检查字符串是否有 1 个数字、1 个字母，长度在 5-15 个字符之间

Question

我使用以下扩展名来确保字符串至少有 1 个数字、1 个字母和 5-15 个字符的长度，我觉得它可以更有效率。有什么建议吗？

func checkPassword(password： String) -> Bool{

if password.characters.count > 15 || password.characters.count < 5 {
return false
}

let capitalLetterRegEx  = ".*[A-Za-z]+.*"
let texttest = NSPredicate(format:"SELF MATCHES %@", capitalLetterRegEx)
let capitalresult = texttest.evaluate(with: password)
let numberRegEx  = ".*[0-9]+.*"
let texttest1 = NSPredicate(format:"SELF MATCHES %@", numberRegEx)
let numberresult = texttest1.evaluate(with: password)
let specialRegEx  = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLKMNOPQRSTUVWXYZ0123456789"
let texttest2 = NSPredicate(format:"SELF MATCHES %@", specialRegEx)
let specialresult = !texttest2.evaluate(with: password)


if !capitalresult || !numberresult || !specialresult  {
return false
}
return true
}

Answer 1

使用 Regex

正则表达式是一种方法，但如果使用它，我们可能会将您的规范合并到单个正则表达式搜索中，利用以下问答中的积极前瞻断言技术：

• 用于验证密码强度的正则表达式

在这里，使用正则表达式：

^(?=.*[A-Za-z])(?=.*[0-9])(?!.*[^A-Za-z0-9]).{5,15}$
// where:
// (?=.*[A-Za-z])     Ensures string has at least one letter.
// (?=.*[0-9])        Ensures string has at least one digit.
// (?!.*[^A-Za-z0-9]) Ensures string has no invalid (non-letter/-digit) chars.
// .{5,15}            Ensures length of string is in span 5...15.

我还包含一个负面的前瞻断言 (?!...)，以使任何无效字符的密码无效。

我们可以按如下方式实现正则表达式搜索：

extension String {
func isValidPassword() -> Bool {
let regexInclude = try! NSRegularExpression(pattern: "^(?=.*[A-Za-z])(?=.*[0-9])(?!.*[^A-Za-z0-9]).{5,15}$")
return regexInclude.firstMatch(in: self, options: [], range: NSRange(location: 0, length: characters.count)) != nil
}
}
let pw1 = "hs1bés2"  // invalid character
let pw2 = "12345678" // no letters
let pw3 = "shrt"     // short
let pw4 = "A12345"   // ok
print(pw1.isValidPassword()) // false
print(pw2.isValidPassword()) // false
print(pw3.isValidPassword()) // false
print(pw4.isValidPassword()) // true

---

使用Set/CharacterSet

Swift 原生方法是使用显式指定的Character集：

extension String {
private static var numbersSet = Set("1234567890".characters)
private static var alphabetSet = Set("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLKMNOPQRSTUVWXYZ".characters)
func isValidPassword() -> Bool {
return 5...15 ~= characters.count &&
characters.contains(where: String.numbersSet.contains) &&
characters.contains(where: String.alphabetSet.contains)
}
}

或者，类似地，使用Foundation方法rangeOfCharacter(from:)超过CharacterSet的：

extension String {
private static var numbersSet = CharacterSet(charactersIn: "1234567890")
private static var alphabetSet = CharacterSet(charactersIn: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLKMNOPQRSTUVWXYZ")
func isValidPassword() -> Bool {
return 5...15 ~= characters.count &&
rangeOfCharacter(from: String.numbersSet) != nil &&
rangeOfCharacter(from: String.alphabetSet) != nil
}
}

如果您还想拒绝包含不在指定集中的任何字符的密码，则可以在集合的(倒置)并集上添加搜索操作(可能您还允许在此并集中包含一些特殊字符)。 例如，对于CharacterSet示例：

extension String {
private static var numbersSet = CharacterSet(charactersIn: "1234567890")
private static var alphabetSet = CharacterSet(charactersIn: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLKMNOPQRSTUVWXYZ")
func isValidPassword() -> Bool {
return 5...15 ~= characters.count &&
rangeOfCharacter(from: String.numbersSet.union(String.alphabetSet).inverted) == nil &&
rangeOfCharacter(from: String.numbersSet) != nil &&
rangeOfCharacter(from: String.alphabetSet) != nil
}
}
let pw1 = "hs1bés2"  // invalid character
let pw2 = "12345678" // no letter
let pw3 = "shrt"     // too short
let pw4 = "A12345"   // OK
print(pw1.isValidPassword()) // false
print(pw2.isValidPassword()) // false
print(pw3.isValidPassword()) // false
print(pw4.isValidPassword()) // true

---

使用模式匹配

只是为了讨论它，另一种方法是使用本机 Swift 模式匹配：

extension String {
private static var numberPattern = Character("0")..."9"
private static var alphabetPattern = Character("a")..."z"
func isValidPassword() -> Bool {
return 5...15 ~= characters.count &&
characters.contains { String.numberPattern ~= $0 } &&
lowercased().characters.contains { String.alphabetPattern ~= $0 }
}
}
let pw1 = "hs1bs2"
let pw2 = "12345678"
let pw3 = "shrt"
let pw4 = "A12345"
print(pw1.isValidPassword()) // true
print(pw2.isValidPassword()) // false
print(pw3.isValidPassword()) // false
print(pw4.isValidPassword()) // true

请注意，这种方法将允许带有变音符号(和类似符号)的字母作为最小 1 个字母规范通过，例如：

let diacritic: Character = "é"
print(Character("a")..."z" ~= diacritic) // true
let pw5 = "12345é6"
print(pw5.isValidPassword()) // true

由于这些包含Character范围"a"..."z";例如，请参阅以下线程中的出色答案：

• Swift 中的字符串和字符比较不区分区域设置是什么意思？

Answer 2

感谢您的回复。我用它们来创建这个：

extension String {
private static var numbersSet = Set("1234567890".characters)
private static var alphabetSet = Set("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLKMNOPQRSTUVWXYZ".characters)
func isValidPassword() -> Bool {
let characterset = CharacterSet(charactersIn: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLKMNOPQRSTUVWXYZ0123456789")
return 5...15 ~= characters.count &&
characters.contains(where: String.numbersSet.contains) &&
characters.contains(where: String.alphabetSet.contains) &&
self.rangeOfCharacter(from: characterset.inverted) != nil
}
}