import Foundation
let pattern = #"^(?<ip_addr>((\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})|((([\da-fA-F]{1,4}:){7})([\da-fA-F]{1,4})$|(([\da-fA-F]{1,4}:){1,6}:)(([\da-fA-F]{1,4}:){0,4})([\da-fA-F]{1,4}))|(::[\da-fA-F]{1,4})|(([\da-fA-F]{1,4}:){0,7}[\da-fA-F]{1,4}::)))$"#
let regex = try! NSRegularExpression(pattern: pattern, options: .anchorsMatchLines)
let testString = #"""
1.2.3.4
01.02.3.4
01.02.03.04
128.101.1.1
128.1.101.101
987.654.321.0
2607:ea00:101:102:103:104:105:106
2607:eA00:abcd:efab:cDeF:0000:0000:0001
fe80::64
ff02::
::1
2001:468::
2001::468::4
:::1
2001:410:::
123.123.1a3.313
1.2.3.
.2.3.4
"""#
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