Swift 3 - How do I extract captured groups in regular expressions?

Question

I am using Swift 3 and trying to access captured groups.

let regexp = "((ALREADY PAID | NOT ALR | PROVIDER MAY | READY | MAY BILL | BILL YOU | PAID)((.|\\n)*))(( \\d+)(\\.+|-+)(\\d\\d))"  // check if some substring is in the recognized text if let range = stringText.range(of:regexp, options: .regularExpression) {     let result = tesseract.recognizedText.substring(with:range) } 

I want to be able to extract out the last two numbers captured (\d\d) so if the text was: ALREADY PAID asfasdfadsfasdf 39.15, it would extract 15. Here is a regex builder that shows what I want. Normally, I would be able to do $8 to get the 8th group that was extracted but I don't know how to do that in Swift 3.

http://regexr.com/3fh1e

Answer 1

Swift 4, Swift 5

extension String {     func groups(for regexPattern: String) -> [[String]] {     do {         let text = self         let regex = try NSRegularExpression(pattern: regexPattern)         let matches = regex.matches(in: text,                                     range: NSRange(text.startIndex..., in: text))         return matches.map { match in             return (0..<match.numberOfRanges).map {                 let rangeBounds = match.range(at: $0)                 guard let range = Range(rangeBounds, in: text) else {                     return ""                 }                 return String(text[range])             }         }     } catch let error {         print("invalid regex: \(error.localizedDescription)")         return []     } } } 

example:

let res = "1my 2own 3string".groups(for:"(([0-9]+)[a-z]+) ") 

(lldb) po res ▿ 2 elements
▿ 0 : 3 elements

- 0 : "1my "  - 1 : "1my"  - 2 : "1"    

▿ 1 : 3 elements

- 0 : "2own "  - 1 : "2own"  - 2 : "2"