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 ?
transform string scala in an elegant way
Viewed 89
3 Answers
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/$3")
Output
result: String = 1986/08/03
i haven't tested this, but i think the following will work
val expr = raw"(\d{4})(\d{2})(\d{2}) (.*)".r
val formatted = "19860803 000000" match {
case expr(year,month,day,_) =>. s"$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
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("/")
Related