import Foundation
let pattern = #"(^([0-9a-fA-F]{1,4}:){7,7}[0-9a-fA-F]{1,4}$|^([0-9a-fA-F]{1,4}:){6}:(?:[0-9a-fA-F]{1,4}){0,1}$|^([0-9a-fA-F]{1,4}:){5}(?:(:[0-9a-fA-F]{1,4}){1,2}|:)$|^([0-9a-fA-F]{1,4}:){4}(?:(:[0-9a-fA-F]{1,4}){1,3}|:)$|^([0-9a-fA-F]{1,4}:){3}(?:(:[0-9a-fA-F]{1,4}){1,4}|:)$|^([0-9a-fA-F]{1,4}:){2}(?:(:[0-9a-fA-F]{1,4}){1,5}|:)$|^[0-9a-fA-F]{1,4}:(?:(:[0-9a-fA-F]{1,4}){1,6}|:)$|^:((:[0-9a-fA-F]{1,4}){1,7}|:)$)|(^(?:(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])\.){3,3}(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])$)"#
let regex = try! NSRegularExpression(pattern: pattern, options: .caseInsensitive)
let testString = #""#
let stringRange = NSRange(location: 0, length: testString.utf16.count)
let matches = regex.matches(in: testString, range: stringRange)
var result: [[String]] = []
for match in matches {
var groups: [String] = []
for rangeIndex in 1 ..< match.numberOfRanges {
let nsRange = match.range(at: rangeIndex)
guard !NSEqualRanges(nsRange, NSMakeRange(NSNotFound, 0)) else { continue }
let string = (testString as NSString).substring(with: nsRange)
groups.append(string)
}
if !groups.isEmpty {
result.append(groups)
}
}
print(result)
Please keep in mind that these code samples are automatically generated and are not guaranteed to work. If you find any syntax errors, feel free to submit a bug report. For a full regex reference for Swift 5.2, please visit: https://developer.apple.com/documentation/foundation/nsregularexpression