By the way, what you're trying to do in your second method called getBitmapFromInputStream most likely cannot work. You are trying to read from the same stream twice, first of all for determining the bitmap size, the second time for decoding the actual bitmap. In between these two read accesses the stream should be reset to its beginning. However, this is not guaranteed to work. Some stream implementations are "one way only". Once it is read there is no way back.
On Monday, December 10, 2012 9:42:37 AM UTC-6, snowdream wrote:
Hi,everyone.--if i want to get a large image (eg.a image which size is 2M or larger.)from url, i use the folllowing code ,and get the exception ."Out of Memory".Any good idea?Thank you.private Bitmap getBitmapFromUrl(String url) {Bitmap bitmap = null;try {URL imagePath = new URL(url);HttpURLConnection conn = (HttpURLConnection)imagePath.openConnection(); conn.setConnectTimeout(CONNECT_TIMEOUT); conn.setReadTimeout(READ_TIMEOUT); conn.connect();InputStream is = (InputStream) conn.getContent();if (is != null) {// bitmap = getBitmapFromInputStream(is, width, height);bitmap = BitmapFactory.decodeStream((InputStream) conn.getContent()); }} catch(Exception e) {e.printStackTrace();}return bitmap;}public static Bitmap getBitmapFromInputStream(InputStream is, int width, int height){ Bitmap bitmap = null;if (null != is ) {BufferedInputStream bis = new BufferedInputStream(is);BitmapFactory.Options opts = null;if (width > 0 && height > 0) {opts = new BitmapFactory.Options();opts.inJustDecodeBounds = true;BitmapFactory.decodeStream(is, null, opts);// 计算图片缩放比例final int minSideLength = Math.min(width, height);opts.inSampleSize = BitmapUtils.computeSampleSize(opts.outWidth, opts.outHeight, minSideLength, width * height);opts.inJustDecodeBounds = false;opts.inInputShareable = true;opts.inPurgeable = true;}try {bitmap = BitmapFactory.decodeStream(bis, null, opts); } catch (OutOfMemoryError e) {e.printStackTrace();}}return bitmap;}--
杨辉
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