用制表符替换行开头的所有空格 [英] Replace all spaces at the start of the line with tabs

问题描述

我想用制表符替换行首的所有空格.以下代码段有效，但仅适用于第一个缩进级别.

I would like to replace all spaces at the start of the line with tabs. The below snippet works, but only for the first indentation level.

我如何使其适用于 1到∞的缩进级别?以便用3个制表符替换12个空格(假设tabstop为4)?

How do I make it work for 1 to ∞ indentation levels? So that it replaces 12 spaces with 3 tabs (assuming a tabstop of 4)?

fun! Retab()
    let l:spaces = repeat(' ', &tabstop)
    silent! execute '%s/^' . l:spaces . '/\t/g'
endfun

请注意，在这里似乎不是使用:retab的选项，因为:retab不仅会更改缩进，而且还会更改文件中所有repeat(' ', &tabstop)出现的位置 .

Note that using :retab doesn't seem to be an option here, since :retab doesn't just change indentation, but also changes all repeat(' ', &tabstop) occurrences everywhere in the file.

也不能选择使用=重新缩进文件，因为Vim&有时候我对应该缩进哪个级别有不同的看法(即，它有太多的副作用).

Re-indenting the file (with =) is also not an option, since Vim & I sometimes have different opinions on what should be indented at which level (ie. it has too many side-effects).

我还考虑过使用expand& unexpand程序，但我希望不要依赖外部实用程序.

I also thought about using the expand & unexpand programs, but I would prefer not to rely on external utilities.

推荐答案

您的尝试朝着正确的方向前进，但是您需要:help sub-replace-expr来计算匹配空格的数量并将其转换为相应数量的制表符: /p>

Your attempt goes into the right direction, but you need :help sub-replace-expr to count the number of matched spaces and convert this into a corresponding number of tab characters:

silent! execute '%substitute#^\%(' . l:spaces . '\)\+#\=repeat("\t", len(submatch(0)) / &tabstop)#e'

要执行相反操作(将制表符替换为空格)，您可以执行以下操作:

To do the reverse (replaces tabs to spaces), you can do:

silent! execute '%substitute#^\%(\t\)\+#\=repeat("' . l:spaces . '", len(submatch(0)))#e'

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