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名单我有这样的代码:你如何解构在斯卡拉
val host:String = Play.configuration.getString("auth.ldap.directory.host").get
val port:java.lang.Integer = Play.configuration.getString("auth.ldap.directory.port").get.toInt
val userDNFormat:String = Play.configuration.getString("auth.ldap.userDNFormat").get
到我需要它添加了十多种配置选项,所以我希望将它重构的东西,如:
val params = Seq("auth.ldap.directory.host", "auth.ldap.directory.port", "auth.ldap.userDNFormat")
params.map(Play.configuration.getString) match {
case host~port~userDNFormat => foo(host, port, userDNFormat)
}
我做了那个代码。什么是适当的语法来做到这一点?在地图/匹配行我得到这个错误,我不明白:
error: type mismatch;
found : (String, Option[Set[String]]) => Option[String]
required: java.lang.String => ?