Swift extract regex matches

I want to extract substrings from a string that match a regex pattern.

So I'm looking for something like this:

func matchesForRegexInText(regex: String!, text: String!) -> [String] {
   ???
}

So this is what I have:

func matchesForRegexInText(regex: String!, text: String!) -> [String] {

    var regex = NSRegularExpression(pattern: regex, 
        options: nil, error: nil)

    var results = regex.matchesInString(text, 
        options: nil, range: NSMakeRange(0, countElements(text))) 
            as Array<NSTextCheckingResult>

    /// ???

    return ...
}

The problem is, that matchesInString delivers me an array of NSTextCheckingResult, where NSTextCheckingResult.range is of type NSRange.

NSRange is incompatible with Range<String.Index>, so it prevents me of using text.substringWithRange(...)

Any idea how to achieve this simple thing in swift without too many lines of code?

Even if the matchesInString() method takes a String as the first argument, it works internally with NSString, and the range parameter must be given using the NSString length and not as the Swift string length. Otherwise it will fail for "extended grapheme clusters" such as "flags".

As of Swift 4 (Xcode 9), the Swift standard library provides functions to convert between Range<String.Index> and NSRange.

func matches(for regex: String, in text: String) -> [String] {
    
    do {
        let regex = try NSRegularExpression(pattern: regex)
        let results = regex.matches(in: text,
                                    range: NSRange(text.startIndex..., in: text))
        return results.map {
            String(text[Range($0.range, in: text)!])
        }
    } catch let error {
        print("invalid regex: \(error.localizedDescription)")
        return []
    }
}

Example:

let string = "🇩🇪€4€9"
let matched = matches(for: "[0-9]", in: string)
print(matched)
// ["4", "9"]

Note: The forced unwrap Range($0.range, in: text)! is safe because the NSRange refers to a substring of the given string text. However, if you want to avoid it then use

		return results.flatMap {
			Range($0.range, in: text).map { String(text[$0]) }
		}

instead.

---

(Older answer for Swift 3 and earlier:)

So you should convert the given Swift string to an NSString and then extract the ranges. The result will be converted to a Swift string array automatically.

(The code for Swift 1.2 can be found in the edit history.)

Swift 2 (Xcode 7.3.1) :

func matchesForRegexInText(regex: String, text: String) -> [String] {
    
    do {
        let regex = try NSRegularExpression(pattern: regex, options: [])
        let nsString = text as NSString
        let results = regex.matchesInString(text,
                                            options: [], range: NSMakeRange(0, nsString.length))
        return results.map { nsString.substringWithRange($0.range)}
    } catch let error as NSError {
        print("invalid regex: \(error.localizedDescription)")
        return []
    }
}

Example:

let string = "🇩🇪€4€9"
let matches = matchesForRegexInText("[0-9]", text: string)
print(matches)
// ["4", "9"]

---

Swift 3 (Xcode 8)

func matches(for regex: String, in text: String) -> [String] {
    
    do {
        let regex = try NSRegularExpression(pattern: regex)
        let nsString = text as NSString
        let results = regex.matches(in: text, range: NSRange(location: 0, length: nsString.length))
        return results.map { nsString.substring(with: $0.range)}
    } catch let error {
        print("invalid regex: \(error.localizedDescription)")
        return []
    }
}

Example:

let string = "🇩🇪€4€9"
let matched = matches(for: "[0-9]", in: string)
print(matched)
// ["4", "9"]

My answer builds on top of given answers but makes regex matching more robust by adding additional support:

• Returns not only matches but returns also all capturing groups for each match (see examples below)
• Instead of returning an empty array, this solution supports optional matches
• Avoids do/catch by not printing to the console and makes use of the guard construct
• Adds matchingStrings as an extension to String

Swift 4.2

//: Playground - noun: a place where people can play

import Foundation

extension String {
    func matchingStrings(regex: String) -> [[String]] {
        guard let regex = try? NSRegularExpression(pattern: regex, options: []) else { return [] }
        let nsString = self as NSString
        let results  = regex.matches(in: self, options: [], range: NSMakeRange(0, nsString.length))
        return results.map { result in
            (0..<result.numberOfRanges).map {
                result.range(at: $0).location != NSNotFound
                    ? nsString.substring(with: result.range(at: $0))
                    : ""
            }
        }
    }
}

"prefix12 aaa3 prefix45".matchingStrings(regex: "fix([0-9])([0-9])")
// Prints: [["fix12", "1", "2"], ["fix45", "4", "5"]]

"prefix12".matchingStrings(regex: "(?:prefix)?([0-9]+)")
// Prints: [["prefix12", "12"]]

"12".matchingStrings(regex: "(?:prefix)?([0-9]+)")
// Prints: [["12", "12"]], other answers return an empty array here

// Safely accessing the capture of the first match (if any):
let number = "prefix12suffix".matchingStrings(regex: "fix([0-9]+)su").first?[1]
// Prints: Optional("12")

