java 正则表达式 i_java-用于验证用户名的正则表达式

谁借莪１个温暖的怀抱￠ 2022-11-02 15:27 227阅读 0赞

java-用于验证用户名的正则表达式

我正在尝试创建一个正则表达式来根据以下条件验证用户名：

仅包含字母数字字符，下划线和点。

下划线和点不能在结尾或用户名的开头(例如user\_\_name/user..name/.username/username.)。

下划线和点不能彼此相邻(例如user\_\_name)。

下划线或点不能连续使用多次(例如user\_\_name/user..name)。

字符数必须在8到20之间。

到目前为止，这是我所做的； 听起来它强制执行所有准则规则，但第5条规则除外。 我不知道如何添加第五条规则：

^\[a-zA-Z0-9\]+(\[.\_\]?\[a-zA-Z0-9\]+)\*$

8个解决方案

181 votes

^(?=.\{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

Ωmega answered 2020-08-06T08:25:30Z

8 votes

对Phillip的答案稍作修改即可解决最新要求

^\[a-zA-Z0-9\](\[.\_\](?!\[.\_\])|\[a-zA-Z0-9\])\{6,18\}\[a-zA-Z0-9\]$

Patrick McDonald answered 2020-08-06T08:25:50Z

7 votes

我想您必须在这里使用Lookahead表达式。 \[http://www.regular-expressions.info/lookaround.html\]

尝试

\[a-zA-Z0-9\]

\[a-zA-Z0-9\]一个字母数字的THEN(

\[a-zA-Z0-9\] a \_后面没有。 要么

\[a-zA-Z0-9\]一个。 不跟\_ OR

\[a-zA-Z0-9\]字母数字)FOR

\[a-zA-Z0-9\]最小6到最大18倍THEN

\[a-zA-Z0-9\]一个字母数字

(第一个字符为字母数字，然后为6至18个字符，最后一个字符为字母数字，6 + 2 = 8，18 + 2 = 20)

PhilMasterG answered 2020-08-06T08:26:45Z

2 votes

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");

isurie answered 2020-08-06T08:27:00Z

1 votes

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，但是不需要维护人员打开expresso即可理解。

当然，您可以评论一个长的正则表达式，但是读过它的人必须依靠信任...。

James Kyburz answered 2020-08-06T08:27:33Z

1 votes

^ \[a-z0-9 \_-\] \{3,15\} $

^＃行首

\[a-z0-9\_-\]＃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，最大长度为15

$＃行尾

Mehul Jariwala answered 2020-08-06T08:28:11Z

0 votes

这应该可以解决问题：

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

ridgerunner answered 2020-08-06T08:28:31Z

0 votes

很抱歉，我是从我自己的库中生成的，并且它使用了对Dart / Javascript / Java / Python有效的语法，但是无论如何，这里是这样：

(?:^)(?:(?:\[abcdefghijklmnopqrstuvwxyzABCDEFGHIJKlMNOPQRSTUVWXYZ0123456789\])\{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()