在我的vim插件中，我有两个文件：
myplugin/plugin.vim myplugin/plugin_helpers.py
我想从plugin.vim(使用vim python支持)导入plugin_helpers，所以我相信我首先需要把我的插件的目录在python的sys.path。
我如何(在vimscript)获得当前执行脚本的路径？在python中，这是__file__。在ruby中，它是__FILE__。我找不到类似的vim通过谷歌，可以做到吗？
注意：我不是在寻找当前编辑的文件(“％：p”和朋友)。
" 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')
我使用最后一个经常，因为我的〜/ .vimrc是一个符号链接到Git存储库中的脚本。
