Question

I want to prepend all my classNames with o- without having to adjust every className by hand. I use Vim.

I know a substitution will do the job so I came up with this, which is obviously not working (and the reason I am here).

:%s/class="[^"]*"/\='class="'.substitute(submatch(0), '[^o-]*', 'o-'.submatch(1), 'g').'"'/g

Explanation:

class="[^"] - matches all instances of class="foo bar baz"
\='class="'.substitute(subexp).'"' - replaces the found instances class="subexp"
subexp in two should repace each space separated class with the original className prepended with o-

All in all, in procedural terms, for each class="foo bar baz", replace each className with the className prepended with o-.

Thanks in advance.

(BONUS) EDIT: How can this be written to ignore or cope with classNames that already begin with o-, when encountered, so that o-o- is not a resulting edit.

Answer 1

示例

class="foo bar baz"

此行有效：

%s/class="\zs[^"]*\ze"/\=join( map(split(submatch(0)),"'o-'.v:val"), ' ')/

所以有嵌套函数调用：

我没有使用\<边界，因为如果你的类名中有一些“特殊”字符，它就会失败。例如。 # - or @。我不知道你的语言是否属实。
submatch（0）是"foo bar baz"
split()使（每个类）成为列表
map()为每个类名

o-

join()将修改后的列表转回字符串

执行此命令后，您应该看到：

class="o-foo o-bar o-baz"

编辑“奖金”要求：

我们只需要检查每个元素（classname）。检查下面的代码，它应该适合你：

%s/class="\zs[^"]*/\=join(map(split(submatch(0)),"(v:val=~'^o-'?'':'o-').v:val"))/

我们有：

(v:val=~'^o-'?'':'o-').v:val

如果元素以o-开头，则我们不再添加其他o-。

Answer 2

一种方法是利用\zs...\ze，以及submatch(0)解析为\zs...\ze之间匹配的字符串的事实：

:%s/\mclass="\zs.\{-}\ze"/\=substitute(submatch(0), '\<', 'o-', 'g')/g

如果您不想依赖它，您仍然可以使用显式分组：

:%s/\mclass="\zs\([^"]*\)\ze"/\=substitute(submatch(1), '\<', 'o-', 'g')/g

In Vim, How to substitute within a subexpression?

2 个答案:

编辑“奖金”要求：