import Foundation
let pattern = #"([\'\"]?[\[])(.*)([\]][\'\"]?)"#
let regex = try! NSRegularExpression(pattern: pattern)
let testString = #"""
["a", "b", "c", "1"]
"["a", "b", "c", "1"]"
'["a", "b", "c", "1"]'
['a', 'b', 'c', '1']
"['a', 'b', 'c', '1']"
'['a', 'b', 'c', '1']'
['a', 'b', 'c', 1]
"['a', 'b', 'c', 1]"
'['a', 'b', 'c', 1]'
['a', 'b', 'c', 1]
"['a', 'b', 'c', 1]"
'['a', 'b', 'c', 1]'
['a', 'b', c, 1]
"['a', 'b', c, 1]"
'['a', 'b', c, 1]'
[a, b, c, 1]
"[a, b, c, 1]"
'[a, b, c, 1]'
['a', 'b', 'c', 1]
"['a', 'b', 'c', 1]"
'['a', 'b', 'c', 1]'
"""#
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