import Foundation
let pattern = #"^(?!^[0-9])(?![0-9][AZ])(?![AZ0-9*]*\*\*)(?![AZ0-9*]*\*[AZ])((?=[AZ0-9]{6})|(?=[AZ0-9*]*\*))(?![AZ0-9*]*[0-9][A-Z])[AZ0-9*]{1,6}$"#
let regex = try! NSRegularExpression(pattern: pattern, options: .anchorsMatchLines)
let testString = #"""
^(?![0-9])(?![0-9][AZ])(?![AZ0-9*]*\*\*)(?![AZ0-9*]*\*[AZ0])(?=[AZ0-9*]*\*)(?![AZ0-9*]*[0-9][A-Z])[AZ0-9*]{1,6}$
Broken down:
^(?![0-9])
(?![0-9][AZ])
(?![AZ0-9*]*\*\*)
(?![AZ0-9*]*\*[AZ0])
(?=[AZ0-9*]*\*)
(?![AZ0-9*]*[0-9][A-Z])
[AZ0-9*]{1,6}$
Check these for patterns
Following would be valid
*
*1
*1*
*12345
A*
A*2
A*2*
A1*3
A1*3*
A12*4
A*2
A*23
A*234
A*2345
A12*45
A1234*
A*2345
A12*4*
A*23*5
A123*5
*12345
*1234*
*1*3*5
*12*4*
*1*3*5
*123*5
A12345
Examples of invalid entries: <Any ** entry> - <Any * A-Z character> - <Any value longer than 6>
A1
A12
A123
A1234
*A1234
B*
B1*
BA*45
B*2345
**
*1A
*1A1*
*A2345
*123456
1234*
A
**12
*1**
*1*3**
*12**
*1*3*56
*123**
*12*45*
A*23456
"""#
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