Android - 有效地从互联网上显示大型位图

时间:2012-07-10 18:22:51

标签: android android-layout android-intent android-emulator android-widget

鉴于图片网址,我需要下载并在ImageView中显示 一切正常,除了图像非常大的情况,并且抛出 OutOfMemoryException

希望Android文档提供此问题的解决方案:http://developer.android.com/training/displaying-bitmaps/load-bitmap.html

我尝试调整剪切的代码来接受InputStream,而不是资源 但是,似乎我在那里遗漏了一些东西,因为图像没有显示,也没有异常,但我在LogCat中看到: SkImageDecoder :: Factory返回null

以下是我如何解码图像并将其缩小(基于Android文档):

public static Bitmap decodeBitmapFromInputStream(InputStream inputStream,
            int reqWidth, int reqHeight) {
        BitmapFactory.Options options = new BitmapFactory.Options();
        options.inJustDecodeBounds = true;
        BitmapFactory.decodeStream(inputStream, null, options);

        options.inSampleSize = calculateInSampleSize(options, reqWidth, reqHeight);

        options.inJustDecodeBounds = false;
        return BitmapFactory.decodeStream(inputStream, null, options);
    }

    public static int calculateInSampleSize(BitmapFactory.Options options,
            int reqWidth, int reqHeight) {
        final int height = options.outHeight;
        final int width = options.outWidth;
        int inSampleSize = 1;

        if (height > reqHeight || width > reqWidth) {
            if (width > height) {
                inSampleSize = Math.round((float) height / (float) reqHeight);
            } else {
                inSampleSize = Math.round((float) width / (float) reqWidth);
            }
        }
        return inSampleSize;
    }

现在,在我的位图下载器方法中,我这样称呼它:

inputStream=entity.getContent();
return decodeBitmapFromInputStream(inputStream, 320, 480);

以下是完整的方法,以防万一:

static Bitmap downloadBitmap(String url){
        AndroidHttpClient client=AndroidHttpClient.newInstance("Android");
        HttpGet getRequest=new HttpGet(url);

        try{
            HttpResponse response=client.execute(getRequest);
            int statusCode=response.getStatusLine().getStatusCode();

            if(statusCode!=HttpStatus.SC_OK){
                Log.d("GREC", "Error "+statusCode+" while retrieving bitmap from "+url);
                return null;
            }

            HttpEntity entity=response.getEntity();
            if(entity!=null){
                InputStream inputStream=null;
                try{
                    inputStream=entity.getContent();
                    return decodeBitmapFromInputStream(inputStream, 320, 480);
                }catch (Exception e) {
                    Log.d("GREC", "Exception occured in BitmapDownloader");
                    e.printStackTrace();
                }
                finally{
                    if(inputStream!=null){
                        inputStream.close();
                    }
                    entity.consumeContent();
                }
            }
        }catch (Exception e) {
            getRequest.abort();
            Log.d("GREC", "Error while retriving bitmap from "+url+", "+e.toString());
        }finally{
            if(client!=null){
                client.close();
            }
        }
        return null;
    }

1 个答案:

答案 0 :(得分:0)

问题在于,一旦您使用了来自HttpUrlConnection的InputStream,就无法再次回放并再次使用相同的InputStream。因此,您必须为图像的实际采样创建新的InputStream。否则我们必须中止http请求。 见decodeStream returns null