Swift 3

//: Playground - noun: a place where people can play

import Foundation

extension String {
    func matchingStrings(regex: String) -> [[String]] {
        guard let regex = try? NSRegularExpression(pattern: regex, options: []) else { return [] }
        let nsString = self as NSString
        let results  = regex.matches(in: self, options: [], range: NSMakeRange(0, nsString.length))
        return results.map { result in
            (0..<result.numberOfRanges).map {
                result.rangeAt($0).location != NSNotFound
                    ? nsString.substring(with: result.rangeAt($0))
                    : ""
            }
        }
    }
}

"prefix12 aaa3 prefix45".matchingStrings(regex: "fix([0-9])([0-9])")
// Prints: [["fix12", "1", "2"], ["fix45", "4", "5"]]

"prefix12".matchingStrings(regex: "(?:prefix)?([0-9]+)")
// Prints: [["prefix12", "12"]]

"12".matchingStrings(regex: "(?:prefix)?([0-9]+)")
// Prints: [["12", "12"]], other answers return an empty array here

// Safely accessing the capture of the first match (if any):
let number = "prefix12suffix".matchingStrings(regex: "fix([0-9]+)su").first?[1]
// Prints: Optional("12")

Swift 2

extension String {
    func matchingStrings(regex: String) -> [[String]] {
        guard let regex = try? NSRegularExpression(pattern: regex, options: []) else { return [] }
        let nsString = self as NSString
        let results  = regex.matchesInString(self, options: [], range: NSMakeRange(0, nsString.length))
        return results.map { result in
            (0..<result.numberOfRanges).map {
                result.rangeAtIndex($0).location != NSNotFound
                    ? nsString.substringWithRange(result.rangeAtIndex($0))
                    : ""
            }
        }
    }
}

The fastest way to return all matches and capture groups in Swift 5

extension String {
	func match(_ regex: String) -> [[String]] {
		let nsString = self as NSString
		return (try? NSRegularExpression(pattern: regex, options: []))?.matches(in: self, options: [], range: NSMakeRange(0, nsString.length)).map { match in
			(0..<match.numberOfRanges).map { match.range(at: $0).location == NSNotFound ? "" : nsString.substring(with: match.range(at: $0)) }
		} ?? []
	}
}

Returns a 2-dimentional array of strings:

"prefix12suffix fix1su".match("fix([0-9]+)su")

returns...

[["fix12su", "12"], ["fix1su", "1"]]

// First element of sub-array is the match
// All subsequent elements are the capture groups

If you want to extract substrings from a String, not just the position, (but the actual String including emojis). Then, the following maybe a simpler solution.

extension String {
  func regex (pattern: String) -> [String] {
    do {
      let regex = try NSRegularExpression(pattern: pattern, options: NSRegularExpressionOptions(rawValue: 0))
      let nsstr = self as NSString
      let all = NSRange(location: 0, length: nsstr.length)
      var matches : [String] = [String]()
      regex.enumerateMatchesInString(self, options: NSMatchingOptions(rawValue: 0), range: all) {
        (result : NSTextCheckingResult?, _, _) in
        if let r = result {
          let result = nsstr.substringWithRange(r.range) as String
          matches.append(result)
        }
      }
      return matches
    } catch {
      return [String]()
    }
  }
} 

Example Usage:

"someText 👿🏅👿⚽️ pig".regex("👿⚽️")

Will return the following:

["👿⚽️"]

Note using "\w+" may produce an unexpected ""

"someText 👿🏅👿⚽️ pig".regex("\\w+")

Will return this String array

["someText", "️", "pig"]

I found that the accepted answer's solution unfortunately does not compile on Swift 3 for Linux. Here's a modified version, then, that does:

import Foundation

func matches(for regex: String, in text: String) -> [String] {
    do {
        let regex = try RegularExpression(pattern: regex, options: [])
        let nsString = NSString(string: text)
        let results = regex.matches(in: text, options: [], range: NSRange(location: 0, length: nsString.length))
        return results.map { nsString.substring(with: $0.range) }
    } catch let error {
        print("invalid regex: \(error.localizedDescription)")
        return []
    }
}

The main differences are:

1. Swift on Linux seems to require dropping the NS prefix on Foundation objects for which there is no Swift-native equivalent. (See Swift evolution proposal #86.)

2. Swift on Linux also requires specifying the options arguments for both the RegularExpression initialization and the matches method.

3. For some reason, coercing a String into an NSString doesn't work in Swift on Linux but initializing a new NSString with a String as the source does work.

This version also works with Swift 3 on macOS / Xcode with the sole exception that you must use the name NSRegularExpression instead of RegularExpression.

