CODEWITHSUNDEEP
regexscala
CSnerd
CSnerd

Reputation: 2229

How to use regular expression to match whole String

I have the following code:

def main(args: Array[String]) {
      val it = ("\\b" + "'as" + "\\b").r.findAllMatchIn("'as you are a's".toLowerCase());
      val lst = it.map(_.start).toList
      print(lst)
}

I expected the answer would be List(0) (because it matched 'as and index should be 0), but it gave me List()

Also,

def main(args: Array[String]) {
      val it = ("\\b" + "as" + "\\b").r.findAllMatchIn("'as you are a's".toLowerCase());
      val lst = it.map(_.start).toList
      print(lst)
  }

This gave me the answer List(1) but I expected the answer to be List() because I want to match the whole thing (need exactly match 'as ), that is why I use \b here

But this worked well:

def main(args: Array[String]) {
      val it = ("\\b" + "a's" + "\\b").r.findAllMatchIn("'as you are a's".toLowerCase());
      val lst = it.map(_.start).toList
      print(lst)
  }

it returned List(12) which is what I want (because it matched a's and index should be 12).

I did not understand why it did not work when I put ' at the front of word. How can I do this?

Upvotes: 1

Views: 95

Answers (1)

Gábor Bakos
Gábor Bakos

Reputation: 9100

The problem is \b does not match if the first character after that is not a letter or other word character. So it will not match when it is followed by a '. See: http://www.regular-expressions.info/wordboundaries.html

Edit:

val it = ("(?:\\b|')" + "as" + "\\b").r.findAllMatchIn("'as you are a's".toLowerCase())

Upvotes: 1

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