Question

我的iPhone应用程序中有1 UITextfield个密码。

我想通过以下验证来验证此文本字段。

必须至少10个字符
必须至少包含一个小写字母，一个大写字母，一个数字和一个特殊字符
有效的特殊字符为 - @#$%^&+=^.*(?=.{10,})(?=.*d)(?=.*[a-z])(?=.*[A-Z])(?=.*[@#$%^&+=]).*$

如何根据上述要求限制UITextField？

Answer 1

我就是这样做的。验证应在用户输入密码时结束时进行，而不是介于两者之间。我不会使用NSRegularExpression。

-(void)textFieldDidEndEditing:(UITextField *)textField{
   int numberofCharacters = 0;
   BOOL lowerCaseLetter,upperCaseLetter,digit,specialCharacter = 0;
   if([textField.text length] >= 10)
   {
      for (int i = 0; i < [textfield.text length]; i++) 
      {
         unichar c = [textfield.text characterAtIndex:i];
         if(!lowerCaseLetter)
         {
           lowerCaseLetter = [[NSCharacterSet lowercaseLetterCharacterSet] characterIsMember:c];
         }
         if(!upperCaseLetter)
         {
           upperCaseLetter = [[NSCharacterSet uppercaseLetterCharacterSet] characterIsMember:c];
         }
         if(!digit)
         {
            digit = [[NSCharacterSet decimalDigitCharacterSet] characterIsMember:c];
         }
         if(!specialCharacter)
         {
            specialCharacter = [[NSCharacterSet symbolCharacterSet] characterIsMember:c];
         }
      }

      if(specialCharacter && digit && lowerCaseLetter && upperCaseLetter)
      {
         //do what u want
      }
      else
      { 
         UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Error"
                                                        message:@"Please Ensure that you have at least one lower case letter, one upper case letter, one digit and one special character"
                                                       delegate:nil cancelButtonTitle:@"OK" otherButtonTitles:nil];
        [alert show];
      }

   }
   else
   {
      UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Error"
                                                        message:@"Please Enter at least 10 password"
                                                       delegate:nil cancelButtonTitle:@"OK" otherButtonTitles:nil];
        [alert show];
   }
}

希望这会有所帮助......

Answer 2

您也可以使用Regex执行此操作。以下是我为您提供的一些示例：

// *** Validation for Password ***

    // "^(?=.*[A-Za-z])(?=.*\\d)[A-Za-z\\d]{8,}$" --> (Minimum 8 characters at least 1 Alphabet and 1 Number)
    // "^(?=.*[A-Za-z])(?=.*\\d)(?=.*[$@$!%*#?&])[A-Za-z\\d$@$!%*#?&]{8,16}$" --> (Minimum 8 and Maximum 16 characters at least 1 Alphabet, 1 Number and 1 Special Character)
    // "^(?=.*[a-z])(?=.*[A-Z])(?=.*\\d)[a-zA-Z\\d]{8,}$" --> (Minimum 8 characters at least 1 Uppercase Alphabet, 1 Lowercase Alphabet and 1 Number)
    // "^(?=.*[a-z])(?=.*[A-Z])(?=.*\\d)(?=.*[$@$!%*?&])[A-Za-z\\d$@$!%*?&]{8,}" --> (Minimum 8 characters at least 1 Uppercase Alphabet, 1 Lowercase Alphabet, 1 Number and 1 Special Character)
    // "^(?=.*[a-z])(?=.*[A-Z])(?=.*\\d)(?=.*[$@$!%*?&])[A-Za-z\\d$@$!%*?&]{8,10}" --> (Minimum 8 and Maximum 10 characters at least 1 Uppercase Alphabet, 1 Lowercase Alphabet, 1 Number and 1 Special Character)

您可以从列表中找到

第四 ，以下代码段显示了如何使用它：

-(BOOL)isValidPassword:(NSString *)passwordString
{
    NSString *stricterFilterString = @"^(?=.*[a-z])(?=.*[A-Z])(?=.*\\d)(?=.*[$@$!%*?&])[A-Za-z\\d$@$!%*?&]{10,}";
    NSPredicate *passwordTest = [NSPredicate predicateWithFormat:@"SELF MATCHES %@", stricterFilterString];
    return [passwordTest evaluateWithObject:passwordString];
}

使用方法：

if(![self isValidPassword:txtPassword.text]) {
    /* Show alert: "Password must be minimum 10 characters,
       at least 1 Uppercase Alphabet, 1 Lowercase Alphabet, 
       1 Number and 1 Special Character" */
}
else {
    // Password is valid
}

