import Foundation
let pattern = #"^((((2(0[1-9]|[1-9][0-9]))|([3-6][0-9]{2})|(7([0-6][0-9]|7[0-5])))[0-9]{6})|((0|8)((2(0[1-9]|[1-9][0-9]))|([3-6][0-9]{2})|(7([0-6][0-9]|7[0-5])))[0-9]{5})|((00|99|98)((2(0[1-9]|[1-9][0-9]))|([3-6][0-9]{2})|(7([0-6][0-9]|7[0-5])))[0-9]{4})|(111((2(0[1-9]|[1-9][0-9]))|([3-6][0-9]{2})|(7([0-6][0-9]|7[0-5])))[0-9]{3})|((970|972|974)[0-9]{6}))$"#
let regex = try! NSRegularExpression(pattern: pattern, options: .anchorsMatchLines)
let testString = #"""
000000000
1193046
100000000
111111111
123456789
222222222
999999999
201000000
200123456
775123456
776123456
7751234562
111456000
077500000
877500000
002010000
995642323
988885687
986885687
970123456
972000000
974999999
975999999
"""#
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