RegEx：在另一个模式之前出现的最后一次出现的模式

Question

我想采用在另一个文本模式之前最后出现的文本模式。

例如我有这样的文字：

code 4ab6-7b5
Another lorem ipsum
Random commentary.

code f6ee-304
Lorem ipsum text 
Dummy text

code: ebf6-649
Other random text
id-x: 7662dd41-29b5-9646-a4bc-1f6e16e8095e

code: abcd-ebf
Random text
id-x: 7662dd41-29b5-9646-a4bc-1f6e16e8095e

我想取最后一个出现在id-x 之前的code（这意味着我想获取代码ebf6-649）

如何使用正则表达式做到这一点？

Answer 1

如果您的正则表达式支持lookaheads，您可以使用这样的解决方案

^code:[ ]([0-9a-f-]+)(?:(?!^code:[ ])[\s\S])*id-x

您可以在捕获号1 中找到您的结果。

它是如何工作的？

^code:[ ]           # match "code: " at the beginning of a line, the square 
                    # brackets are just to aid readability. I recommend always
                    # using them for literal spaces.

(                   # capturing group 1, your key
  [0-9a-f-]+        # match one or more hex-digits or hyphens
)                   # end of group 1

(?:                 # start a non-capturing group; each "instance" of this group
                    # will match a single arbitrary character that does not start
                    # a new "code: " (hence this cannot go beyond the current
                    # block)

  (?!               # negative lookahead; this does not consume any characters,
                    # but causes the pattern to fail, if its subpattern could
                    # match here

    ^code:[ ]       # match the beginning of a new block (i.e. "code: " at the
                    # beginning of another line

  )                 # end of negative lookahead, if we've reached the beginning
                    # of a new block, this will cause the non-capturing group to
                    # fail. otherwise just ignore this.

  [\s\S]            # match one arbitrary character
)*                  # end of non-capturing group, repeat 0 or more times
id-x                # match "id-x" literally

(?:(?!stopword)[\s\S])* 模式让您尽可能匹配，而不会超出stopword 的另一次出现。

请注意，您可能必须对^ 使用某种形式的多行模式来匹配行首。如果您的random text 包含open:，则^ 对于避免误报很重要。

Working demo（使用 Ruby 的正则表达式，因为我不确定你最终会使用哪一个）