在scala中使用java.util.base64时使用非法的base64字符“a”

qhhrdooz  于 2021-07-06  发布在  Java
关注(0)|答案(2)|浏览(714)

假设我从github api调用到文件中有以下base64编码字符串:

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ICogICAgICAgICAgICBXaWRnZXQoCiAgICogICAgICAgICAgICAgIExpbmVH
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ICogICAgICAgICAgICAgICAgUHJvcGVydHkoYmFja2dyb3VuZF9jb2xvcixM
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dXQKfQ==

据我所知,这种编码是正确的,只包含base64编码的标准字符表。如果我在这里解码这个编码,我会得到一个正确的翻译。然而,我尝试了各种方法以编程方式对其进行解码,但还没有找到解决方案。
contentEncoded 是包含编码文件的字符串。我尝试了以下方法:

java.util.Base64.getDecoder.decode(contentEncoded)
java.util.Base64.getDecoder.decode(contentEncoded.getBytes)
java.util.Base64.getDecoder.decode(contentEncoded.getBytes(StandardCharsets.UTF_8))
java.util.Base64.getUrlDecoder.decode(contentEncoded))
java.util.Base64.getUrlDecoder.decode(contentEncoded.getBytes(StandardCharsets.UTF_8))
java.util.Base64.getMimeDecoder.decode(contentEncoded.replaceAll("\\n", "").replaceAll("\\r", ""))

但是,所有这些都会导致错误消息: java.lang.IllegalArgumentException: Illegal base64 character a .
我的问题是:我没有看到明显的东西吗?是否有一些隐藏的控制字符?有没有人有过类似的问题并且能够解决?

8dtrkrch

8dtrkrch1#

只要去掉断线就行了。

contentEncoded.replace("\n", "")
0tdrvxhp

0tdrvxhp2#

以下代码段正确地解码了编码:

val decodedWithMime = java.util.Base64.getMimeDecoder.decode(contentEncoded)
val convertedByteArray = decodedWithMime.map(_.toChar).mkString

正如评论所指出的,错误消息 Illegal Base64 character a 对应于换行符的十六进制值 \n . 使用mime解码器可以解码字符串,而无需事先删除换行符。

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