如何让组件用正则表达式分隔，也用分隔符？

Question

我有一个字符串扩展名：

extension String {
    func split(usingRegex pattern: String) -> [String] {
        let regex = try! NSRegularExpression(pattern: pattern)
        let matches = regex.matches(in: self, range: NSRange(0..<utf16.count))
        let ranges = [startIndex..<startIndex] + matches.map{Range($0.range, in: self)!} + [endIndex..<endIndex]
        return (0...matches.count).map {String(self[ranges[$0].upperBound..<ranges[$0+1].lowerBound])}
    }
}

我是这样使用它的：

var string = "Hello45playground23today"
var output = string.split(usingRegex: "[0-9]+")

输出是：

["Hello", "playground", "today"]

但我需要的是：

["Hello", "45", "playground", "23", "today"]

有没有办法在 Swift 中实现这一点？

Answer 1

您的代码仅将匹配项之间的子字符串（以及第一个匹配项之前和最后一个匹配项之后）添加到结果中。您还需要匹配项本身的子字符串。这可以通过创建一个包含匹配开始或结束的所有索引的数组来完成，然后在连续索引之间获取所有子字符串：

extension String {
    func split(usingRegex pattern: String) -> [String] {
        let regex = try! NSRegularExpression(pattern: pattern)
        let matches = regex.matches(in: self, range: NSRange(startIndex..., in: self))
        let splits = [startIndex]
            + matches
                .map { Range($0.range, in: self)! }
                .flatMap { [ $0.lowerBound, $0.upperBound ] }
            + [endIndex]

        return zip(splits, splits.dropFirst())
            .map { String(self[$0 ..< $1])}
    }
}

例子：

let string = "Hello45playground23today"
let output = string.split(usingRegex: "[0-9]+")
print(output) // ["Hello", "45", "playground", "23", "today"]

显式循环也可以做到这一点（不太复杂，但可读性更好）：

extension String {
    func split(usingRegex pattern: String) -> [String] {
        let regex = try! NSRegularExpression(pattern: pattern)
        let matches = regex.matches(in: self, range: NSRange(startIndex..., in: self))
        var result: [String] = []
        var pos = startIndex
        for match in matches {
            let range = Range(match.range, in: self)!
            result.append(String(self[pos..<range.lowerBound]))
            result.append(String(self[range]))
            pos = range.upperBound
        }
        result.append(String(self[pos...]))
        return result
    }
}