"如何在 .vimrc 中获取文件"

Question

I am trying a simple means of sourcing a file based on the filetype of the new opened file. For example, I want to source python.vimrc when a new file is *.py. This code (in .vimrc)

function LoadFileTypeDefaults()
  let vimrcfile = &filetype . '.vimrc' 
  if filereadable(vimrcfile)
    echo vimrcfile
    source vimrcfile
  endif
endfunction

gives the following error when I do vi new.py

python.vimrc

Error detected while processing function LoadFileTypeDefault:

line 4:

E484: Can't open file vimrcfile

My python.vimrc is in my runpath. Moreover, if I replace

source vimrcfile

with

source python.vimrc 

everything works as required.

What am I missing?

Answer 1

The source command expects a file name, not an expression. You gave it vimrcfile so it's looking for the file called vimrcfile . Use execute to string together a command from one or more expressions.

exe 'source' vimrcfile

(If you pass multiple arguments they will be joined with spaces before executing.)

---

What you probably really want is to just add

autocmd Filetype python set expandtab " etc

in your .vimrc .

---

Or if you really have a lot of settings, put them in ~/.vim/after/ftplugin/python.vim .

Answer 2

Shouldn't it be execute "source " . vimrcfile execute "source " . vimrcfile ?

Answer 3

Vim has a system for dealing with this kind of thing, called filetype plugins, or ftplugins.

In your ~/.vim (or whatever is standard on your OS), go to the ftplugins directory (create if it doesn't exist) and create a python directory under that. Any .vim file under that directory will automatically be sourced when a .py file is loaded.

Note that you need to have filetype plugin on .

Alternatively you can do this with autocommands, though it is a bit cumbersome:

autocmd BufNewFile,BufRead *.py source python.vimrc

Put that in your normal vimrc, and it'll do what you want.

Answer 4

作为 vim 命令：

:so ~/.vimrc