我想从匹配正则表达式模式的字符串中提取子字符串。

所以我在寻找这样的东西:

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

这就是我得到的:

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 ...
}

问题是,matchesInString提供给我一个NSTextCheckingResult数组,其中NSTextCheckingResult。range类型为NSRange。

NSRange与Range<String不兼容。索引>,它阻止我使用text。substringwithrange(…)

你知道如何在没有太多行代码的情况下实现这个简单的事情吗?


当前回答

非常感谢Lars Blumberg用Swift 4捕获小组和完整比赛的答案,这对我帮助很大。我还为那些确实希望出现错误的人添加了一些内容。当他们的正则表达式无效时localizedDescription响应:

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 []
        }
    }
}

对于我来说,localizedDescription作为错误有助于理解转义时发生了什么错误,因为它显示了swift试图实现的最终正则表达式。

其他回答

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)!])
        }
    }
}

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

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。

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))
                    : ""
            }
        }
    }
}

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)")
}