import Foundation
let pattern = #"^(?:(?:(?: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])(?:-(?:(?: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])|\/(?:[0-9]|[1-2][0-9]|3[0-2]))?|(?:[a-z0-9]+(?:-[a-z0-9]+)*\.)+[a-z]{2,})$"#
let regex = try! NSRegularExpression(pattern: pattern, options: .anchorsMatchLines)
let testString = #"""
Valid
1.2.3.4
255.255.255.255
1.2.3.4-1.2.3.4
1.2.3.4/24
1.2.3.4/8
www.example.com
www.example.lv
www.example.com
Invalid
1.2.3
1.2.3.
.1.2.3
1.2.3.4/33
www.-example.com
1.2.3.4-1.2.3.256
1.2.3.4.5.6.7.8 (use anchors ^ and $ to skip these if needed)
999.999.999.999 (use a complex regex to skip these if needed)
299.299.299.299
001.002.003.004 (these use octal notation, not decimal)
"""#
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