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.