IP地址的Scala正则表达式模式匹配 [英] Scala regex pattern match of ip address

问题描述

我不明白为什么这段代码返回 false:

I can't understand why this code returns false:

      val reg = """.*(\d{1,3})\.(\d{1,3})\.(\d{1,3})\.(\d{1,3}).*""".r
      "ttt20.30.4.140ttt" match{
        case reg(one, two, three, four) =>
          if (host == one + "." + two + "." + three + "." + four) true else false
        case _ => false
      }

并且仅当我将其更改为:

and only if I change it to:

  val reg = """.*(\d{1,3})\.(\d{1,3})\.(\d{1,3})\.(\d{1,3}).*""".r
  "20.30.4.140" match{
    case reg(one, two, three, four) =>
      if (host == one + "." + two + "." + three + "." + four) true else false
    case _ => false
  }

确实匹配

推荐答案

你的变体

def main( args: Array[String] ) : Unit = {
  val regex = """.*(\d{1,3})\.(\d{1,3})\.(\d{1,3})\.(\d{1,3}).*""".r
  val x = "ttt20.30.4.140ttt"

  x match {
    case regex(ip1,ip2,ip3,ip4) => println(ip1, ip2, ip3, ip4)
    case _ => println("No match.")
  }
}

匹配，但不是你想要的.结果将是 (0,30,4,140) 而不是 (20,30,4,140).正如你所看到的 .* 是贪婪的，所以尽可能多地消耗输入.

matches, but not as you intend. Result will be (0,30,4,140) instead of (20,30,4,140). As you can see .* is greedy, so consumes as much input as it can.

例如ab12 可以通过 .*(\d{1,3}) 分隔成

e.g. ab12 could be separated via .*(\d{1,3}) into

• ab 和 12
• ab1 和 2 .... 这是选择的变体，因为 .* 消耗尽可能多的输入

• ab and 12
• ab1 and 2 .... this is the variant chosen, as .* consumes as much input as it can

1. 让 .* 不情愿(而不是贪婪)，也就是 .*? 所以总的来说

1. Make .* reluctant (and not greedy), that is .*? so in total

""".*?(\d{1,3})\.(\d{1,3})\.(\d{1,3})\.(\d{1,3}).*""".r

在第一个数字之前精确定义模式，例如如果这些只是字符，请执行

Precisely define the pattern before the first number, e.g. if these are only characters, do

"""[a-zA-Z]*(\d{1,3})\.(\d{1,3})\.(\d{1,3})\.(\d{1,3}).*""".r

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