如何在 Scala 中将 InputStream 转换为 base64 字符串？
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How do you convert an InputStream to a base64 string in Scala?
提问by Dax Fohl
Trying to pull an image off of Amazon S3 (returns S3ObjectInputStream) and send it to the mandrill email api(takes a base64-encoded string). How can this be done in Scala?
尝试从 Amazon S3 中提取图像（返回S3ObjectInputStream）并将其发送到mandrill 电子邮件 API（采用 base64 编码的字符串）。这如何在 Scala 中完成？
回答by Dax Fohl
I also managed to do it just using the Apache commons; not sure which approach is better, but figured I'd leave this answer for the record:
我还设法仅使用 Apache 公共资源来做到这一点；不确定哪种方法更好，但我想我会把这个答案留作记录：
import org.apache.commons.codec.binary.Base64
import org.apache.commons.io.IOUtils
val bytes = IOUtils.toByteArray(stream)
val bytes64 = Base64.encodeBase64(bytes)
val content = new String(bytes64)
回答by Lomig Mégard
Here is one solution, there are probably others more efficient.
这是一种解决方案，可能还有其他更有效的解决方案。
val is = new ByteArrayInputStream(Array[Byte](1, 2, 3)) // replace by your InputStream
val stream = Stream.continually(is.read).takeWhile(_ != -1).map(_.toByte)
val bytes = stream.toArray
val b64 = new sun.misc.BASE64Encoder().encode(bytes)
You could (and should) also replace the sun.miscencoder by the apache commons Base64for a better compatibility.
您也可以（并且应该）用sun.miscapache commons Base64替换编码器以获得更好的兼容性。
val b64 = org.apache.commons.codec.binary.Base64.encodeBase64(bytes)
回答by Mark Lister
Here's a simple encoder/decoderI wrote that you can include as source. So, no external dependencies.
这是我编写的一个简单的编码器/解码器，您可以将其包含为源。所以，没有外部依赖。
The interface is a bit more scala-esque:
界面有点像 Scala 风格：
import io.github.marklister.base64.Base64._
// Same as Lomig Mégard's answer
val b64 = bytes.toBase64
import io.github.marklister.base64.Base64._
// Same as Lomig Mégard's answer
val b64 = bytes.toBase64
