Like source files in .vimrc

Question

I am trying to use a simple way to search for a file based on the file type of a new open file. For example, I want the source python.vimrc when the new * .py file. 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 processing function LoadFileTypeDefault:

line 4:

E484: cannot open vimrcfile file

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

 source vimrcfile 

with

 source python.vimrc 

everything works as needed.

What am I missing?

Answer 1

The source command expects a file name, not an expression. You gave it a vimrcfile , so it searches for a file named vimrcfile . Use execute to combine a command from one or more expressions.

 exe 'source' vimrcfile 

(If you pass multiple arguments, they will be concatenated with spaces before execution.)

---

What you probably really want is just to add

 autocmd Filetype python set expandtab " etc 

in .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 similar things called file plugins or ftplugins.

In your ~/.vim (or something else standard for your OS), go to the ftplugins directory (create if it does not exist) and create a python directory in it. Any .vim file in this directory will automatically load when the .py file is downloaded.

Note that you need to have filetype plugin on .

Alternatively, you can do this with auto-commands, although this is a bit cumbersome:

 autocmd BufNewFile,BufRead *.py source python.vimrc 

Put this in your regular vimrc and it will do what you want.