解析 – 轻松解析Key = Value对的字符串到Scala案例类
发布时间:2020-12-16 18:53:45 所属栏目:安全 来源:网络整理
导读:有没有办法轻松地将一串键值对解析成 scala案例类? 例如,来自以下字符串 "consumer_key=1234ABC,consumer_secret=12345ABC" 成 case class Auth( consumerKey:String,consumerSecret:String) 甚至 case class Auth(consumer_key:String,consumer_secret:Str
有没有办法轻松地将一串键值对解析成
scala案例类?
例如,来自以下字符串 "consumer_key=1234ABC,consumer_secret=12345ABC" 成 case class Auth( consumerKey:String,consumerSecret:String) 甚至 case class Auth(consumer_key:String,consumer_secret:String) 解决方法
您可以使用正则表达式和模式匹配:
scala> val R = "consumer_key=(.*),consumer_secret=(.*)".r R: scala.util.matching.Regex = consumer_key=(.*),consumer_secret=(.*) scala> "consumer_key=1234ABC,consumer_secret=12345ABC" match { | case R(k,v) => Auth(k,v) | } res0: Auth = Auth(1234ABC,12345ABC) 使用JavaTokenParsers进行更灵活的解析: import scala.util.parsing.combinator._ case class Auth( consumerKey: String,consumerSecret: Option[String]) class AuthParser extends JavaTokenParsers { def auth: Parser[Auth] = key ~ opt("," ~> secret) ^^ { case k ~ s => Auth(k,s)} def key: Parser[String] = value("consumer_key") def secret: Parser[String] = value("consumer_secret") def value(k: String): Parser[String] = k ~ "=" ~> "[^,]*".r def apply(s: String) = parseAll(auth,s) } 用法: scala> val p = new AuthParser p: AuthParser = AuthParser@433b9799 scala> p("consumer_key=1234ABC,consumer_secret=12345ABC").get res0: Auth = Auth(1234ABC,Some(12345ABC)) scala> p("consumer_key=1234ABC").get res1: Auth = Auth(1234ABC,None) (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |