import Foundation
let pattern = #"\(([^()]*)\)"#
let regex = try! NSRegularExpression(pattern: pattern)
let testString = #"(((119.7500182 25.3, 119.7730556 25.3, 119.7730556 25.3833333, 119.9672222 25.3833333, 119.9672222 25.1583333, 119.6125 25.1583333, 119.6125 25.3, 119.7500182 25.3)), ((119.8672912 25.8163889, 119.9672222 25.8163889, 119.9672222 25.6916667, 119.8561111 25.6916667, 119.8561111 25.8163889, 119.8672912 25.8163889)), ((119.4675 25.6008704, 119.4675 25.8163889, 119.5633333 25.8163889, 119.5633333 25.65, 119.5833333 25.65, 119.5833333 25.4805556, 119.6033333 25.4805556, 119.6033333 25.4316667, 119.4675 25.4316667, 119.4675 25.6008704)))"#
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