@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"]

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。

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"]

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)")
}

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