如何匹配不在两个特殊字符之间的正则表达式? [英] How to match something with regex that is not between two special characters?

问题描述

我有一个这样的字符串:

I have a string like this:

a b c a b " a b " b a " a "

a b c a b " a b " b a " a "

如何匹配不是由 " 分隔的字符串的一部分的每个 a?我想匹配这里粗体的所有内容:

How do I match every a that is not part of a string delimited by "? I want to match everything that is bold here:

a bc a b " ab " b a " a "

a bc a b " ab " b a " a "

我想替换那些匹配项(或者更确切地说，通过用空字符串替换它们来删除它们)，因此删除引用的部分进行匹配将不起作用，因为我希望它们保留在字符串中.我正在使用 Ruby.

I want to replace those matches (or rather remove them by replacing them with an empty string), so removing the quoted parts for matching won't work, because I want those to remain in the string. I'm using Ruby.

推荐答案

假设引号正确平衡并且没有转义引号，那么很简单:

Assuming the quotes are correctly balanced and there are no escaped quotes, then it's easy:

result = subject.gsub(/a(?=(?:[^"]*"[^"]*")*[^"]*\Z)/, '')

当且仅当匹配的 a 前面有偶数个引号时，这会将所有 a 替换为空字符串.

This replaces all the as with the empty string if and only if there is an even number of quotes ahead of the matched a.

说明:

a        # Match a
(?=      # only if it's followed by...
 (?:     # ...the following:
  [^"]*" #  any number of non-quotes, followed by one quote
  [^"]*" #  the same again, ensuring an even number
 )*      # any number of times (0, 2, 4 etc. quotes)
 [^"]*   # followed by only non-quotes until
 \Z      # the end of the string.
)        # End of lookahead assertion

如果你可以在引号内转义引号(a "length: 2\"")，它仍然是可能的，但会更复杂:

If you can have escaped quotes within quotes (a "length: 2\""), it's still possible but will be more complicated:

result = subject.gsub(/a(?=(?:(?:\\.|[^"\\])*"(?:\\.|[^"\\])*")*(?:\\.|[^"\\])*\Z)/, '')

这与上面的正则表达式本质上是一样的，只是用(?:\\.|[^"\\])代替[^"]:

This is in essence the same regex as above, only substituting (?:\\.|[^"\\]) for [^"]:

(?:     # Match either...
 \\.    # an escaped character
|       # or
 [^"\\] # any character except backslash or quote
)       # End of alternation

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