javascript 如何使用正则表达式匹配任意字母组合？
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How to match any combination of letters using regex?
提问by John_Sheares
How can I match letters a,b,c once in any combination and varying length like this:
如何以任意组合和不同长度匹配字母 a,b,c 一次，如下所示：
The expression should match these cases:
该表达式应与以下情况匹配：
abc
bc
a
b
bca
but should not match these ones:
但不应与这些匹配：
abz
aab
cc
x
回答by ?mega
Use regex pattern
使用正则表达式
\b(?!(?:.\B)*(.)(?:\B.)*)[abc]+\b
You can use this pattern with any set and size, just replace [abc]with desired set...
您可以将此图案与任何套装和尺寸一起使用，只需替换[abc]为所需的套装...
示例：
(above output is from myregextester)
（以上输出来自myregextester）
回答by phant0m
^(?=([^a]*a?[^a]*)$)(?=([^b]*b?[^b]*)$)(?=([^c]*c?[^c]*)$)[abc]{1,3}$
This workswith lookaheads.
这适用于lookaheads。
It includes this pattern in three variations: (?=([^a]*a?[^a]*)$)
它以三种变体形式包含此模式： (?=([^a]*a?[^a]*)$)
It says: There needs to be at most one afrom here (the beginning) until the end.
它说：a从这里（开始）到结束最多需要一个。
Combining lookaheadsand backreferences:
^([abc])((?!)([abc])((?!)(?!)[abc])?)?$
回答by Alan Moore
Just to round out the collection:
只是为了完善集合：
^(?:([abc])(?!.*))+$
Want to handle a larger set of characters? No problem:
想要处理更大的字符集？没问题：
^(?:([abcdefgh])(?!.*))+$
EDIT:Apparently I misread the question; you're not validating individual strings like "abc"and "ba", you're trying to find whole-word matches in a larger string. Here's how I would do that:
编辑：显然我误读了这个问题；您不是在验证像"abc"and这样的单个字符串"ba"，而是试图在更大的字符串中找到全字匹配。这是我将如何做到这一点：
\b(?:([abc])(?![abc]*))+\b
The tricky part is making sure the lookahead doesn't look beyond the end of the word that's currently being matched. For example, if I had left the lookahead as (?!.*\1), it would fail to match the abcin abc zabecause the lookahead would incorrectly flag the ain zaas a duplicate of the ain abc. Allowing the lookahead to look only at valid characters ([abc]*) keeps it on a sufficiently short leash. And if there are invalid characters in the current word, it's not the lookahead's job to spot them anyway.
棘手的部分是确保前瞻不会超出当前匹配的单词的末尾。例如，如果我离开了超前的(?!.*\1)，它会失败，以配合abc中abc za，因为先行将错误旗的a中za作为的副本a中abc。允许前瞻只查看有效字符 ( [abc]*) 使其保持足够短的皮带。如果当前单词中存在无效字符，则无论如何发现它们不是前瞻的工作。
(Thanks to Honest Abefor bringing this back to my attention.)
（感谢Honest Abe让我重新注意到这一点。）
回答by Kirill Polishchuk
回答by Bohemian
^(?=(.*a.*)?$)(?=(.*b.*)?$)(?=(.*c.*)?$)[abc]{,3}$
The anchored look-aheads limit the number of occurrences of each letter to one.
锚定前瞻将每个字母的出现次数限制为一个。
回答by Aimon Bustardo
I linked it in comment (this is sort of a dupe of How can I find repeated characters with a regex in Java?).. but to be more specific.. the regex:
我在评论中链接了它（这有点像如何在 Java 中找到带有正则表达式的重复字符？）..但更具体地说..正则表达式：
(\w)+
Will match any two or more of the same character. Negate that and you have your regex.
将匹配任意两个或多个相同字符。否定它，你就有了你的正则表达式。
