感谢您的帮助。

我想获得一个FileInputStream对象，用户从图片库中选择一个图像。这是android.provider.MediaStore.Images.Media.INTERNAL_CONTENT_URI返回android的URI

内容：//媒体/外部/图片/媒体/ 3

当我试图从该对象​​建立一个java URI对象，我得到异常的说明预计文件方案，URI一个IllegalArgumentException：内容：//媒体/外部/图片/媒体/ 3 而android的URI显示计划为内容

从未发现原来的问题的解决方案。但是，如果你想要在图片库的图像的字节流，这片code能做到这一点。

 位图位= Media.getBitmap（getContentResolver（），imageUri）;
ByteArrayOutputStream字节=新ByteArrayOutputStream（）;
bitmap.com preSS（Bitmap.Com pressFormat.JPEG，40个字节）;
ByteArrayInputStream进行的FileInputStream =新ByteArrayInputStream的（bytes.toByteArray（））;
 

解决方案

您可以在java URI的字符串基于构造的组合使用android的URI的toString方法

  android.net.URI AURI =新android.net.URI（什么都）;
 java.net.URI中的树里=新java.net.URI中（auri.toString（））;
 

Android的URI | Java的URI

Thanks for any help.

I am trying to get a FileInputStream object on an image that the user selects from the picture gallery. This is the android URI returned by android.provider.MediaStore.Images.Media.INTERNAL_CONTENT_URI

content://media/external/images/media/3

When I try to construct a java URI object from this object, I get an IllegalArgumentException with the exception description Expected file scheme in URI: content://media/external/images/media/3 whereas the android URI shows the scheme as content

Never found a solution for the original question. But if you want the byte stream of an image in the pictures gallery, this piece of code will do that.

Bitmap bitmap = Media.getBitmap(getContentResolver(), imageUri);
ByteArrayOutputStream bytes = new ByteArrayOutputStream();
bitmap.compress(Bitmap.CompressFormat.JPEG, 40, bytes);
ByteArrayInputStream fileInputStream = new ByteArrayInputStream(bytes.toByteArray());

You could use the toString method of the android URI in combination of the String based constructor of the java URI

 android.net.URI auri = new android.net.URI(what ever);
 java.net.URI juri = new java.net.URI(auri.toString());

Android URI | Java URI