vim 正则表达式 - 在行尾匹配任意数量的空格，除了 2

Question

I want to write a fixer for ale to remove all whitespace at the end of a line except for a double whitespace - in markdown this is used to create a linebreak.

I need to match "at end of row, 1 or more white space AND not 2 white space"

kind of like \s\+$\&\s\{^2}$ except that ^ is not negation inside curly brackets. Some googling reveals that negating a count of a meta character seems to be a particularly niche problem.

Answer 1

You can use

:%s/\v(\s)@<!\s(\s{2,})?$//g

Details

• % - search on all lines
• s - substitute
• \v - very magic mode
• (\s)@<! - location not immediately preced with a whitespace
• \s - a whitespace
• (\s{2,})? - an optional occurrence of two or more whitespaces
• $ - end of line
• g - all occurrences on the line.

This is how this regex works (translated into PCRE).

Answer 2

(This really should be a comment but formatting is important, here)

How do you want the following snippet to look after you have "fixed" it (spaces marked with _ )?

First line, without trailing spaces
Second line, with one trailing space_
Third line, with two trailing spaces__
Fourth line, with more than two trailing spaces______