Answer 3

条件：密码应包含至少8个字符，1个大写和1个数字

Swift 3中的解决方案

您可以像这样编写字符串扩展，

extension String {
    func isValidPassword() -> Bool {
        let regularExpression = "^(?=.*[a-z])(?=.*[A-Z])(?=.*\\d)(?=.*[$@$!%*?&])[A-Za-z\\d$@$!%*?&]{8,}"
        let passwordValidation = NSPredicate.init(format: "SELF MATCHES %@", regularExpression)

        return passwordValidation.evaluate(with: self)
    }
}

//Example 1

var password = "@Abcdef011" //string from UITextField (Password)
password.isValidPassword() // -> true

//Example 2

var password = "Abcdef011" //string from UITextField 
password.isValidPassword() // -> false

或者你可以写这样的函数，

func validate(password: String) -> Bool
{
    let regularExpression = "^(?=.*[a-z])(?=.*[A-Z])(?=.*\\d)(?=.*[$@$!%*?&])[A-Za-z\\d$@$!%*?&]{8,}"

    let passwordValidation = NSPredicate.init(format: "SELF MATCHES %@", regularExpression)

    return passwordValidation.evaluate(with: password)
}

这会给你相同的结果。

Answer 4

您可以使用以下函数验证密码验证，只需传递密码字符串，这将返回BOOL值。

-(BOOL) isPasswordValid:(NSString *)pwd {

    NSCharacterSet *upperCaseChars = [NSCharacterSet characterSetWithCharactersInString:@"ABCDEFGHIJKLKMNOPQRSTUVWXYZ"];
    NSCharacterSet *lowerCaseChars = [NSCharacterSet characterSetWithCharactersInString:@"abcdefghijklmnopqrstuvwxyz"];

    //NSCharacterSet *numbers = [NSCharacterSet characterSetWithCharactersInString:@"0123456789"];

    if ( [pwd length]<6 || [pwd length]>20 )
        return NO;  // too long or too short
    NSRange rang;
    rang = [pwd rangeOfCharacterFromSet:[NSCharacterSet letterCharacterSet]];
    if ( !rang.length )
        return NO;  // no letter
    rang = [pwd rangeOfCharacterFromSet:[NSCharacterSet decimalDigitCharacterSet]];
    if ( !rang.length )
        return NO;  // no number;
    rang = [pwd rangeOfCharacterFromSet:upperCaseChars];
     if ( !rang.length )
         return NO;  // no uppercase letter;
    rang = [pwd rangeOfCharacterFromSet:lowerCaseChars];
    if ( !rang.length )
        return NO;  // no lowerCase Chars;
    return YES;
}

Answer 5

Swift 3

检查密码是否强大？

长度大于等于8
小写
大写
十进制数字
特殊字符，如！@＃$％^＆amp; *（）_- +是可选的

为什么我不使用正则表达式？

因为很难在常规中支持保留字符表达式语法。

 func isValidated(_ password: String) -> Bool {
     var lowerCaseLetter: Bool = false
     var upperCaseLetter: Bool = false
     var digit: Bool = false
     var specialCharacter: Bool = false

     if password.characters.count  >= 8 {
         for char in password.unicodeScalars {
             if !lowerCaseLetter {
                 lowerCaseLetter = CharacterSet.lowercaseLetters.contains(char)
             }
             if !upperCaseLetter {
                 upperCaseLetter = CharacterSet.uppercaseLetters.contains(char)
             }
             if !digit {
                 digit = CharacterSet.decimalDigits.contains(char)
             }
             if !specialCharacter {
                 specialCharacter = CharacterSet.punctuationCharacters.contains(char)
             }
         }
         if specialCharacter || (digit && lowerCaseLetter && upperCaseLetter) {
             //do what u want
             return true
         }
         else {
             return false
         }
     }
     return false
 }
 let isVaildPass:Bool = isValidated("Test**00+-")
 print(isVaildPass)

Answer 6

使用正则表达式（NSRegularExpression类有关于如何编写模式本身的文档）然后：

- (BOOL)textField:(UITextField *)theTextField shouldChangeCharactersInRange:(NSRange)range replacementString:(NSString *)string {
    //delete
    if (string.length == 0) {
        return YES;
    }

    if (self.regEx) {
        NSMutableString* check = [NSMutableString stringWithString:theTextField.text];
        [check replaceCharactersInRange:range withString:string];
        NSTextCheckingResult* match = [self.regEx firstMatchInString:check options:0 range:NSMakeRange(0, [check length])];
        if (match.range.length != check.length) {
            return NO;
        }
    }
}

