Escape string to be passed as regex

Question

I would like to create a function that creates regex matching an arbitrary string given at the input. For example, when I feed it with 123$ it should match literally "123$" and not 123 at the end of the string.

def convert( xs: String ) = (xs map ( x => "\\"+x)).mkString                   val text = """ 123 \d+ 567 """                                                 val x = """\d+"""                                                             val p1 = x.r                                                                  val p2 = convert(x).r                                                          println( p1.toString )                                                          \d+ // regex to match number                                                 println( ( p1 findAllIn text ).toList )                                         List(123, 567) // ok, numbers are matched                                    println( p2.toString )                                                          \\\d\+ // regex to match "backshash d plus"                                  println( ( p2 findAllIn text ).toList )                                         List() // nothing matched :(                                                

So the last findAllIn should find \d+ in text, but it doesn't. What's wrong here?

Answer 1

You can use Java's Pattern class to escape strings as regular expressions. See http://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html#quote%28java.lang.String%29

For example:

scala> Pattern.quote("123$").r.findFirstIn("123$") res3: Option[String] = Some(123$) 

Answer 2

Just to bring more attention to Harold L's comment above, if you want to do this with a Scala library you can use:

import scala.util.matching.Regex Regex.quote("123$").r.findFirstIn("123$")