我写一些可以用来创建Linux用户帐户的Web UI页面。此Web UI将用于CentOS 6(源自RHEL 6)。我发现关于什么构成有效的Linux用户名的不一致和不完整的信息。我去源，检查一个Linux shadow-utils源包，但我不能确保我正在看的版本实际上是作为CentOS 6的一部分相同。
原文链接：https://www.f2er.com/centos/374817.html下面是我目前使用的代码片段，其中包括从shadow-utils包4.1.4.3中复制/粘贴注释，加上一些我自己的注释，以及一个Java正则表达式搜索， utils源。
在chkname.c中引用的“is_valid_name()”检查显然不是从Linux上的useradd命令使用的，因为注释(和C代码源)不允许以数字开头的名称。但是，useradd允许创建一个类似“1234”的帐户。
我希望帮助调整从我现在到什么是更正确的，以及关于如何实现useradd.c与一些稍微不同的is_valid_name函数的信息。
谢谢！
艾伦
/** * Define constants for use in isNameLinuxCompatible(...) method. * * The source for the Linux compatible user name rule is is_valid_name(...) a function in the "shadow" package * for Linux. The source file for that function has a comment as follows: * User/group names must match [a-z_][a-z0-9_-]*[$] * That expression is a little loose/sloppy since * (1) the trailing $ sign is optional,and * (2) uppercase A-Z is also ok (and case is significant,'A' != 'a'). * * We deal with (1) by using the [$]? form where the ? means zero or more characters (aka "greedy"). * We deal with (2) by using the CASE_INSENSITIVE option. * * Another way to express this is: * 1st character: a-z_ required at least one char * chars other than first and last: a-z0-9_- optional * last character: $ optional * Max length is 31. Min length is 1. * * NOTE: The initial ^ and final $ below are important since we need the entire string to satisfy the rule,* from beginning to end. * * See http://download.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html for reference info on pattern matching. */ private static final String LINUX_USERNAME_REGEX = "^[a-z_][a-z0-9_-]*[$]?$"; private static final Pattern LINUX_USERNAME_PATTERN = Pattern.compile(LINUX_USERNAME_REGEX,Pattern.CASE_INSENSITIVE); private static final int LINUX_USERNAME_MINLENGTH = 1; private static final int LINUX_USERNAME_MAXLENGTH = 31; /** * See if username is compatible with standard Linux rules for usernames,in terms of length and * in terms of content. * * @param username the name to be checked for validity * @return true if Linux compatible,else false */ public boolean isNameLinuxCompatible (final String username) { boolean nameOK = false; if (username != null) { int len = username.length(); if ((len >= LINUX_USERNAME_MINLENGTH) && (len <= LINUX_USERNAME_MAXLENGTH)) { Matcher m = LINUX_USERNAME_PATTERN.matcher(username); nameOK = m.find(); } } return (nameOK); }
