Reputation: 10629
I wondering if it's possible to get the MatchData generated from the matching regular expression in the grammar below.
object DateParser extends JavaTokenParsers {
....
val dateLiteral = """(\d{4}[-/])?(\d\d[-/])?(\d\d)""".r ^^ {
... get MatchData
}
}
One option of course is to perform the match again inside the block, but since the RegexParser has already performed the match I'm hoping that it passes the MatchData to the block, or stores it?
Upvotes: 12
Views: 4146
Reputation: 7119
I ran into a similar issue using scala 2.8.1 and trying to parse input of the form "name:value" using the RegexParsers
class:
package scalucene.query
import scala.util.matching.Regex
import scala.util.parsing.combinator._
object QueryParser extends RegexParsers {
override def skipWhitespace = false
private def quoted = regex(new Regex("\"[^\"]+"))
private def colon = regex(new Regex(":"))
private def word = regex(new Regex("\\w+"))
private def fielded = (regex(new Regex("[^:]+")) <~ colon) ~ word
private def term = (fielded | word | quoted)
def parseItem(str: String) = parse(term, str)
}
It seems that you can grab the matched groups after parsing like this:
QueryParser.parseItem("nameExample:valueExample") match {
case QueryParser.Success(result:scala.util.parsing.combinator.Parsers$$tilde, _) => {
println("Name: " + result.productElement(0) + " value: " + result.productElement(1))
}
}
Upvotes: 0
Reputation: 297205
Here is the implicit definition that converts your Regex
into a Parser
:
/** A parser that matches a regex string */
implicit def regex(r: Regex): Parser[String] = new Parser[String] {
def apply(in: Input) = {
val source = in.source
val offset = in.offset
val start = handleWhiteSpace(source, offset)
(r findPrefixMatchOf (source.subSequence(start, source.length))) match {
case Some(matched) =>
Success(source.subSequence(start, start + matched.end).toString,
in.drop(start + matched.end - offset))
case None =>
Failure("string matching regex `"+r+"' expected but `"+in.first+"' found", in.drop(start - offset))
}
}
}
Just adapt it:
object X extends RegexParsers {
/** A parser that matches a regex string and returns the Match */
def regexMatch(r: Regex): Parser[Regex.Match] = new Parser[Regex.Match] {
def apply(in: Input) = {
val source = in.source
val offset = in.offset
val start = handleWhiteSpace(source, offset)
(r findPrefixMatchOf (source.subSequence(start, source.length))) match {
case Some(matched) =>
Success(matched,
in.drop(start + matched.end - offset))
case None =>
Failure("string matching regex `"+r+"' expected but `"+in.first+"' found", in.drop(start - offset))
}
}
}
val t = regexMatch("""(\d\d)/(\d\d)/(\d\d\d\d)""".r) ^^ { case m => (m.group(1), m.group(2), m.group(3)) }
}
Example:
scala> X.parseAll(X.t, "23/03/1971")
res8: X.ParseResult[(String, String, String)] = [1.11] parsed: (23,03,1971)
Upvotes: 21
Reputation: 8590
No, you can't do this. If you look at the definition of the Parser used when you convert a regex to a Parser, it throws away all context and just returns the full matched string:
You have a couple of other options, though:
The first would look like
val separator = "-" | "/"
val year = ("""\d{4}"""r) <~ separator
val month = ("""\d\d"""r) <~ separator
val day = """\d\d"""r
val date = ((year?) ~ (month?) ~ day) map {
case year ~ month ~ day =>
(year.getOrElse("2009"), month.getOrElse("11"), day)
}
The <~
means "require these two tokens together, but only give me the result of the first one.
The ~
means "require these two tokens together and tie them together in a pattern-matchable ~ object.
The ?
means that the parser is optional and will return an Option.
The .getOrElse
bit provides a default value for when the parser didn't define a value.
Upvotes: 3
Reputation: 26486
When a Regex is used in a RegexParsers instance, the implicit def regex(Regex): Parser[String] in RegexParsers is used to appoly that Regex to the input. The Match instance yielded upon successful application of the RE at the current input is used to construct a Success in the regex() method, but only its "end" value is used, so any captured sub-matches are discarded by the time that method returns.
As it stands (in the 2.7 source I looked at), you're out of luck, I believe.
Upvotes: 1