警告：以这种方式限制输入对用户来说确实很困惑。您输入并输入类型，而您输入的字符不会出现！

我可能会选择测试区旁边的小红色（！），但我总是允许输入！

Answer 7

我有一个优雅的表格（如注册）解决方案，你有很多验证

我在我的自定义UITextField中有出口：

@IBInspectable var regexpValidation: String? = nil

在故事板中，我可以通过属性检查器访问它并将regexp字符串放入（对于电子邮件）：

[a-z0-9!#$%&'*+/=?^_ {|}〜 - ] +（？：！？[A-Z0-9＃$％＆安培;'* + / = ^ _ {|}~-]+)*@(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?

然后在我的子类中我有这个计算的var：

@IBInspectable var regexpValidation：String？ = nil //可选，在InterfaceBuilder中设置

var inputIsValid: Bool {
    get {
        if let expr = regexpValidation {
           return (text.rangeOfString(expr, options: NSStringCompareOptions.RegularExpressionSearch, range: nil, locale: nil) != nil)
        } else {
            return true
        }

    }
}

可以像这样使用：

   override func resignFirstResponder() -> Bool {
        if (inputIsValid) {
            return super.resignFirstResponder()
        }
        else {
            text = ""
            return false
        }
    }

Answer 8

对我来说最好的方法是使用NSPredicate和正则表达式。这是你的案例的正则表达式：^(?=.{10,})(?=.*[0-9])(?=.*[a-zA-Z])([@#$%^&=a-zA-Z0-9_-]+)$

目标C代码：

NSString *regex = @"^(?=.{10,})(?=.*[0-9])(?=.*[a-zA-Z])([@#$%^&=a-zA-Z0-9_-]+)$";

NSPredicate *passwordTest = [NSPredicate predicateWithFormat:@"SELF MATCHES %@", regex];

BOOL isValid = [passwordTest evaluateWithObject:yourTextfield.text];

Answer 9

您需要使用UITextField

的委托方法编写验证代码

- (BOOL)textField:(UITextField *)textField shouldChangeCharactersInRange:(NSRange)range replacementString:(NSString *)string

您可能希望参考实施的链接很少

how to use regular expression in iOS sdk

iOS TextField Validation

Answer 10

使用control:isValidObject:协议的NSTextFieldDelegate方法，该方法允许您验证NSTextField的值。假设您已正确配置了所有接口构建器位和部分，您可以执行以下操作：

@interface PreferencesController : NSWindowController <NSTextFieldDelegate> {
    IBOutlet NSTextField *username, *password;
}

@end

@implementation PreferencesController

- (BOOL)control:(NSControl *)control isValidObject:(id)object
{
    if (control == password) {
      // Perform validation and return YES or NO
    }

    return YES;
}

@end

Answer 11

SWIFT 5使用RXSWIFT，这是一种更好，整洁且反应迅速的方法。

验证密码功能将如下所示。很明显，您可以根据需要添加任意多个条件。

 func validatePassword(password: String) -> (Bool, String) {
  //Minimum 8 characters at least 1 Alphabet and 1 Number:
var tuple: (Bool, String) = (true, "")
var string = "Requires atleast"

if(password.rangeOfCharacter(from: CharacterSet.letters) == nil){
    string = "uppercase"
    tuple = (false, string)

}
if(password.rangeOfCharacter(from: CharacterSet.decimalDigits) == nil){
    string += ", number"
       tuple = (false, string)

   }
if(password.count < 8 ){
    string += ", 8 chars"
       tuple = (false, string)

   }
return tuple }


 func isPasswordValid(in string: String) -> Observable<(Bool, String)> {
    return Observable.create { observer -> Disposable in

        let tuple = self.validation.validatePasswordForSymbol(password: string)

        observer.onNext(tuple)


        return Disposables.create()
    }
}

您可以根据您的体系结构在viewModel或VC中使用上述功能。然后在您的VC中调用以下相同的函数。

passwordTextField.rx.text
            .orEmpty //1
            .filter { $0.count >= 1 } //2
            .flatMap { self.isPasswordValid(in: $0) }
            .subscribe(onNext: { result in
                print("Valid password", result)
                //update UI here 
                // result will be like (false, "Requires atleast, 8 chars, number")
            }).disposed(by: disposeBag)

iOS中UITextField中的密码验证

11 个答案: