regex - 需要正则表达式来验证用户名

Question

需要一个正则表达式来验证用户名:

应该允许尾随空格但不允许字符之间的空格

必须至少包含一个字母，可以包含字母和数字

最多 7-15 个字符(字母数字)

不能包含特殊字符

允许下划线

不知道如何做到这一点。任何帮助表示赞赏。谢谢你。

这是我使用的，但它允许字符之间有空格

"(?=.*[a-zA-Z])[a-zA-Z0-9_]{1}[_a-zA-Z0-9\\s]{6,14}"

示例:用户名
用户名中不允许有空格

Answer 1

尝试这个:

foundMatch = Regex.IsMatch(subjectString, @"^(?=.*[a-z])\w{7,15}\s*$", RegexOptions.IgnoreCase);

也允许使用 _因为您在尝试中允许这样做。

所以基本上我使用三个规则。一个检查是否至少存在一个字母。
另一个检查字符串是否仅包含字母加上 _最后我接受尾随空格和至少 7 个最多 15 个 alpha 的空格。你在一个很好的轨道上。坚持下去，你也会在这里回答问题:)

故障:

    "
^           # Assert position at the beginning of the string
(?=         # Assert that the regex below can be matched, starting at this position (positive lookahead)
   .        # Match any single character that is not a line break character
      *     # Between zero and unlimited times, as many times as possible, giving back as needed (greedy)
   [a-z]    # Match a single character in the range between “a” and “z”
)
\w          # Match a single character that is a “word character” (letters, digits, etc.)
   {7,15}   # Between 7 and 15 times, as many times as possible, giving back as needed (greedy)
\s          # Match a single character that is a “whitespace character” (spaces, tabs, line breaks, etc.)
   *        # Between zero and unlimited times, as many times as possible, giving back as needed (greedy)
$           # Assert position at the end of the string (or before the line break at the end of the string, if any)
"