正则表达式匹配任何空行或以指定字符开头的任何行 [英] Regular expression to match any empty line or any line starting with a specified character

问题描述

给出以下输入...

;
; comment
; another comment
;

data
data

我正在寻找正则表达式可以用来剥去空白行并仅返回包含数据的两行（但不保留换行符）。

I am looking for a regular expression that can be used to strip the blank lines and return only the two lines containing the "data" (but leaving the line breaks intact).

谢谢。

推荐答案

编辑

等等，我想我理解您的意思是：您只想保留数据行之后的换行符。如果是这样，请尝试：

Wait, I think I understand what you mean: you only want to preserve the line breaks after your "data" lines. If so, try:

(?m)^([ \t]*|;.*)(\r?\n|$)

一个小解释：

(?m)          # enable multi-line option
^             # match the beginning of a line
(             # start capture group 1
  [ \t]*      #   match any character from the set {' ', '\t'} and repeat it zero or more times
  |           #   OR
  ;           #   match the character ';'
  .*          #   match any character except line breaks and repeat it zero or more times
)             # end capture group 1
(             # start capture group 2
  \r?         #   match the character '\r' and match it once or none at all
  \n          #   match the character '\n'
  |           #   OR
  $           #   match the end of a line
)             # end capture group 2

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