如果您实际上不需要更改tabstops并且只需插入正确数量的空格就可以逃脱,我建议您编写脚本.这是一个快速而肮脏的版本,可能会做你想要的：
原文链接：https://www.f2er.com/bash/384248.htmllet s:tabstops = [0,5,80]
fun! Find_next(pos)
if a:pos > min(s:tabstops) && a:pos < max(s:tabstops)
let my_count = 0
while my_count < len(s:tabstops) - 1
if a:pos > get(s:tabstops,my_count) && a:pos < get(s:tabstops,my_count+1)
return get(s:tabstops,my_count+1)
endif
let my_count = my_count + 1
endwhile
return -1
endif
return -1
endfun
fun! Tabbing()
let pos = col('.')
let next_stop = Find_next(pos)
let the_command = "normal i"
let my_count = 0
while my_count < next_stop - pos
let the_command = the_command . " "
let my_count = my_count + 1
endwhile
let the_command = the_command . ""
execute the_command
endfun
imap <TAB> j<ESC>:call Tabbing()<CR>lxi
