字母在 abcd 范围内的正则表达式至少出现一次
您将如何编写一个正则表达式来返回包含每个字母序列（例如 abcd）至少一次的所有单词（除了所需的子序列之外还可能包含其他字母）？
abcd 必须是该单词的子序列。
谢谢！
Ps 与 lex 一起使用
%{
%}
delim [ \t\n]
ws {delim}+
lc [a-z]
%%
{ws} {/* no action taken */}
(?={lc}*a)(?={lc}*b)(?={lc}*c)(?={lc}*d) { /* some code */ }
%%
How would you write a regular expression that will return all words that contain each of a sequence of letters (such as abcd) at least once (may also contain other letters besides the required subsequence)?
abcd must be a subsequence of the word.
Thanks!
P.s. for use with lex
%{
%}
delim [ \t\n]
ws {delim}+
lc [a-z]
%%
{ws} {/* no action taken */}
(?={lc}*a)(?={lc}*b)(?={lc}*c)(?={lc}*d) { /* some code */ }
%%
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评论（3）
使用前瞻检查单词中所有字母是否存在的简单模式：
您可能需要
^...$而不是\b...\b如果您想要验证该单词，而不是捕获它，并且您可能希望将\w更改为您可接受的字母表。每个
(?=\w*a)标记都是一个先行 - 它检查前面是否有字母和a，但它不会前进 - 下一个条件 < code>b，再次从头开始检查。在我们检查完所有字母后，最后一个\w+是实际捕获单词的内容。工作示例：http://rubular.com/r/L00DTpE813
另请参阅：前向和后向零宽度断言
A simple pattern to check the existence of all letters in a word, using lookaheads:
You may want
^...$instead of\b...\bif you want to validate the word, rather than capture it, and you may want to change\wto your acceptable alphabet.Each
(?=\w*a)token is a lookahead - it checks there are letters and anaahead, but it doesn't advance - the next condition,b, is checked from the start again. The last\w+is what actually captures the word, after we've checked all letters are there.Working example: http://rubular.com/r/L00DTpE813
See also: Lookahead and Lookbehind Zero-Width Assertions
此正则表达式匹配包含
abcd字符序列的所有单词。This regex matches all words containing
abcdcharacter sequence..如果
a、b、c和d必须按此顺序出现，则以下模式将匹配：
如果没有这样的顺序约束，正如 Kobi 回答的那样，前瞻将是
需要。
虽然flex有
尾随上下文，有限制，
并且无法满足复杂的前瞻目的。
If
a,b,canddhave to appear in this order, the following patternwill match:
If there isn't such order constraint, as Kobi answered, look-ahead will be
needed.
Though flex has
trailing context, there are limitations,
and won't meet complex look-ahead purpose.