How to de-escape string in scala?

Question

I found many posts about escaping string but no about de-escaping one.

Using Scala Play, my controller accept a JSON as a request. I extract a string from it via:

val text: play.api.libs.json.JsValue = request.body.\("source")

If I print text.toString I get e.g.

"Hello\tworld\nmy name is \"ABC\""

How can I transform this escaped text into normal one? The result should look like

Hello    world
my name is "ABC"

Up to this point, I've tried an approach like it follows:

replaceAll("""\\t""", "\t")

However, creating all possible escaping rules may be too complicated. So my question is: How to do that easily? Possibly using standard library. Java solutions are also possible.

Answer 1

There are interpolations which allow you to transform Strings into formatted and/or escaped sequences. These interpolations like s"..." or f"..." are handled in StringContext.

It also offers a reverse function:

val es = """Hello\tworld\nmy name is \"ABC\""""
val un = StringContext treatEscapes es

Answer 2

This isn't an answer for unescaping strings in general, but specifically for handling JsValue:

text match { 
  case JsString(value) => value
  case _ => // what do you want to do for other cases?
}

You can implement unescape in this way:

def unescapeJson(s: String) = JSON.parse('"' + s + '"') match {
  case JsString(value) => value
  case _ => ??? // can't happen
}

or use StringEscapeUtils from Apache Commons Lang (though this is a rather large dependency).