Scala regex pattern match of ip address

问题

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
      }

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
  }

it does match

回答1:

Your variant

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.")
  }
}

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.

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

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

Solutions

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

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

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
```

回答2:

You should use reluctant quantifier rather than greedy quantifier:

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

来源：https://stackoverflow.com/questions/33008914/scala-regex-pattern-match-of-ip-address

标签

regex

scala