判断邮箱是否合法的代码

来源:互联网 发布:轴承数据库 编辑:程序博客网 时间:2024/05/01 16:25

  1. 1...代码用了断言, 如果是正则达人,那更容易理解。

  2. BOOL NSStringIsValidEmail(NSString *checkString)    
  3. {    
  4.     NString *stricterFilterString = @"[A-Z0-9a-z._%+-]+@[A-Za-z0-9.-]+//.[A-Za-z]{2,4}";    
  5.     NSString *laxString = @".+@.+/.[A-Za-z]{2}[A-Za-z]*";    
  6.     NSString *emailRegex = stricterFilter ? stricterFilterString : laxString;    
  7.     NSPredicate *emailTest = [NSPredicate predicateWithFormat:@"SELF MATCHES %@", emailRegex];    
  8.     return [emailTest evaluateWithObject:checkString];    
  9. }   


2..介绍另一种方法,通过字符串操作来检查邮箱合法性


-(BOOL)validateEmail:(NSString*)email{
    
    if( (0 != [email rangeOfString:@"@"].length) &&  (0 != [email rangeOfString:@"."].length) )
    {
        NSMutableCharacterSet *invalidCharSet = [[[[NSCharacterSet alphanumericCharacterSet] invertedSet]mutableCopy]autorelease];
        [invalidCharSet removeCharactersInString:@"_-"];
        
        NSRange range1 = [email rangeOfString:@"@" options:NSCaseInsensitiveSearch];
        
        // If username part contains any character other than "."  "_" "-"
        
        NSString *usernamePart = [email substringToIndex:range1.location];
        NSArray *stringsArray1 = [usernamePart componentsSeparatedByString:@"."];
        for (NSString *string in stringsArray1) {
            NSRange rangeOfInavlidChars=[string rangeOfCharacterFromSet: invalidCharSet];
            if(rangeOfInavlidChars.length !=0 || [string isEqualToString:@""])
                return NO;
        }
        
        NSString *domainPart = [email substringFromIndex:range1.location+1];
        NSArray *stringsArray2 = [domainPart componentsSeparatedByString:@"."];
        
        for (NSString *string in stringsArray2) {
            NSRange rangeOfInavlidChars=[string rangeOfCharacterFromSet:invalidCharSet];
            if(rangeOfInavlidChars.length !=0 || [string isEqualToString:@""])
                return NO;
        }
        
        return YES;
    }
    else // no ''@'' or ''.'' present
        return NO;
}

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