Swift 4 without NSString.

extension String {
    func matches(regex: String) -> [String] {
        guard let regex = try? NSRegularExpression(pattern: regex, options: [.caseInsensitive]) else { return [] }
        let matches  = regex.matches(in: self, options: [], range: NSMakeRange(0, self.count))
        return matches.map { match in
            return String(self[Range(match.range, in: self)!])
        }
    }
}

@p4bloch if you want to capture results from a series of capture parentheses, then you need to use the rangeAtIndex(index) method of NSTextCheckingResult, instead of range. Here's @MartinR 's method for Swift2 from above, adapted for capture parentheses. In the array that is returned, the first result [0] is the entire capture, and then individual capture groups begin from [1]. I commented out the map operation (so it's easier to see what I changed) and replaced it with nested loops.

func matches(for regex: String!, in text: String!) -> [String] {
    
    do {
        let regex = try NSRegularExpression(pattern: regex, options: [])
        let nsString = text as NSString
        let results = regex.matchesInString(text, options: [], range: NSMakeRange(0, nsString.length))
        var match = [String]()
        for result in results {
            for i in 0..<result.numberOfRanges {
                match.append(nsString.substringWithRange( result.rangeAtIndex(i) ))
            }
        }
        return match
        //return results.map { nsString.substringWithRange( $0.range )} //rangeAtIndex(0)
    } catch let error as NSError {
        print("invalid regex: \(error.localizedDescription)")
        return []
    }
}

An example use case might be, say you want to split a string of title year eg "Finding Dory 2016" you could do this:

print ( matches(for: "^(.+)\\s(\\d{4})" , in: "Finding Dory 2016"))
// ["Finding Dory 2016", "Finding Dory", "2016"]

Most of the solutions above only give the full match as a result ignoring the capture groups e.g.: ^\d+\s+(\d+)

To get the capture group matches as expected you need something like (Swift4) :

public extension String {
    public func capturedGroups(withRegex pattern: String) -> [String] {
        var results = [String]()

        var regex: NSRegularExpression
        do {
            regex = try NSRegularExpression(pattern: pattern, options: [])
        } catch {
            return results
        }
        let matches = regex.matches(in: self, options: [], range: NSRange(location:0, length: self.count))

        guard let match = matches.first else { return results }

        let lastRangeIndex = match.numberOfRanges - 1
        guard lastRangeIndex >= 1 else { return results }

        for i in 1...lastRangeIndex {
            let capturedGroupIndex = match.range(at: i)
            let matchedString = (self as NSString).substring(with: capturedGroupIndex)
            results.append(matchedString)
        }

        return results
    }
}

This is how I did it, I hope it brings a new perspective how this works on Swift.

In this example below I will get the any string between []

var sample = "this is an [hello] amazing [world]"

var regex = NSRegularExpression(pattern: "\\[.+?\\]"
, options: NSRegularExpressionOptions.CaseInsensitive 
, error: nil)

var matches = regex?.matchesInString(sample, options: nil
, range: NSMakeRange(0, countElements(sample))) as Array<NSTextCheckingResult>

for match in matches {
   let r = (sample as NSString).substringWithRange(match.range)//cast to NSString is required to match range format.
    println("found= \(r)")
}

This is a very simple solution that returns an array of string with the matches

Swift 3.

internal func stringsMatching(regularExpressionPattern: String, options: NSRegularExpression.Options = []) -> [String] {
        guard let regex = try? NSRegularExpression(pattern: regularExpressionPattern, options: options) else {
            return []
        }
        
        let nsString = self as NSString
        let results = regex.matches(in: self, options: [], range: NSMakeRange(0, nsString.length))

        return results.map {
            nsString.substring(with: $0.range)
        }
    }

Big thanks to Lars Blumberg his answer for capturing groups and full matches with Swift 4, which helped me out a lot. I also made an addition to it for the people who do want an error.localizedDescription response when their regex is invalid:

extension String {
    func matchingStrings(regex: String) -> [[String]] {
        do {
            let regex = try NSRegularExpression(pattern: regex)
            let nsString = self as NSString
            let results  = regex.matches(in: self, options: [], range: NSMakeRange(0, nsString.length))
            return results.map { result in
                (0..<result.numberOfRanges).map {
                    result.range(at: $0).location != NSNotFound
                        ? nsString.substring(with: result.range(at: $0))
                        : ""
                }
            }
        } catch let error {
            print("invalid regex: \(error.localizedDescription)")
            return []
        }
    }
}

For me having the localizedDescription as error helped understand what went wrong with escaping, since it's displays which final regex swift tries to implement.

update @Mike Chirico's to Swift 5

extension String{



  func regex(pattern: String) -> [String]?{
    do {
        let regex = try NSRegularExpression(pattern: pattern, options: NSRegularExpression.Options(rawValue: 0))
        let all = NSRange(location: 0, length: count)
        var matches = [String]()
        regex.enumerateMatches(in: self, options: NSRegularExpression.MatchingOptions(rawValue: 0), range: all) {
            (result : NSTextCheckingResult?, _, _) in
              if let r = result {
                    let nsstr = self as NSString
                    let result = nsstr.substring(with: r.range) as String
                    matches.append(result)
              }
        }
        return matches
    } catch {
        return nil
    }
  }
}

basic phone number matching

let phoneNumbers = ["+79990001101", "+7 (800) 000-11-02", "+34 507 574 147 ", "+1-202-555-0118"]

let match: (String) -> String = {
    $0.replacingOccurrences(of: #"[^\d+]"#, with: "", options: .regularExpression)
}

print(phoneNumbers.map(match))
// ["+79990001101", "+78000001102", "+34507574147", "+12025550118"]