如何对字符串值格式的约束进行编码 [英] How to encode a constraint on the format of String values

问题描述

正如我经常观察到的以及我经常实现 name 属性的方式，就是将其简单地建模为 String.

As I frequently observe and how I often implement a name attribute, is to simply model it as String.

现在，如果名称必须遵循某种语法，即格式，该怎么办?在 Java 中，我可能会定义一个构造函数并检查其参数，例如:

What now, if the name has to follow a certain syntax, i.e. format? In Java I probably would define a constructor with a check on its arguments, something like:

public Name(str: String) {
    if (str == null) throw new IllegalArgumentException("Str must not be null.");
    if (!str.matches("name format expressed as regex")) throw new IllegalArgumentException("Str must match 'regex' but was " + str);
    this.str = str;
}

在 Scala 中，我想出了以下解决方案:

In Scala I came up with the following solution:

import StdDef.Str
import StdDef.Bol
import StdDef.?

import scala.util.parsing.combinator.RegexParsers

final case class Name private (pfx: ?[Str] = None, sfx: Str) {

  override def toString = pfx.mkString + sfx

}

object Name extends RegexParsers {

  implicit def apply(str: Str): Name = parseAll(syntax, str) match {
    case Success(res, _) => Name(res._1, res._2)
    case rej: NoSuccess => error(rej.toString)
  }

  lazy val syntax = (prefix ?) ~! suffix

  lazy val prefix = (("x" | "X") ~! hyph) ^^ { case a ~ b => a + b }

  lazy val suffix = alpha ~! (alpha | digit | hyph *) ^^ { case a ~ b => a + b.mkString }

  lazy val alpha: Parser[Str] = """\p{Alpha}""".r

  lazy val digit: Parser[Str] = """\p{Digit}""".r

  lazy val hyph: Parser[Str] = "-"

  override lazy val skipWhitespace = false

}

我的意图是:

1. 从其自然表示中组合一个 Name，即一个 String 值
2. 在构建时检查其自然表示是否形成有效的Name.
3. 除了通过工厂方法apply:(str:Str)Str之外，禁止任何其他构造.
4. 使构造从其自然表示隐含，例如val a: Name = "ISBN 978-0-9815316-4-9".
5. 根据语法元素将 Name 分解成各个部分.
6. 在消息中抛出错误，例如:

1. Compose a Name from its natural representation, i.e. a String value
2. Check whether its natural representation forms a valid Name at construction time.
3. Disallow any other construction than through the factory method apply:(str:Str)Str.
4. Make the construction from its natural representation implicit, e.g. val a: Name = "ISBN 978-0-9815316-4-9".
5. Decompose a Name into its parts according to its syntactical elements.
6. Have errors being thrown with messages, such as:

===

--
^
[1.3] error: string matching regex `\p{Alpha}' expected but end of source found

我想知道您想出了什么解决方案.

在对主题进行了更多思考之后，我目前正在采取以下方法.

After giving the topic some more thoughts, I am currently taking the following approach.

Token.scala:

Token.scala:

abstract class Token {
  val value: Str
}
object Token {
  def apply[A <: Token](ctor: Str => A, syntax: Regex) = (value: Str) => value match {
    case syntax() => ctor(value)
    case _ => error("Value must match '" + syntax + "' but was '" + value + "'.")
  }
}

Tokens.scala:

Tokens.scala:

final case class Group private (val value: Str) extends Token
final case class Name private (val value: Str) extends Token
trait Tokens {
  import foo.{ bar => outer }
  val Group = Token(outer.Group, """(?i)[a-z0-9-]++""".r)
  val Name = Token(outer.Name, """(?i)(?:x-)?+[a-z0-9-]++""".r)
}

推荐答案

鉴于您已经习惯在 Java 中使用正则表达式，然后尝试在 Scala 中使用解析器解决同样的问题似乎有些过分.

Given that you'd be comfortable using a regex in Java, it seems like overkill to then try and solve the same problem with a parser in Scala.

>

坚持你在这里所知道的，但添加一个 Scala 扭曲来清理解决方案.Scala 中的正则表达式还定义了提取器，允许在模式匹配中使用它们:

Stick with what you know here, but add a Scala twist to clean up the solution a bit. Regexes in Scala also define extractors, allowing them to be used in a pattern match:

//triple-quote to make escaping easier, the .r makes it a regex
//Note how the value breaks normal naming conventions and starts in uppercase
//This is to avoid backticks when pattern matching

val TestRegex = """xxyyzz""".r

class Name(str: String) {
  str match {
    case Null => throw new IllegalArgumentException("Str must not be null")
    case TestRegex => //do nothing
    case _ => throw new IllegalArgumentException(
      "Str must match 'regex' but was " + str)
  }
}

免责声明:我没有实际测试这段代码，它可能包含错别字

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