Re: Recommended way to get file directory

[Gaspard Bucher <gaspard@...>]

I have found a better solution with debug.getinfo:

-- Find the source of the current file.

function lk.source(level)

  local level = level or 0

  return debug.getinfo(level + 2).source

end

-------------------------------- lk.dir()

-- Find the directory of the current file.

function lk.dir(level)

  local level = level or 0

  local source = string.gsub(lk.source(level+1), '/[^/]+$', '')

  return string.match(source, '^@(.*)$')

end

On Thu, Feb 24, 2011 at 11:06 AM, Gaspard Bucher <gaspard@teti.ch> wrote:

Hi list,

To ease installation and improve stability when moving scripts around, I need to load Lua files relative to the current file. What I basically need is a local FILE_DIR variable set to the file's directory.

My solution involves inserting a "local FILE_DIR" at the top of the files on file load but I wonder if there is a better way to solve this problem...

Gaspard

Typical usage script:

================ foo_test.lua

require 'lubyk'

FILE_DIR = FILE_DIR or lfs.currentdir() -- in case the file is run alone

...

function should.do_this_and_that()

  local fixture = fixture.readall(FILE_DIR, 'xxx.txt')

end

================

================ lk.dofile

function lk.dofile(filepath)

  local abs_path = filepath

  local cur_path = lfs.currentdir()

  if not string.match(abs_path, '^/') then

    abs_path = string.format('%s/%s', cur_path, filepath)

  end

  ----------- FILE_DIR hack -----------

  local code_str = string.format("local FILE_DIR = '%s';%s", string.gsub(abs_path, '/[^/]+$', ''), lk.readall(abs_path))

  local code = assert(loadstring(code_str))

  code()

end

================