Cannot figure out how to get the path of the current file for a lua script.

[Rabcor]

I've been looking for a way for actual hours, i've seen other scripts do it but when I try to replicate it their way it always fails...

local path = utils.getcwd() is the best i've got but it only works when opened from a terminal because the file manager (for some reason) sets the launch path to my home directory, no matter which directory i'm actually in.

I tried various things. Furthest I gat was this.

    local path = mp.get_property("path", "")
    local dir, filename = utils.split_path(path)

which returns a '.' for dir and nothing at all for filename, doesn't actually give me a proper path. Most other things i tried just return 'nil'. I've seen this used in a bunch of other scripts and it seems to work there, but not in mine for no discernible reason (i'm literally running this at the start of the script)

For instance:

local directory, filename = utils.split_path(mp.get_property('path'))
mp.msg.warn("Path:", directory)

Gives me Lua error: /home/rabcor/.config/mpv/scripts/script.lua:33: bad argument #1 to 'split_path' (string expected, got nil)

local test = mp.get_property_native('playlist')[1].filename gives me filename but no path...

How do i get the real path of the current file? or just filename with path?

[guidocella]

I assume you want the directory of the file instead of the path:

local utils = require 'mp.utils'

mp.register_event('file-loaded', function()
    local directory, _ = utils.split_path(utils.join_path(mp.get_property('working-directory'), mp.get_property('path')))

    print(directory)
end)

On master:

local utils = require 'mp.utils'

mp.register_event('file-loaded', function()
    local directory, _ = utils.split_path(mp.command_native({'normalize-path', mp.get_property('path')}))

    print(directory)
end)

[Rabcor]

Thanks! That was it!