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

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"

Answer 2

but I don't know how to do that in Swift 3.
When you receive a match from NSRegularExpression, what you get is an NSTextCheckingResult. You call rangeAt to get a specific capture group.
Example:
let s = "hey ho ha"
let pattern = "(h).*(h).*(h)"
// our goal is capture group 3, "h" in "ha"
let regex = try! NSRegularExpression(pattern: pattern)
let result = regex.matches(in:s, range:NSMakeRange(0, s.utf16.count))
let third = result[0].rangeAt(3) // <-- !!
third.location // 7
third.length // 1

Answer 3

As ever, a simple extension seems to be the way around swift's bizarre overcomplication...
extension NSTextCheckingResult {
func groups(testedString:String) -> [String] {
var groups = [String]()
for i in 0 ..< self.numberOfRanges
{
let group = String(testedString[Range(self.range(at: i), in: testedString)!])
groups.append(group)
}
return groups
}
}
Use it like this:
if let match = myRegex.firstMatch(in: someString, range: NSMakeRange(0, someString.count)) {
let groups = match.groups(testedString: someString)
//... do something with groups
}

Answer 4

A slightly altered version based on #Vyacheslav's answer with different error handling approach:
enum ParsingError: Error {
// You can pass more info here with parameter(s) if you want, e.g. `case let invalidRange(originalString, failedAtRange)`
case invalidRange
}
protocol StringUtilityRequired {
var stringUtility: StringUtility { get }
}
extension StringUtilityRequired {
var stringUtility: StringUtility { StringUtility() }
}
enum StringUtility {
func groups(_ str: String, pattern: String) throws -> [[String]] {
let regex = try NSRegularExpression(pattern: pattern)
let matches = regex.matches(in: str, range: NSRange(str.startIndex..., in: str))
return try matches.map { match throws in
return try (0 ..< match.numberOfRanges).map { range throws in
let rangeBounds = match.range(at: range)
guard let range = Range(rangeBounds, in: str) else {
throw ParsingError.invalidRange
}
return String(str[range])
}
}
}
// This component is stateless; it doesn't have any side effect
case pure
init() { self = .pure }
}
Usage:
struct MyComponent: StringUtilityRequired {
func myFunc() throws {
let groups = try stringUtility.groups("Test 123", pattern: "(.+)\s(.+)")
print(groups)
}
}