快速获取字符串中子字符串的所有范围

Question

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 {
    ...
}

Answer 1

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
    }
}

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

Answer 2

Swift 5:

The improved version of the most popular answer:

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

Answer 3

I would suggest such a solution:

import Foundation

extension String {

    func rangesOfString(s: String) -> [Range<Index>] {
        let re = try! NSRegularExpression(pattern: NSRegularExpression.escapedPatternForString(s), options: [])
        return re.matchesInString(self, options: [], range: nsRange(startIndex ..< endIndex)).flatMap { range($0.range) }
    }

    func range(nsRange : NSRange) -> Range<Index>? {
        let utf16from = utf16.startIndex.advancedBy(nsRange.location, limit: utf16.endIndex)
        let utf16to   = utf16from.advancedBy(nsRange.length, limit: utf16.endIndex)

        if let from = String.Index(utf16from, within: self),
           let to   = String.Index(utf16to,   within: self)
        {
            return from ..< to
        } else {
            return nil
        }
    }

    func nsRange(range : Range<Index>) -> NSRange {
        let utf16from = String.UTF16View.Index(range.startIndex, within: utf16)
        let utf16to   = String.UTF16View.Index(range.endIndex,   within: utf16)
        return NSRange(location: utf16.startIndex.distanceTo(utf16from), length: utf16from.distanceTo(utf16to))
    }

}

print("[^x]? [^x]? [^x]?".rangesOfString("[^x]?")) // [Range(0..<5), Range(6..<11), Range(12..<17)]

Aside the main question, this code also shows the way to convert NSRange to and from Range<String.Index> (based on this post ).

Answer 4

You are using regular expressions, so you need to take care about characters that have special meaning - . is only one of them.

If you're doing a search for substrings, I suggest to use the good old rangeOf... methods instead:

func rangeOfString(_ searchString: String,
           options mask: NSStringCompareOptions,
             range searchRange: NSRange) -> NSRange

Just keep calling that method on your string (and adjust the searchRange ), until no further matches are found.

Answer 5

You can get occurance count for particular string by following code:

let str: NSMutableString = "ab ad adk fda kla kad ab ab kd"
let count = str.replaceOccurrencesOfString("ab", withString: "ab", options: NSStringCompareOptions.LiteralSearch, range: NSMakeRange(0, str.length))