在Scala中使用解析器組合器時，如何將任意數量的項目轉換為對象？

Question

我一直在與Scala的組合器和解析器一起玩，並且有一個問題可能太簡單了（如果是的話，很抱歉）。 我已經在這段代碼中寫了出來，以使問題易於理解，而我的問題就在結尾。

import scala.util.parsing.combinator._
// First, I create a case class
case class Human(fname: String, lname: String, age: Int)
// Now, I create a parser object
object SimpleParser extends RegexParsers {
 def fname: Parser[String] = """[A-Za-z]+""".r ^^ {_.toString}
 def lname: Parser[String] = """[A-Za-z]+""".r ^^ {_.toString}
 def age: Parser[Int] = """[1-9][0-9]{0,2}""".r ^^ {_.toInt}
 def person: Parser[Human] = fname ~ lname ~ age ^^ {case f ~ l ~ a => Human(f, l, a)}
// Now, I need to read a list of these, not just one. 
// How do I do this? Here is what I tried, but can't figure out what goes inside 
// the curly braces to the right of ^^
// def group: Parser[List[Human]] = rep(person) ^^ {}
// def group: Parser[List[Human]] = (person)+ ^^ {}
}

// Here is what I did to test things:
val s1 = "Bilbo Baggins 123"
val r = SimpleParser.parseAll(SimpleParser.person, s1)
println("First Name: " + r.get.fname)
println("Last Name: " + r.get.lname)
println("Age: " + r.get.age) 
// So that worked well; I could read these things into an object and examine the object,
// and can do things with the object now.
// But how do I read either of these into, say, a List[Human] or some collection?
val s2 = "Bilbo Baggins 123 Frodo Baggins 40 John Doe 22"
val s3 = "Bilbo Baggins 123; Frodo Baggins 40; John Doe 22"

如果有什么很明顯的我想念的地方，請告訴我。 謝謝！

Answer 1

你很親近 對於以空格分隔的版本， rep是您所需要的：

lazy val people: Parser[List[Human]] = rep(person)

對於帶有分號的版本，可以使用repsep ：

lazy val peopleWithSemicolons: Parser[List[Human]] = repsep(person, ";")

請注意，在兩種情況下rep*返回所需的結果-無需使用^^映射結果。 對於fname和lname也是這種情況，其中正則表達式將隱式轉換為Parser[String] ，這意味着_.toString映射_.toString不會更改任何內容。