import Foundation
// WARNING: You included a flag that Swift doesn't support: u
// When this flag is set, it makes the pattern and subject strings to be treated as unicode.
// Swift already treats the pattern and subject strings as unicode by default, so including this flag is redundant.
let pattern = ##"([0-9#][\x{20E3}])|[\x{00ae}\x{00a9}\x{203C}\x{2047}\x{2048}\x{2049}\x{3030}\x{303D}\x{2139}\x{2122}\x{3297}\x{3299}][\x{FE00}-\x{FEFF}]?|[\x{2190}-\x{21FF}][\x{FE00}-\x{FEFF}]?|[\x{2300}-\x{23FF}][\x{FE00}-\x{FEFF}]?|[\x{2460}-\x{24FF}][\x{FE00}-\x{FEFF}]?|[\x{25A0}-\x{25FF}][\x{FE00}-\x{FEFF}]?|[\x{2600}-\x{27BF}][\x{FE00}-\x{FEFF}]?|[\x{2900}-\x{297F}][\x{FE00}-\x{FEFF}]?|[\x{2B00}-\x{2BF0}][\x{FE00}-\x{FEFF}]?|[\x{1F000}-\x{1F6FF}][\x{FE00}-\x{FEFF}]?"##
let regex = try! NSRegularExpression(pattern: pattern)
let testString = #"""
⎛⎝ The Zone ⎠⎞
"""#
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