我是一个vim用户,我希望能够在替换时循环一系列子串.我如何使用一些vim魔法来从这样的一组行：
Afoo Bfoo Cfoo Dfoo
至
Abar Bbar Cbaz Dbaz
？
我想从头开始搜索我的文件,以便下一次出现foo,并用bar替换前两个实例,用baz替换后两个实例.使用for循环是最好的选择吗？如果是这样,那么如何在替换命令中使用循环变量？
我会使用一个具有状态的函数,并从％s调用此函数.就像是：
原文链接：https://www.f2er.com/bash/386930.html" untested code
function! InitRotateSubst()
let s:rs_idx = 0
endfunction
function! RotateSubst(list)
let res = a:list[s:rs_idx]
let s:rs_idx += 1
if s:rs_idx == len(a:list)
let s:rs_idx = 0
endif
return res
endfunction
并使用它们：
:call InitRotateSubst() :%s/foo/\=RotateSubst(['bar','bar','baz','baz'])/
如果您愿意,可以将对这两个命令的调用封装到单个命令中.
编辑：这是一个集成为命令的版本：
>根据需要接受尽可能多的替换,所有替换都需要用分隔符分隔;
>支持反向引用;
>只能替换N个第一次出现,N ==如果命令调用被绑定指定的替换次数(带！)
>不支持像g,i(：h：s_flags)这样的通常标志 – 为此,我们可能会强制执行命令调用,最后总是以/(或任何分隔符)结束,如果不是最后一个文本被解释为标志.
这是命令定义：
:command! -bang -nargs=1 -range RotateSubstitute <line1>,<line2>call s:RotateSubstitute("<bang>",<f-args>)
function! s:RotateSubstitute(bang,repl_arg) range
let do_loop = a:bang != "!"
" echom "do_loop=".do_loop." -> ".a:bang
" reset internal state
let s:rs_idx = 0
" obtain the separator character
let sep = a:repl_arg[0]
" obtain all fields in the initial command
let fields = split(a:repl_arg,sep)
" prepare all the backreferences
let replacements = fields[1:]
let max_back_ref = 0
for r in replacements
let s = substitute(r,'.\{-}\(\\\d\+\)','\1','g')
" echo "s->".s
let ls = split(s,'\\')
for d in ls
let br = matchstr(d,'\d\+')
" echo '##'.(br+0).'##'.type(0) ." ~~ " . type(br+0)
if !empty(br) && (0+br) > max_back_ref
let max_back_ref = br
endif
endfor
endfor
" echo "max back-ref=".max_back_ref
let sm = ''
for i in range(0,max_back_ref)
let sm .= ','. 'submatch('.i.')'
" call add(sm,)
endfor
" build the action to execute
let action = '\=s:DoRotateSubst('.do_loop.',' . string(replacements) . sm .')'
" prepare the :substitute command
let args = [fields[0],action ]
let cmd = a:firstline . ',' . a:lastline . 's' . sep . join(args,sep)
" echom cmd
" and run it
exe cmd
endfunction
function! s:DoRotateSubst(do_loop,list,replaced,...)
" echom string(a:000)
if ! a:do_loop && s:rs_idx == len(a:list)
return a:replaced
else
let res0 = a:list[s:rs_idx]
let s:rs_idx += 1
if a:do_loop && s:rs_idx == len(a:list)
let s:rs_idx = 0
endif
let res = ''
while strlen(res0)
let ml = matchlist(res0,'\(.\{-}\)\(\\\d\+\)\(.*\)')
let res .= ml[1]
let ref = eval(substitute(ml[2],'\\\(\d\+\)','a:\1',''))
let res .= ref
let res0 = ml[3]
endwhile
return res
endif
endfunction
可以这样使用：
:%RotateSubstitute#foo#bar#bar#baz#baz#
甚至,考虑到初始文本：
AfooZ BfooE CfooR DfooT
命令
%RotateSubstitute/\(.\)foo\(.\)/\2bar\1/\1bar\2/
会产生：
ZbarA BbarE RbarC DbarT
