CODEWITHSUNDEEP
regexstringscalasplit
scalacode
scalacode

Reputation: 1106

transform string scala in an elegant way

I have the following input string: val s = 19860803 000000

I want to convert it to 1986/08/03 I tried this s.split(" ").head, but this is not complete

is there any elegant scala coding way with regex to get the expected result ?

Upvotes: 1

Views: 101

Answers (3)

hughw
hughw

Reputation: 43

An alternative, without a regular expression, by using slice and take.

val s = "19860803 000000"
val year = s.take(4)
val month = s.slice(4,6)
val day = s.slice(6,8)
val result = s"$year/$month/$day"

Or as a one liner

val result = Seq(s.take(4), s.slice(4,6), s.slice(6,8)).mkString("/")

Upvotes: 0

The fourth bird
The fourth bird

Reputation: 163277

You can use a date like pattern using 3 capture groups, and match the following space and the 6 digits.

In the replacement use the 3 groups in the replacement with the forward slashes.

val s = "19860803 000000"
val result = s.replaceAll("^(\\d{4})(\\d{2})(\\d{2})\\h\\d{6}$", "$1/$2/$3")

Output

result: String = 1986/08/03

Upvotes: 3

Jason
Jason

Reputation: 15931

i haven't tested this, but i think the following will work

val expr = raw"(\d{4})(\d{2})(\d{2}) (.*)".r

val formatted = "19860803 000000" match {
  case expr(year,month,day,_) =>. s"$year/$month/$day"
}

scala docs have a lot of good info https://www.scala-lang.org/api/2.13.6/scala/util/matching/Regex.html

Upvotes: 1

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