import Foundation
let pattern = #"^(?:(?:(?:[A-F0-9]{1,4}:){6}|(?=(?:[A-F0-9]{0,4}:){0,6}(?:[0-9]{1,3}\.){3}[0-9]{1,3}$)(([0-9A-F]{1,4}:){0,5}|:)((:[0-9A-F]{1,4}){1,5}:|:)|::(?:[A-F0-9]{1,4}:){5})(?:(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])\.){3}(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])|(?:[A-F0-9]{1,4}:){7}[A-F0-9]{1,4}|(?=(?:[A-F0-9]{0,4}:){0,7}[A-F0-9]{0,4}$)(([0-9A-F]{1,4}:){1,7}|:)((:[0-9A-F]{1,4}){1,7}|:)|(?:[A-F0-9]{1,4}:){7}:|:(:[A-F0-9]{1,4}){7})$"#
let regex = try! NSRegularExpression(pattern: pattern, options: .anchorsMatchLines)
let testString = #"""
1762:0:0:0:0:B03:1:AF18
1762:0:0:0:0:B03:127.32.67.15
1762::B03:1:AF18
1762::B03::1:AF18
1762::B03:127.32.67.15
1762::B03::127.32.67.15
"""#
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