快速获取字符串中子字符串的所有范围 [英] get all ranges of a substring in a string in swift

问题描述

我有一个字符串，例如"ab ad adk fda kla kad ab ab kd".我想获取所有范围的ab.(这里ab位于3位置，所以我应该获得3范围).在正常情况下，我的代码工作正常，但是如果搜索文本为."，那么我得到错误的结果

I have a string for example "ab ad adk fda kla kad ab ab kd". I want to get all range of ab.(Here ab is present at 3 position so I should get 3 range).In normal scenarion my code is working fine, but if search text is ".",then I am getting wrong result

do {
    let regEx = try NSRegularExpression(pattern: searchText, options: NSRegularExpressionOptions.CaseInsensitive)

    let matchesRanges = regEx.matchesInString(attributedText.string, options:[], range: NSMakeRange(0, attributedText.string.length))

    for rng in matchesRanges {
        let wordRange = rng.rangeAtIndex(0)
    }
} catch {
    ...
}

推荐答案

以下解决方案使用本机Swift 4函数range(of:, options:, range:, locale:)):

The following solution uses the native Swift 4 function range(of:, options:, range:, locale:)):

extension String {
    func ranges(of substring: String, options: CompareOptions = [], locale: Locale? = nil) -> [Range<Index>] {
        var ranges: [Range<Index>] = []
        while ranges.last.map({ $0.upperBound < self.endIndex }) ?? true,
            let range = self.range(of: substring, options: options, range: (ranges.last?.upperBound ?? self.startIndex)..<self.endIndex, locale: locale)
        {
            ranges.append(range)
        }
        return ranges
    }
}

(然后Swift 4提供了本机API，可以从Range<Index>转换为NSRange)

(Swift 4 then provides native API to convert from Range<Index> to NSRange)

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