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 ?

CodePudding user response：

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

CodePudding user response：

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