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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
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 {
–
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)
–
–
–
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
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).
–
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.
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))
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