我有一个多行JSON文件,其中的记录包含编码为十六进制的特殊字符。以下是单个JSON记录的示例:
{\x22value\x22:\x22\xC4\xB1arines Bint\xC4\xB1\xC3\xA7 Ramu\xC3\xA7lar\x22}
此记录应为{"value":"ıarines Bintıç Ramuçlar"}
,例如'''字符被替换为相应的十六进制\ x22,其他特殊Unicode字符被替换为一个或两个十六进制(例如\ xC3 \ xA7编码ç等)。
我需要将类似的字符串转换为Scala中的常规Unicode字符串,因此在打印时它会生成{"value":"ıarines Bintıç Ramuçlar"}
而不使用十六进制。
在Python中,我可以使用一行代码轻松解码这些记录:
>>> a = "{\x22value\x22:\x22\xC4\xB1arines Bint\xC4\xB1\xC3\xA7 Ramu\xC3\xA7lar\x22}"
>>> a.decode("utf-8")
u'{"value":"\u0131arines Bint\u0131\xe7 Ramu\xe7lar"}'
>>> print a.decode("utf-8")
{"value":"ıarines Bintıç Ramuçlar"}
但是在Scala中我无法找到解码它的方法。我没有尝试像这样转换它:
scala> val a = """{\x22value\x22:\x22\xC4\xB1arines Bint\xC4\xB1\xC3\xA7 Ramu\xC3\xA7lar\x22}"""
scala> print(new String(a.getBytes(), "UTF-8"))
{\x22value\x22:\x22\xC4\xB1arines Bint\xC4\xB1\xC3\xA7 Ramu\xC3\xA7lar\x22}
我也试过了URLDecoder,就像我在解决方案中找到的类似问题一样(但是有URL):
scala> val a = """{\x22value\x22:\x22\xC4\xB1arines Bint\xC4\xB1\xC3\xA7 Ramu\xC3\xA7lar\x22}"""
scala> print(java.net.URLDecoder.decode(a.replace("\\x", "%"), "UTF-8"))
{"value":"ıarines Bintıç Ramuçlar"}
它为此示例生成了所需的结果,但对于通用文本字段似乎不安全,因为它旨在使用URL并且需要将字符串中的所有\x
替换为%
。
Scala有更好的方法来处理这个问题吗?
我是Scala的新手,感谢任何帮助
更新:
我使用javax.xml.bind.DatatypeConverter.parseHexBinary
制作了自定义解决方案。它现在有效,但看起来很麻烦而且一点也不优雅。我认为应该有一种更简单的方法来做到这一点。
以下是代码:
import javax.xml.bind.DatatypeConverter
import scala.annotation.tailrec
import scala.util.matching.Regex
def decodeHexChars(string: String): String = {
val regexHex: Regex = """\A\\[xX]([0-9a-fA-F]{1,2})(.*)""".r
def purgeBuffer(buffer: String, acc: List[Char]): List[Char] = {
if (buffer.isEmpty) acc
else new String(DatatypeConverter.parseHexBinary(buffer)).reverse.toList ::: acc
}
@tailrec
def traverse(s: String, acc: List[Char], buffer: String): String = s match {
case "" =>
val accUpdated = purgeBuffer(buffer, acc)
accUpdated.foldRight("")((str, b) => b + str)
case regexHex(chars, suffix) =>
traverse(suffix, acc, buffer + chars)
case _ =>
val accUpdated = purgeBuffer(buffer, acc)
traverse(s.tail, s.head :: accUpdated, "")
}
traverse(string, Nil, "")
}
答案 0 :(得分:0)
问题是编码确实特定于python(我认为)。这样的事情可能有用:
val s = """{\x22value\x22:\x22\xC4\xB1arines Bint\xC4\xB1\xC3\xA7 Ramu\xC3\xA7lar\x22}"""
"""\\x([A-F0-9]{2})""".r.replaceAllIn(s, (x: Regex.Match) =>
new String(BigInt(x.group(1), 16).toByteArray, "UTF-8")
)
答案 1 :(得分:0)
Each reactor.stop()
encodes one byte, like multiprocessing
encodes threading
and \x??
encodes \x22
. But in UTF-8 some characters are encoded using multiple bytes, so you need to transform "
to \x5C
symbol and so on.
\
is really nice, but it might eat your slashes. So, if you don't use groups (like \xC4\xB1
) in a replaced string, ı
is a recommended way to escape replaceAllIn
and \1
symbols.
quoteReplacement
P.S. Does anyone know the difference between \
and $
?