import Foundation
let pattern = #"^(?=(?:\+|0{2})?(?:(?:[\(\-\)\t\f ]*\d){7,10})?(?:[\t\f ]?\d{2,3})(?:[\-\s]?[ext]{1,3}[\-\t\f ]?\d{1,4})?$)((?:\+|0{2})\d{0,3})?(?:[\-\t\f ]?)(\(0?\d{1,3}\)|\d{0,3})(?:[\-\t\f ]{0,2}\d){3,}(?:[\-\s]?(?:x|ext)[\-\t\f ]?(\d{1,4}))?$"#
let regex = try! NSRegularExpression(pattern: pattern, options: [.anchorsMatchLines, .caseInsensitive])
let testString = #"""
123456789
1234567890123456
1234567890123
0091234567890
+9 123-4567890-x321
+9-123-456-7890
+9 123 4 5 6-7890x1234
009-(123)-456-7890 ext4321
0091234567890x1234
+9-123-456-7890x12345
+9-123-456-7890 ext 1234
0027-123-456-7890
+27-123-456-7890
+9(123)4567890
911
(0)123456789
+34123456789
+34 (0)123456789
+340123456789
+34 123 456 789
+34 1 23 45 67 89
001230123456789
ext1234
++34123456789
00-0-----000
01 555012345
001-555012345
+012345678
112
12(34567890
123)456789012345
"""#
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