My program is:
val pattern = "[*]prefix_([a-zA-Z]*)_[*]".r
val outputFieldMod = "TRASHprefix_target_TRASH"
var tar =
outputFieldMod match {
case pattern(target) => target
}
println(tar)
Basically, I try to get the "target" and ignore "TRASH" (I used *). But it has some error and I am not sure why..
Simple and straight forward standard library function (unanchored)
Use Unanchored
Solution one
Use unanchored on the pattern to match inside the string ignoring the trash
val pattern = "prefix_([a-zA-Z]*)_".r.unanchored
unanchored will only match the pattern ignoring all the trash (all the other words)
val result = str match {
case pattern(value) => value
case _ => ""
}
Example
Scala REPL
scala> val pattern = """foo\((.*)\)""".r.unanchored
pattern: scala.util.matching.UnanchoredRegex = foo\((.*)\)
scala> val str = "blahblahfoo(bar)blahblah"
str: String = blahblahfoo(bar)blahblah
scala> str match { case pattern(value) => value ; case _ => "no match" }
res3: String = bar
Solution two
Pad your pattern from both sides with .*. .* matches any char other than a linebreak character.
val pattern = ".*prefix_([a-zA-Z]*)_.*".r
val result = str match {
case pattern(value) => value
case _ => ""
}
Example
Scala REPL
scala> val pattern = """.*foo\((.*)\).*""".r
pattern: scala.util.matching.Regex = .*foo\((.*)\).*
scala> val str = "blahblahfoo(bar)blahblah"
str: String = blahblahfoo(bar)blahblah
scala> str match { case pattern(value) => value ; case _ => "no match" }
res4: String = bar
This will work, val pattern = ".*prefix_([a-z]+).*".r, but it distinguishes between target and trash via lower/upper-case letters. Whatever determines real target data from trash data will determine the real regex pattern.
