Ruby正则表达式过滤掉带有“字符串”后缀的单词结尾

Posted 2021-04-07

我想提出一个与以下字符串匹配的Ruby Regex：

MAINT: Refactor something
STRY-1: Add something
STRY-2: Update something

但不应该匹配以下内容：

MAINT: Refactored something
STRY-1: Added something
STRY-2: Updated something

MAINT: Refactoring something
STRY-3: Adding something
STRY-4: Updating something

基本上，:之后的第一个词不应该以ed或ing结束

这就是我目前所拥有的：

^(MAINT|(STRY|PRB)-d+):s([A-Z][a-z]+)s([a-zA-Z0-9._-].*)

我已经尝试过[^ed]和[^ing]，但他们不会在这里工作，因为我的目标不仅仅是单个角色。

我无法找到适当的解决方案来实现这一目标。

答案

你可以用

^[-w]+:s*(?:(?!(?:ed|ing))w)+.+

见a demo on regex101.com。

---

Broken down this says:

^                     # start of the line/string
[-w]+:s*            # match - and word characters, 1+ then :
(?:                   # non-capturing group
    (?!(?:ed|ing))  # neg. lookahead: no ed or ing followed by a word boundary
    w                # match a word character
)+                  # as long as possible, followed by a boundary
.*                    # match the rest of the string, if any

---

I have no experience in Ruby but I guess you could alternatively do a split and check if the second word ends with ed or ing. The latter approach might be easier to handle for future programmers/colleagues.

另一答案

r = /
    A             # match beginning of string
    (?:            # begin a non-capture group
      MAINT        # match 'MAINT'
      |            # or
      STRY-d+    # match 'STRY-' followed by one or more digits
    )              # end non-capture group
    :[ ]           # match a colon followed by a space
    [[:alpha:]]+   # match one or more letters
    (?<!           # begin a negative lookbehind
      ed           # match 'ed'
      |            # or
      ing          # match 'ing'
    )              # end negative lookbehind
    [ ]            # match a space
    /x             # free-spacing regex definition mode

   "MAINT: Refactor something".match?(r)   #=> true
   "STRY-1: Add something".match?(r)       #=> true
   "STRY-2: Update something".match?(r)    #=> true

   "MAINT: Refactored something".match?(r) #=> false
   "STRY-1: Added something".match?(r)     #=> false
   "STRY-2: Updated something".match?(r)   #=> false

   "A MAINT: Refactor something".match?(r) #=> false
   "STRY-1A: Add something".match?(r)      #=> false

该正则表达式通常如下编写。

r = /A(?:MAINT|STRY-d+): [[:alpha:]]+(?<!ed|ing) /

以这种方式表达，两个空格可以各自表示空格字符。但是，在自由间隔模式下，字符类之外的所有空格都被删除，这就是我需要将每个空格括在一个字符类中的原因。

另一答案

（代表作者提问）。

这是我最终使用的：

^(MAINT|(STRY|PRB)-d+):s(?:(?!(?:ed|ing))[A-Za-z])+s([a-zA-Z0-9._-].*)