CODEWITHSUNDEEP
vimself-reference
gfxmonk
gfxmonk

Reputation: 8746

How to get path to the current vimscript being executed

In my vim plugin, I have two files:

myplugin/plugin.vim
myplugin/plugin_helpers.py

I would like to import plugin_helpers from plugin.vim (using the vim python support), so I believe I first need to put the directory of my plugin on python's sys.path.

How can I (in vimscript) get the path to the currently executing script? In python, this is __file__. In ruby, it's __FILE__. I couldn't find anything similar for vim by googling, can it be done?

Note: I am not looking for the currently edited file ("%:p" and friends).

Upvotes: 73

Views: 19393

Answers (4)

Masquue
Masquue

Reputation: 322

If you really want to get the script path inside a function (which is what I'd like to), you can still use <sfile>'s second semantic, or its equivalent <stack> inside expand().

   <sfile>    ...
              When executing a legacy function, is replaced with the call
              stack, as with <stack>
              ...
                                                   :<stack> <stack>
   <stack>    is replaced with the call stack, using
              "function {function-name}[{lnum}]" for a function line
              and "script {file-name}[{lnum}]" for a script line, and
              ".." in between items.  E.g.:
              "function {function-name1}[{lnum}]..{function-name2}[{lnum}]"
              If there is no call stack you get error E489 .

However you possibly don't want to use it in a plugin, as you can use autoload functions in plugin, using this relative#path#to#plugin#root#script notation.

I use this for sourcing purpose:

function! s:SourceLocal(script)
  let l:callstack = expand("<stack>")
  let l:list = split(l:callstack, '\.\.')
  " list[-1] is SourceLocal function itself
  " list[-2] is the calling script
  let l:script_name = matchstr(l:list[-2], '^\(script \)\=\zs.\+\ze\[\d\+\]$')
  let l:script_path = fnamemodify(l:script_name, ":p:h")
  " l:script_path is the path where the script calling this function resides
  execute printf("source %s/%s", l:script_path, a:script)
endfunction

command! -nargs=1 SourceLocal :call s:SourceLocal(<f-args>)

Then you can SourceLocal inside any script to source another script relative to it.

Upvotes: 0

edthedev
edthedev

Reputation: 187

It is worth mentioning that the above solution will only work outside of a function.

This will not give the desired result:

function! MyFunction()
let s:current_file=expand('<sfile>:p:h')
echom s:current_file
endfunction

But this will:

let s:current_file=expand('<sfile>')
function! MyFunction()
echom s:current_file
endfunction

Here's a full solution to OP's original question:

let s:path = expand('<sfile>:p:h')

function! MyPythonFunction()
import sys
import os
script_path = vim.eval('s:path')

lib_path = os.path.join(script_path, '.') 
sys.path.insert(0, lib_path)                                       

import vim
import plugin_helpers
plugin_helpers.do_some_cool_stuff_here()
vim.command("badd %(result)s" % {'result':plugin_helpers.get_result()})

EOF
endfunction

Upvotes: 18

Zenexer
Zenexer

Reputation: 19611

" Relative path of script file:
let s:path = expand('<sfile>')

" Absolute path of script file:
let s:path = expand('<sfile>:p')

" Absolute path of script file with symbolic links resolved:
let s:path = resolve(expand('<sfile>:p'))

" Folder in which script resides: (not safe for symlinks)
let s:path = expand('<sfile>:p:h')

" If you're using a symlink to your script, but your resources are in
" the same directory as the actual script, you'll need to do this:
"   1: Get the absolute path of the script
"   2: Resolve all symbolic links
"   3: Get the folder of the resolved absolute file
let s:path = fnamemodify(resolve(expand('<sfile>:p')), ':h')

I use that last one often because my ~/.vimrc is a symbolic link to a script in a git repository.

Upvotes: 114

gfxmonk
gfxmonk

Reputation: 8746

Found it:

let s:current_file=expand("<sfile>")

Upvotes: 39

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