import Foundation
let pattern = #"^((401178|401179|431274|438935|451416|457393|457631|457632|504175|627780|636297|636369|(506699|5067[0-6]\d|50677[0-8])|(50900\d|5090[1-9]\d|509[1-9]\d{2})|65003[1-3]|(65003[5-9]|65004\d|65005[0-1])|(65040[5-9]|6504[1-3]\d)|(65048[5-9]|65049\d|6505[0-2]\d|65053[0-8])|(65054[1-9]|6505[5-8]\d|65059[0-8])|(65070\d|65071[0-8])|65072[0-7]|(65090[1-9]|65091\d|650920)|(65165[2-9]|6516[6-7]\d)|(65500\d|65501\d)|(65502[1-9]|6550[3-4]\d|65505[0-8]))[0-9]{10,12})"#
let regex = try! NSRegularExpression(pattern: pattern, options: [.anchorsMatchLines, .caseInsensitive])
let testString = #"""
5090402132456747
6500311231321321
6500321231321321
6500331231321321
6500341231321321
6500351231321321
6500391231321321
6500401231321321
6500451231564654
6500511231564654
6504051231321331
6504851231321321
6505411231321313
6507001321321313
6507201231321133
6509011235646432
5066998786245463
6516527887324132
6550007896242321
"""#
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