import Foundation
let pattern = #"(\/?[S])|(\,)|([A-Z]\~[A-Z])|([A-Z]\-[A-Z]$)|([A-Z]\/[A-Z])|([A-Z]{2,}$)"#
let regex = try! NSRegularExpression(pattern: pattern, options: .anchorsMatchLines)
let testString = #"""
60-B-1230S
60-B-1230ABCDEF
60-B-1230A-F
60-B-1230A/F
60-B-1230A~H,0290A~H
35-K-0101A/B,0201A/B,0301A/B
35-E-0102A~C, 0202A~C, 0302A~C
35-E-0103A~H, 0203A~H, 0303A~H
35-E-0101A/B, 0201A/B, 0301A/B
35-E-0104A/B, 0204A/B, 0304A/B
35-P-0101A/B, 0201A/B, 0301A/B
35-P-0102A/B, 0202A/B, 0302A/B
35-P-0103A/B, 0203A/B, 0303A/B
35-P-0104A/B, 0204A/B, 0304A/B
35-P-0105A/B, 0205A/B, 0305A/B
35-P-0106/S, 0206, 0306
35-P-0107/S, 0207, 0307
35-P-0109A/B, 0209A/B, 0309A/B
35-P-0111A/B
35-P-0401A/B
92-P-0101A/B
35-V-0108, 0208, 0308
"""#
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