Swift提取正则表达式匹配

问题描述 投票：144回答：10

我想从匹配正则表达式模式的字符串中提取子字符串。

所以我正在寻找这样的东西：

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

所以这就是我所拥有的：

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为我提供了一系列NSTextCheckingResult，其中NSTextCheckingResult.range是NSRange类型。

NSRange与Range<String.Index>不相容，所以它阻止我使用text.substringWithRange(...)

有没有想过如何在没有太多代码的情况下在swift中实现这个简单的事情？

ios regex string swift
10个回答
269
投票

即使matchesInString()方法采用String作为第一个参数，它在内部使用NSString，并且range参数必须使用NSString长度而不是Swift字符串长度。否则它将失败“扩展的字形集群”，如“标志”。

从Swift 4（Xcode 9）开始，Swift标准库提供了在Range<String.Index>和NSRange之间进行转换的功能。

func matches(for regex: String, in text: String) -> [String] {

    do {
        let regex = try NSRegularExpression(pattern: regex)
        let results = regex.matches(in: text,
                                    range: NSRange(text.startIndex..., in: text))
        return results.map {
            String(text[Range($0.range, in: text)!])
        }
    } catch let error {
        print("invalid regex: \(error.localizedDescription)")
        return []
    }
}

例：

let string = "🇩🇪€4€9"
let matched = matches(for: "[0-9]", in: string)
print(matched)
// ["4", "9"]

注意：强制展开Range($0.range, in: text)!是安全的，因为NSRange指的是给定字符串text的子字符串。但是，如果你想避免它，那么使用

        return results.flatMap {
            Range($0.range, in: text).map { String(text[$0]) }
        }

代替。

（Swift 3及更早版本的旧答案:)

因此，您应该将给定的Swift字符串转换为NSString，然后提取范围。结果将自动转换为Swift字符串数组。

（可以在编辑历史中找到Swift 1.2的代码。）

Swift 2（Xcode 7.3.1）：

func matchesForRegexInText(regex: String, 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))
        return results.map { nsString.substringWithRange($0.range)}
    } catch let error as NSError {
        print("invalid regex: \(error.localizedDescription)")
        return []
    }
}

例：

let string = "🇩🇪€4€9"
let matches = matchesForRegexInText("[0-9]", text: string)
print(matches)
// ["4", "9"]

Swift 3（Xcode 8）

func matches(for regex: String, in text: String) -> [String] {

    do {
        let regex = try NSRegularExpression(pattern: regex)
        let nsString = text as NSString
        let results = regex.matches(in: text, range: NSRange(location: 0, length: nsString.length))
        return results.map { nsString.substring(with: $0.range)}
    } catch let error {
        print("invalid regex: \(error.localizedDescription)")
        return []
    }
}

例：

let string = "🇩🇪€4€9"
let matched = matches(for: "[0-9]", in: string)
print(matched)
// ["4", "9"]
0
投票

没有NSString的Swift 4。

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)!])
        }
    }
}
51
投票

我的答案建立在给定答案之上，但通过添加额外支持使正则表达式匹配更加健壮：

  • 返回不仅匹配，还返回每个匹配的所有捕获组（请参阅下面的示例）
  • 此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并使用do/catch构造来避免guard
  • 添加matchingStrings作为String的扩展

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")

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))
                    : ""
            }
        }
    }
}
10
投票

如果要从String中提取子字符串，而不仅仅是位置，（但实际的字符串包括emojis）。然后，以下可能是一个更简单的解决方案。

extension String {
  func regex (pattern: String) -> [String] {
    do {
      let regex = try NSRegularExpression(pattern: pattern, options: NSRegularExpressionOptions(rawValue: 0))
      let nsstr = self as NSString
      let all = NSRange(location: 0, length: nsstr.length)
      var matches : [String] = [String]()
      regex.enumerateMatchesInString(self, options: NSMatchingOptions(rawValue: 0), range: all) {
        (result : NSTextCheckingResult?, _, _) in
        if let r = result {
          let result = nsstr.substringWithRange(r.range) as String
          matches.append(result)
        }
      }
      return matches
    } catch {
      return [String]()
    }
  }
}

用法示例：

"someText 👿🏅👿⚽️ pig".regex("👿⚽️")

将返回以下内容：

["👿⚽️"]

注意使用“\ w +”可能会产生意外的“”

"someText 👿🏅👿⚽️ pig".regex("\\w+")

将返回此String数组

["someText", "️", "pig"]
9
投票

我发现，接受答案的解决方案很遗憾无法在Swift 3 for Linux上编译。这是一个修改后的版本，它确实：

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 []
    }
}

主要区别是：

  1. Linux上的Swift似乎需要在基础对象上删除NS前缀，而没有Swift本机等价物。 （见Swift evolution proposal #86。）
  2. Linux上的Swift还需要为options初始化和RegularExpression方法指定matches参数。
  3. 出于某种原因，将String强制转换为NSString并不适用于Linux上的Swift，而是使用NSString初始化一个新的String，因为它确实有效。

此版本也适用于macOS / Xcode上的Swift 3，唯一的例外是您必须使用名称NSRegularExpression而不是RegularExpression。

5
投票

@ 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rangeAtIndex(index)的NSTextCheckingResult方法，而不是range。这是@MartinR从上面开始的Swift2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（因此更容易看到我改变了什么）并用嵌套循环替换它。

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 []
    }
}

一个示例用例可能是，比如你要分割一串title year例如“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"]
3
投票

上面的大多数解决方案只给出完全匹配，因此忽略了捕获组，例如：^ \ d + \ s +（\ d +）

要按预期获得捕获组匹配，您需要类似（Swift4）的东西：

public extension String {
    public func capturedGroups(withRegex pattern: String) -> [String] {
        var results = [String]()

        var regex: NSRegularExpression
        do {
            regex = try NSRegularExpression(pattern: pattern, options: [])
        } catch {
            return results
        }
        let matches = regex.matches(in: self, options: [], range: NSRange(location:0, length: self.count))

        guard let match = matches.first else { return results }

        let lastRangeIndex = match.numberOfRanges - 1
        guard lastRangeIndex >= 1 else { return results }

        for i in 1...lastRangeIndex {
            let capturedGroupIndex = match.range(at: i)
            let matchedString = (self as NSString).substring(with: capturedGroupIndex)
            results.append(matchedString)
        }

        return results
    }
}
2
投票

这是一个非常简单的解决方案，它返回带有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)
        }
    }
1
投票

这就是我做到的，我希望它为Swift带来了新的视角。

在下面的这个例子中，我将得到[]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)")
}
0
投票

非常感谢Lars Blumberg他的answer用于捕捉团体和与Swift 4完全匹配，这帮助了我很多。我还为那些想要在正则表达式无效时需要error.localizedDescription响应的人添加了它：

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 []
        }
    }
}

对于我将localizedDescription作为错误有助于了解转义出错的原因，因为它显示了最终正则表达式swift试图实现的内容。

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