这是一个非常简单的解决方案，它返回一个包含匹配项的字符串数组

斯威夫特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)
        }
    }

我就是这么做的，我希望它能给斯威夫特带来一个新的视角。

在下面的例子中，我将获得[]之间的任何字符串

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)")
}

不幸的是，我发现公认的答案的解决方案不能在Linux的Swift 3上编译。下面是一个修改后的版本:

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 []
    }
}

主要的区别是:

Linux上的Swift似乎需要在Foundation对象上删除NS前缀，而Swift本地没有对应的对象。(参见快速进化提案#86。) Linux上的Swift还需要为RegularExpression初始化和matches方法指定options参数。 出于某种原因，在Linux上的Swift中，将String强制转换为NSString是行不通的，但在源中初始化一个新的NSString是可行的。

这个版本也适用于macOS / Xcode上的Swift 3，唯一的例外是你必须使用NSRegularExpression这个名字而不是RegularExpression。

在Swift 5中返回所有匹配和捕获组的最快方式

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)) }
        } ?? []
    }
}

返回一个二维字符串数组:

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

返回……

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

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

更新@Mike Chirico到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
    }
  }
}

基本电话号码匹配

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"]