BitmapFactory.decodeStream(InputStream is)为Android上的非null InputStream返回null

Question

我正在开发一个Android应用程序,它的视图包含多个Gallerys.Gallerys(Bitmaps)的内容是互联网上的红色.

对于第一个库,一切正常,但在尝试下载第二个Gallery的第一个图像时,BitmapFactory.decodeStream(InputStream)返回null,而流是非null.

public void loadBitmap() throws IOException {

        for (int i = 0; i < images.size(); ++i) {
            URL ulrn = new URL(images.get(i).getThumbUrl());
            HttpURLConnection con = (HttpURLConnection) ulrn.openConnection();
            InputStream is = con.getInputStream();
            images.get(i).setImage(BitmapFactory.decodeStream(is));
            Log.i("MY_TAG", "Height: " + images.get(i).getImage().getHeight());
        }
}

所述getThumbUrl()返回图像的URL(例如http://mydomain.com/image.jpg)和它抛出一个NullPointerException在线路Log.i("MY_TAG", "Height: ... ) (images为ArrayList我的类,其保持URL和位图也包含对象).

谢谢你的建议!

Answer 1

我遇到过这个.尝试将BufferedHttpEntity与您的输入流一起使用.我发现这可以防止99.9%的问题从decodeStream获取静默空值.

也许不是很重要,但我可靠地使用org.apache.http.client.HttpClient而不是HttpURLConnection,如:

public static Bitmap decodeFromUrl(HttpClient client, URL url, Config bitmapCOnfig)
{
    HttpResponse response=null;
    Bitmap b=null;
    InputStream instream=null;

    BitmapFactory.Options decodeOptions = new BitmapFactory.Options();
    decodeOptions.inPreferredConfig = bitmapCOnfig;
    try
    {
    HttpGet request = new HttpGet(url.toURI());
        response = client.execute(request);
        if (response.getStatusLine().getStatusCode() != 200)
        {
            MyLogger.w("Bad response on " + url.toString());
            MyLogger.w ("http response: " + response.getStatusLine().toString());
            return null;
        }
        BufferedHttpEntity bufHttpEntity = new BufferedHttpEntity(response.getEntity());
        instream = bufHttpEntity.getContent();

        return BitmapFactory.decodeStream(instream, null, decodeOptions);
    }
    catch (Exception ex)
    {
        MyLogger.e("error decoding bitmap from:" + url, ex);
        if (response != null)
        {
            MyLogger.e("http status: " + response.getStatusLine().getStatusCode());
        }
        return null;
    }
    finally
    {
        if (instream != null)
        {
            try {
                instream.close();
            } catch (IOException e) {
                MyLogger.e("error closing stream", e);
            }
        }
    }
}