import Foundation
let pattern = #"^(?!.*?([a-z0-9])(?:.*?\1){2})(?=.*?[a-z])(?=.*?[0-9])(?!.*?(?:0(?:123|987)|1234|2345|3(?:456|210)|4(?:567|321)|5(?:678|432)|6(?:789|543)|7(?:890|654)|8765|9876))(?!.*?(?:abcd|bcde|cdef|d(?:efg|cba)|e(?:fgh|dcb)|f(?:ghi|edc)|g(?:hij|fed)|h(?:ijk|gfe)|i(?:jkl|hgf)|j(?:klm|ihg)|k(?:lmn|jih)|l(?:mno|kji)|m(?:nop|lkj)|nmlk|onml|ponm|rstu|stuv|tuvw|u(?:vwx|tsr)|v(?:wxy|uts)|w(?:vut)|xwvu|yxwv))[a-pr-y0-9]{8}$"#
let regex = try! NSRegularExpression(pattern: pattern, options: [.anchorsMatchLines, .caseInsensitive, .allowCommentsAndWhitespace])
let testString = #"""
12345678
asfkd2ls
abcd12js
ibicid13
I need a regex to validate a password. The constraints are :
-Maximum Embedded Spaces: 0
-Minimum Length: 8
-Maximum Length: 8
-Must not contain letters: Q, q, Z, z
-Maximum Occurrences of one Character/Number: 2
-Maximum Repetitive Character/Number: 2
-Maximum Sequential Character: 3
-Maximum Special Character: 0
-Minimum Alphabet: 1
-Minimum Numbers: 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