import Foundation
let pattern = #"(?:(?:[a-f0-9]*)?:){1,7}[a-f0-9]*"#
let regex = try! NSRegularExpression(pattern: pattern, options: [.anchorsMatchLines, .caseInsensitive])
let testString = #"""
[::1]
[:a:1]
[:bh:lh:1]
[1:1:fEd1:1:abdc:1]
[::ffff:192.0.2.128]
[0000:0000:0000:0000:0000:0000:0000:0001]
[2001:0db8:0000:0000:0000:ff00:0042:8329]
[2001:db8:0:0:0:ff00:42:8329]
[2001:db8::ff00:42:8329]
[2001:0db8:0000:0000:0000:ff00:0042:8329]
::1
:a:1
:bh:lh:1
1:1:fEd1:1:abdc:1
::ffff:192.0.2.128
0000:0000:0000:0000:0000:0000:0000:0001
2001:0db8:0000:0000:0000:ff00:0042:8329
2001:db8:0:0:0:ff00:42:8329
2001:db8::ff00:42:8329
2001:0db8:0000:0000:0000:ff00:0042:8329
United Kingdom , London
IP: 91.149.202.91
IPv6: 2605:e440:8::2:1b
http://91.149.202.91/1000MB.test
http://[2605:e440:8::2:1b]/1000MB.test
"""#
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