我的答案建立在给定答案的基础上，但通过添加额外的支持使正则匹配更加健壮:

不仅返回匹配项，还返回每个匹配项的所有捕获组(参见下面的示例) 这个解决方案支持可选匹配，而不是返回空数组 通过不打印到控制台来避免do/catch，并使用guard构造 将matchingStrings作为String的扩展

斯威夫特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")

斯威夫特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")

斯威夫特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))
                    : ""
            }
        }
    }
}

你可以像这样对字符串使用匹配(regex:):

let array = try "Your String To Search".matching(regex: ".")

使用这个简单的扩展:

public extension String {
    func matching(regex: String) throws -> [String] {
        let regex = try NSRegularExpression(pattern: regex)
        let results = regex.matches(in: self, range: NSRange(startIndex..., in: self))
        return results.map { String(self[Range($0.range, in: self)!]) }
    }
}

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)")
}

@p4bloch如果你想从一系列捕获括号中捕获结果，那么你需要使用NSTextCheckingResult的rangeAtIndex(index)方法，而不是range。下面是@MartinR针对Swift2的方法，适用于捕获括号。在返回的数组中，第一个结果[0]是整个捕获，然后各个捕获组从[1]开始。我注释掉了map操作(因此更容易看到我所更改的内容)，并将其替换为嵌套循环。

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 []
    }
}

举个例子，假设你想拆分一串标题年份，比如“Finding Dory 2016”，你可以这样做:

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

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

Swift 4没有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)!])
        }
    }
}