我正在尝试创建一个正则表达式来根据这些条件验证用户名:
_username
/ username_
/ .username
/ username.
)。user_.name
)。user__name
/ user..name
)。这是我到目前为止所做的事情;听起来它强制执行所有标准规则但第5条规则。我不知道如何添加第五条规则:
^[a-zA-Z0-9]+([._]?[a-zA-Z0-9]+)*$
答案 0 :(得分:156)
^(?=.{8,20}$)(?![_.])(?!.*[_.]{2})[a-zA-Z0-9._]+(?<![_.])$
└─────┬────┘└───┬──┘└─────┬─────┘└─────┬─────┘ └───┬───┘
│ │ │ │ no _ or . at the end
│ │ │ │
│ │ │ allowed characters
│ │ │
│ │ no __ or _. or ._ or .. inside
│ │
│ no _ or . at the beginning
│
username is 8-20 characters long
答案 1 :(得分:8)
Phillip的答案略有修改,修复了最新的要求
^[a-zA-Z0-9]([._](?![._])|[a-zA-Z0-9]){6,18}[a-zA-Z0-9]$
答案 2 :(得分:7)
我猜你必须在这里使用Lookahead表达式。 http://www.regular-expressions.info/lookaround.html
尝试
^[a-zA-Z0-9](_(?!(\.|_))|\.(?!(_|\.))|[a-zA-Z0-9]){6,18}[a-zA-Z0-9]$
[a-zA-Z0-9]
字母数字那么(
_(?!\.)
a _后面没有a。 OR
\.(?!_)
a。没有后跟_ OR
[a-zA-Z0-9]
字母数字)FOR
{6,18}
最少6到最多18倍然后
[a-zA-Z0-9]
字母数字
(第一个字符是alphanum,然后是6到18个字符,最后一个字符是alphanum,6 + 2 = 8,18 + 2 = 20)
答案 3 :(得分:1)
尽管我喜欢正则表达式,但我认为可读性是有限的
所以我建议
new Regex("^[a-z._]+$", RegexOptions.IgnoreCase).IsMatch(username) &&
!username.StartsWith(".") &&
!username.StartsWith("_") &&
!username.EndsWith(".") &&
!username.EndsWith("_") &&
!username.Contains("..") &&
!username.Contains("__") &&
!username.Contains("._") &&
!username.Contains("_.");
它更长,但不需要维护者打开expresso来理解。
当然你可以评论一个很长的正则表达式,但是那个读它的人必须依靠信任.......
答案 4 :(得分:1)
^ [A-Z0-9 _-] {3,15} $
^#行的开头
[a-z0-9_-]#匹配列表中的字符和符号,a-z,0-9,下划线,连字符
{3,15}#长度至少3个字符,最大长度15个
$#行结束
答案 5 :(得分:1)
private static final Scanner scan = new Scanner(System.in);
public static void main(String[] args) {
int n = Integer.parseInt(scan.nextLine());
while (n-- != 0) {
String userName = scan.nextLine();
String regularExpression = "^[[A-Z]|[a-z]][[A-Z]|[a-z]|\\d|[_]]{7,29}$";
if (userName.matches(regularExpression)) {
System.out.println("Valid");
} else {
System.out.println("Invalid");
}
}
}
答案 6 :(得分:0)
对不起,我是从我自己的库生成的,它使用的语法对Dart / Javascript / Java / Python有效,但无论如何,这里有:
(?:^)(?:(?:[abcdefghijklmnopqrstuvwxyzABCDEFGHIJKlMNOPQRSTUVWXYZ0123456789]){1,1})(?!(?:(?:(?:(?:_\.){1,1}))|(?:(?:(?:__){1,1}))|(?:(?:(?:\.\.){1,1}))))(?:(?:(?:(?:[abcdefghijklmnopqrstuvwxyzABCDEFGHIJKlMNOPQRSTUVWXYZ0123456789._]){1,1})(?!(?:(?:(?:(?:_\.){1,1}))|(?:(?:(?:__){1,1}))|(?:(?:(?:\.\.){1,1}))))){6,18})(?:(?:[abcdefghijklmnopqrstuvwxyzABCDEFGHIJKlMNOPQRSTUVWXYZ0123456789]){1,1})(?:$)
我的图书馆代码:
var alphaNumeric = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z", "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "l", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9"];
var allValidCharacters = new List.from(alphaNumeric);
allValidCharacters.addAll([".", "_"]);
var invalidSequence = (r) => r
.eitherString("_.")
.orString("__")
.orString("..");
var regex = new RegExpBuilder()
.start()
.exactly(1).from(alphaNumeric).notBehind(invalidSequence)
.min(6).max(18).like((r) => r.exactly(1).from(allValidCharacters).notBehind(invalidSequence))
.exactly(1).from(alphaNumeric)
.end()
.getRegExp();
答案 7 :(得分:0)
这个应该可以解决问题:
if (Regex.IsMatch(text, @"
# Validate username with 5 constraints.
^ # Anchor to start of string.
# 1- only contains alphanumeric characters , underscore and dot.
# 2- underscore and dot can't be at the end or start of username,
# 3- underscore and dot can't come next to each other.
# 4- each time just one occurrence of underscore or dot is valid.
(?=[A-Za-z0-9]+(?:[_.][A-Za-z0-9]+)*$)
# 5- number of characters must be between 8 to 20.
[A-Za-z0-9_.]{8,20} # Apply constraint 5.
$ # Anchor to end of string.
", RegexOptions.IgnorePatternWhitespace))
{
// Successful match
} else {
// Match attempt failed
}
答案 8 :(得分:0)
^(?=.{4,20}$)(?:[a-zA-Z\d]+(?:(?:\.|-|_)[a-zA-Z\d])*)+$
您可以测试正则表达式here
答案 9 :(得分:-1)
function isUserName(val){
let regUser=/^[a-zA-Z0-9](_(?!(\.|_))|\.(?!(_|\.))|[a-zA-Z0-9]){6,18}[a-zA-Z0-9]$/;
if(!regUser.test(val)){
return 'Name can only use letters,numbers, minimum length is 8 characters';
}
}