import Foundation
let pattern = #"^(?:(?:25[0-5]|2[0-4]\d|1?\d{1,2})(?:\.(?!$)|$)){4}$"#
let regex = try! NSRegularExpression(pattern: pattern, options: .anchorsMatchLines)
let testString = #"""
validates:
192.68.35.35
0.0.0.0
255.0.0.0
192.168.1.0
192.168.0.1
255.255.255.0
1.1.1.1
255.255.255.255
249.249.249.249
200.200.200.200
199.199.199.199
100.100.100.100
99.99.99.99
0.0.0.0
9.9.9.9
10.10.10.10
99.99.99.99
100.100.100.100
109.109.109.109
110.110.110.110
199.199.199.199
200.200.200.200
249.249.249.249
250.250.250.250
255.255.255.255
01.01.01.01
09.09.09.09
192.168.0.1
255.255.255.255
1.1.1.1
should not validate:
256.256.256.260
192.168.0.0/24
192.168..1
192.168.1
1
1.
1.1
1.1.
1.1.1
1.1.1.
1.1.1.1.
1.1.1.1.1
.1.1.1.1
1.0.0.1.0
010.1.1.1
123456
123123123123
.127.0.0.1
192.168.0.1000
300.168.0.1
192.168.0.1.
192.168.0..1
192.16a8.0.1
123.234.345
123.123
11.11.1
.192.168.0.1
.192.168.0.
.192.168.0
....
.......
........
"""#
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