Question

大家好，我遇到了如何在我们的SQLite数据库中保存图像的问题。图片位于特定网址，例如 - http://images.201bestbuy.com/images/small_137385013870957.jpg

现在我想在没有用户交互的情况下从此网址中保存我的sqlit数据库中的图像。为了解决这个问题，我使用了

public class ImageLoader {

    MemoryCache memoryCache = new MemoryCache();
    FileCache fileCache;
    private Map<ImageView, String> imageViews = Collections
        .synchronizedMap(new WeakHashMap<ImageView, String>());
    ExecutorService executorService;

    public ImageLoader(Context context) {
        fileCache = new FileCache(context);
        executorService = Executors.newFixedThreadPool(5);
    }

    final int stub_id = R.drawable.no_image;

    public void DisplayImage(String url, ImageView imageView) {
        imageViews.put(imageView, url);
        Bitmap bitmap = memoryCache.get(url);
        if (bitmap != null){
            imageView.setImageBitmap(bitmap);
        }
        else {
            queuePhoto(url, imageView);
            imageView.setImageResource(stub_id);
        }
    }

    private void queuePhoto(String url, ImageView imageView) {
        PhotoToLoad p = new PhotoToLoad(url, imageView);
        executorService.submit(new PhotosLoader(p));
    }

    private Bitmap getBitmap(String url) {
        File f = fileCache.getFile(url);

        // from SD cache
        Bitmap b = decodeFile(f);
        if (b != null)
            return b;

        // from web
        try {
            Bitmap bitmap = null;
            URL imageUrl = new URL(url);
            HttpURLConnection conn = (HttpURLConnection) imageUrl
                .openConnection();
            conn.setConnectTimeout(30000);
            conn.setReadTimeout(30000);
            conn.setInstanceFollowRedirects(true);
            InputStream is = conn.getInputStream();
            OutputStream os = new FileOutputStream(f);
            Utils.CopyStream(is, os);
            os.close();
            bitmap = decodeFile(f);

            return bitmap;
        } catch (Exception ex) {
            ex.printStackTrace();
            return null;
        }
    }

    // decodes image and scales it to reduce memory consumption
    private Bitmap decodeFile(File f) {
        try {
            // decode image size
            BitmapFactory.Options o = new BitmapFactory.Options();
            o.inJustDecodeBounds = true;
            BitmapFactory.decodeStream(new FileInputStream(f), null, o);
            // Find the correct scale value. It should be the power of 2.
            final int REQUIRED_SIZE = 70;
            int width_tmp = o.outWidth, height_tmp = o.outHeight;
            int scale = 1;
            while (true) {
                if (width_tmp / 2 < REQUIRED_SIZE
                    || height_tmp / 2 < REQUIRED_SIZE)
                    break;
                width_tmp /= 2;
                height_tmp /= 2;
                scale *= 2;
            }

            // decode with inSampleSize
            BitmapFactory.Options o2 = new BitmapFactory.Options();
            o2.inSampleSize = scale;
            return BitmapFactory.decodeStream(new FileInputStream(f), null, o2);
        } catch (FileNotFoundException e) {
    }
    return null;
}

// Task for the queue
private class PhotoToLoad {
    public String url;
    public ImageView imageView;

    public PhotoToLoad(String u, ImageView i) {
        url = u;
        imageView = i;
    }
}

class PhotosLoader implements Runnable {
    PhotoToLoad photoToLoad;

    PhotosLoader(PhotoToLoad photoToLoad) {
        this.photoToLoad = photoToLoad;
    }

    @Override
    public void run() {
        if (imageViewReused(photoToLoad))
            return;
        Bitmap bmp = getBitmap(photoToLoad.url);
        memoryCache.put(photoToLoad.url, bmp);
        if (imageViewReused(photoToLoad))
            return;
        BitmapDisplayer bd = new BitmapDisplayer(bmp, photoToLoad);
        Activity a = (Activity) photoToLoad.imageView.getContext();
        a.runOnUiThread(bd);
    }
}

boolean imageViewReused(PhotoToLoad photoToLoad) {
    String tag = imageViews.get(photoToLoad.imageView);
    if (tag == null || !tag.equals(photoToLoad.url))
        return true;
    return false;
}

// Used to display bitmap in the UI thread
class BitmapDisplayer implements Runnable {
    Bitmap bitmap;
    PhotoToLoad photoToLoad;

    public BitmapDisplayer(Bitmap b, PhotoToLoad p) {
        bitmap = b;
        photoToLoad = p;
    }

    public void run() {
        if (imageViewReused(photoToLoad))
            return;
        if (bitmap != null){
            photoToLoad.imageView.setImageBitmap(bitmap);

        }
        else
            photoToLoad.imageView.setImageResource(stub_id);
    }
}

public void clearCache() {
    memoryCache.clear();
    fileCache.clear();
}

}

现在问题是当我使用

时

    ContentValues userdetailValues = new ContentValues();
    userdetailValues.put("Key_ProductImage",imageView.setImageBitmap(bitmap));

当我使用这些代码在SQLite数据库中保存图像时。它显示了我的错误。 Put（）方法将两个参数都作为字符串，但setImageBitmap（位图）的返回类型为void。如果我的方法完全错误，请建议我正确的方法来解决这个问题。

Answer 1

我觉得这是存储图像的一种不好的方法，但提供了将图像存储到数据库的答案。

您可以使用以下代码将图像位图转换为字节

public static byte[] getBytes(Bitmap bitmap)
{
    ByteArrayOutputStream stream=new ByteArrayOutputStream();
    bitmap.compress(CompressFormat.JPEG,0, stream);
   return stream.toByteArray();
}

您可以按如下方式将数据库作为blob数据插入：

ContentValues cv=new ContentValues();
cv.put(data, image);
return db.insert(TABLE_NAME, null, cv);

此处插入时图像数据为BLOB类型。

Answer 2

将映像放在db中是一种糟糕的处理方式 - 你膨胀db并从中获取数据然后需要转换为正确的格式。而是将它存储到磁盘并将存储它的文件名放在db中。然后从磁盘加载它。

从服务器URL将图像保存在SQLite数据库中

2 个答案: