regex - 不以点开头的单词的正则表达式

Question

我需要为不以点开头的单词创建正则表达式，它可能包含任何字母、空格和点。

例如：sample,sample.test,sample test

正则表达式不应允许 .sample,sample.,sample .test

如何为此生成正则表达式？

Answer 1

这个正则表达式：^[^.][\p{L} .]+$应该匹配你所追求的。

这^是一个锚点，它将指示正则表达式引擎从字符串的最开头开始匹配。[^.]将匹配任何 1 个不是句点 ( .) 的字符。[\p{L} .]+将匹配一个或多个字符，这些字符可以是字母（在此处显示的任何语言中）、空格或句点。最后，$将指示正则表达式在字符串末尾终止匹配。

编辑：根据您的评论问题，类似的东西应该是可测试的：^[^.][a-zA-Z .]+$。

Answer 2

用这个

\b\p{L}[\p{L}\s.]*\b

解释

@"
\b            # Assert position at a word boundary
\p{L}         # A character with the Unicode property “letter” (any kind of letter from any language)
[\p{L}\s.]    # Match a single character present in the list below
                 # A character with the Unicode property “letter” (any kind of letter from any language)
                 # A whitespace character (spaces, tabs, and line breaks)
                 # The character “.”
   *             # Between zero and unlimited times, as many times as possible, giving back as needed (greedy)
\b            # Assert position at a word boundary
"