import Foundation
let pattern = #"^(?:((?:IT|SM)\d{2}[A-Z]{1}\d{22})|(NL\d{2}[A-Z]{4}\d{10})|(LV\d{2}[A-Z]{4}\d{13})|((?:BG|GB|IE)\d{2}[A-Z]{4}\d{14})|(GI\d{2}[A-Z]{4}\d{15})|(RO\d{2}[A-Z]{4}\d{16})|(MT\d{2}[A-Z]{4}\d{23})|(NO\d{13})|((?:DK|FI|FO)\d{16})|((?:SI)\d{17})|((?:AT|EE|LU|LT)\d{18})|((?:HR|LI|CH)\d{19})|((?:DE)\d{20})|((?:CZ|ES|SK|SE)\d{22})|(PT\d{23})|((?:IS)\d{24})|((?:BE)\d{14})|((?:FR|MC|GR)\d{25})|((?:PL|HU|CY)\d{26}))$"#
let regex = try! NSRegularExpression(pattern: pattern, options: [.anchorsMatchLines, .caseInsensitive])
let testString = #"""
SEPA
SM07U0854009803000030174419
SI56011006000005649
ES1321000555370200853027
CY02002001950000357009822416
NOT SEPA
SC74NOVH00000021002035257028SCR
TN5901026067111999766058
UA123052990004149497803982794
"""#
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