk 问题

于 2012-12-04T19:07:50.760 回答