The string is given as
var filedate = parser("test_pb_PP_Quality_2-Report_20200707.csv");
The Parser function to extract "20200707" as below.
def parser(filename: String):String = {
val extractDate = """(\d{8}).*""".r
val extractDate(dd) = filename;
dd;
}
But its return error as Exception in thread "main" scala.MatchError:
By default, a Regex created with the .r method is anchored, which means it must match the whole string. (think of your regex as being enclosed in ^ and $ )
You could use an unanchored regex instead:
def parser(filename: String):String = {
val extractDate = """(\d{8})""".r.unanchored
val extractDate(dd) = filename
dd
}
This works, but is bad practice since "parsing" would throw an exception if your input does not match. More idiomatic would be to return an Option[String] and handle that at the calling site. For example:
def parser(filename: String): Option[String] = {
"""\d{8}""".r.findFirstIn(filename)
}
parser("test_pb_PP_Quality_2-Report_20200707.csv") match {
case Some(datetime) => // do something
case None => // handle this case
}
You could try using Java's String#replaceAll here:
def parser(filename: String):String = {
val extractDate = filename.replaceAll(".*(\\d{8}).*", "$1");
extractDate;
}
println(parser("test_pb_PP_Quality_2-Report_20200707.csv"))
This prints:
20200707
:-)Just allow some text before the numbers by adding .* to the beginning of the regular expression:
def parser(filename: String):String = {
val extractDate = """.*(\d{8}).*""".r
val extractDate(dd) = filename;
dd;
}
You could also use a match-expression to handle the case where the filename does not have the expected format:
def parser(filename: String):String = {
val extractDate = """.*(\d{8}).*""".r
filename match {
case extractDate(dd) => dd
case _ => throw new RuntimeException(s"Invalid filename $filename")
}
}
