import Foundation
let pattern = #"\b(?:(?:0[1-9]|1\d|2[0-8])\/(?:02)\/(?:\d+)|(?:0[1-9]|1\d|2\d)\/(?:02)\/(?:(?:\d*?(?:(?:(?!00)[02468][048]|[13579][26])|(?:(?:[02468][048]|[13579][26])00))|[48]00|[48])(?=\D))|(?:0[1-9]|1\d|2\d|30)\/(?:0[469]|11)\/(?:\d+)|(?:0[1-9]|1\d|2\d|3[01])\/(?:0[13578]|1[02])\/(?:\d+))\ (?:(?:0\d|1[01]):(?:[0-5]\d):(?:[0-5]\d)(?:\.(?:\d{3}))?\ ?(?:(?i)[ap]m|[ap]\.m\.(?-i))|(?:[01]\d|2[0-3]):(?:[0-5]\d):(?:[0-5]\d)(?:\.(?:\d{3}))?)\b"#
let regex = try! NSRegularExpression(pattern: pattern, options: .allowCommentsAndWhitespace)
let testString = #"""
29/02/2016 23:10:59 - valid
29/02/2017 23:10:59 - invalid (not leap year)
30/03/2016 23:10:59 - valid
"""#
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