Swift extract regex matches

Question

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

So I\'m looking for something like this:

func matchesForRegexInText(regex: St         


        

Answer 1

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, count)).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

Answer 2

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.

Answer 3

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)
        }
    }

Answer 4

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))
                    : ""
            }
        }
    }
}

Answer 5

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.