import Foundation
let pattern = #"^\d*(0)1[23456789] | (0|1)2[3456789] | (0|1|2)3[456789] |(0|1|2|3)4[56789] | (0|1|2|3|4)5[6789] | (0|1|2|3|4|5)6[789] |(0|1|2|3|4|5|6)7[89] | (0|1|2|3|4|5|6|7)8(9)\d*$"#
let regex = try! NSRegularExpression(pattern: pattern, options: [.anchorsMatchLines, .allowCommentsAndWhitespace])
let testString = #"""
123
9876456000
123456789
246
91370
3456
111
415263
xyz123xyz
"""#
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