import Foundation
let pattern = #"((((0[1-9]|[1-2][0-9]|3[0-1])\.(0[1]))|((0[1-9]|[1-2][0-8])\.(0[2]))|((0[1-9]|[1-2][0-9]|3[0-1])\.(0[3]))|((0[1-9]|[1-2][0-9]|3[0])\.(0[4]))|((0[1-9]|[1-2][0-9]|3[0-1])\.(0[5]))|((0[1-9]|[1-2][0-9]|3[0])\.(0[6]))|((0[1-9]|[1-2][0-9]|3[0-1])\.(0[7]))|((0[1-9]|[1-2][0-9]|3[0-1])\.(0[8]))|((0[1-9]|[1-2][0-9]|3[0])\.(0[9]))|((0[1-9]|[1-2][0-9]|3[0-1])\.(1[0]))|((0[1-9]|[1-2][0-9]|3[0])\.(1[1]))|((0[1-9]|[1-2][0-9]|3[0-1])\.(1[2])))\.[0-9]{2,4})|(2[9])\.(0[2])\.([0-9][0-9](([13579][26])|([02468][048])))"#
let regex = try! NSRegularExpression(pattern: pattern, options: .anchorsMatchLines)
let testString = #"29.02.201"#
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