I have following folder structure
--folder1
------main.lua
------sub_folder
---------func1.lua
--folder2
------func2.lua
I want to load func1.lua and func2.lua in main.lua by require command
PS:what does the init.lua in a folder do
In vanilla lua, you'd load them like this:
local func1 = loadfile "sub_folder/func1.lua"
local func2 = loadfile "../folder2/func2.lua"
See: Pil Chapter 8
If you're using Lua embedded in some other application (i.e, a game, software, etc.), then you need to let us know what this is, or we can't help you. This goes for your P.S question as well. But generally, init.lua is usually the first file to be executed, i.e, it'll load other files, etc.
The func1.lua is easily loaded via
require 'sub_folder.func1'
because sub_folder is in the same folder as main.lua.
For func2.lua, there is no way to tell require to search "one level up", so you must tell Lua how to find the required modules. This can be done at least two ways:
via LUA_PATH environment variable:
append path/to/folder2/?.lua to it, then main.lua can do require "func2".
OR append path/to/parent-of-folder2/?.lua to it, then main.lua can do require "folder2.func2".
by editing package.path in your script:
package.path = package.path .. ';../?.lua'
require 'func2'
or
package.path = package.path .. ';../../?.lua'
require 'folder2.func2'
The first method is more "permanent" since the setting is in the OS environment; it will work even if you move folder1 to some other place on your system, without moving folder2. The second method is dynamic so it will work regardless of where you place your folder structure, i.e., if folder2 is always sibling folder of folder1 the method 2 works, method 1 fails (or requires that you edit the LUA_PATH).
Related
So i've been trying to figure this out, im doing the following
main.lua:
package.path = io.popen("cd"):read'*all' -- to set the path the same as main.lua (for exporting)
local u = require("utilities") -- this is the line that dosent work
utilities.lua:
local u = {}
function u.GetService(s)
return require(tostring(s))
end
function u.Wait(s)
local ntime = os.clock() + s/10
repeat until os.clock() > ntime
end
function u.NPWait(s)
local ntime = os.time() + s
repeat until os.time() > ntime
end
return u
The error:
module 'utilities' not found:
Then here goes a search that DOES look in the correct folder, however it looks for .dll files
Require has never been something I've been good at, Sometimes I get it to work, sometimes it breaks like this, if someone can help me understand how to fix this then thank you, as this has been annoying me every so often
You're misunderstanding package.path.
Unlike the PATH variable used by windows and linux, in Lua you don't just tell it what directory to look in but also what the files should look like.
What you're doing is erasing the default search paths completely and replacing it with a broken path that won't find any files.
Generally speaking, to add a new directory to your path, you should add both C:\path\to\your\files\?.lua, and C:\path\to\your\files\?\init.lua, which are the two ways Lua loads modules by default.
You also shouldn't replace package.path, because you're keeping modules from being loaded elsewhere.
Your code should look somewhat like this:
local cd = io.popen("cd")
package.path = cd:read('*line') .. "\\?.lua;" .. package.path
cd:close()
The reason why it's still looking for DLL files is that package.cpath is separate, so you're not changing that.
require("utilities.lua") should work.
You have to specify how to get from the required name to the file. For example by adding a path and a file extension to it.
If you want to require c:\folder\script.lua with require("script") your package.path must contain something like c:\folder\?.lua where ? is replaced by the required name.
If you have your module in a folder with its name and let's say a init.lua for that module the path is c:\folder\?\init.lua
Just print the original package.path and see what's inside befor you overwrite it.
Refer to Lua Reference Manual 6.3 Modules: package.searchpath
A path is a string containing a sequence of templates separated by
semicolons. For each template, the function replaces each
interrogation mark (if any) in the template with a copy of name
wherein all occurrences of sep (a dot, by default) were replaced by
rep (the system's directory separator, by default), and then tries to
open the resulting file name.
For instance, if the path is the string
"./?.lua;./?.lc;/usr/local/?/init.lua" the search for the name foo.a will try to open the files ./foo/a.lua, ./foo/a.lc, and
/usr/local/foo/a/init.lua, in that order.
I'm trying to repackage into a Docker container some Lua library that is made of the main module and some helper modules. The helper modules are kept inside a subfolder of the library so that imports from the main file are done as
require 'helpers/SomeHelper'
The problem is: because of the way I want the Docker container to work, it would be extremely helpful if I can invoke this library from a different working folder. That is, my call to the main program would be something like
th /app/main.lua
regardless of the actual working directory I'm standing. Unfortunately, relative imports seem to fail when the working directory is different from the directory where the main file is located.
Is there any way I can configure LUA_PATH or any other mechanism to make these imports work correctly? Note that changing the code of the library itself would be a poor solution, as it wasn't developed by me and I would like to be able to update it to newer versions easily.
If you don't care about the working directory, you can just load lfs / LuaFileSytem and use lfs.chdir( src_dir ) to change to the source directory (potentially saving the current working directory with lfs.currentdir( ) first.)
You can also extend the search path of Lua so that it will search those extra directories. The search is driven by package.searchpath. To add a directory /foo/bar/ to the search in a way that supports all normally supported library layouts, add
/foo/bar/?.lua;/foo/bar/?/init.lua to package.path
/foo/bar/?.so (or .dylib or .dll on other OSen) to package.cpath
You can use several ways to extend the path.
One option that works well is to set the LUA_PATH / LUA_CPATH environment variables. (A ;; sequence in one of them will expand to the full default path.) This can be done from .profile or other setup scripts via an earlier export LUA_PATH="..." or (if started from a wrapper script) inline by setting variables just for that call LUA_PATH="..." lua /foo/bar.lua. (Note that if you export this variable in too broad a scope, other Lua scripts will also get their path extended and may find potentially incompatible Lua libraries.)
(You can also manually modify package.(c)path from LUA_INIT. That way, you won't be able to independently disable LUA_INIT or LUA_PATH, but you can use all of Lua to generate the path dynamically.)
A third option (this may be best in your specific case) is to put the extension of package.path at the top of your main script, as in
do
local dir = (arg[0]:match "^(.*)/$")
if dir then -- else cwd is . which works by default
package.path = dir.."/?.lua;"..dir.."/?/init.lua;"..package.path
package.cpath = dir.."/?.so;"..package.cpath
end
end
-- rest of your program goes here
When running a script with the Lua interpreter, arg[0] is the script. So this extends the path to include the program's directory no matter where it is located, and it will only affect the search path of this particular script / program.
You should not forget about that not all modules are loaded from FS directly.
E.g. to improve perfomance it is possible read/compile file to memory and then
use preload table to provide way to load module from memory.
Basic example
--- preload code. It can be done by host application.
local FooUtils = function()
return {
print = function(...)
print("foo", ...)
end
}
end
local Foo = function()
local Utils = require "foo.utils"
return {
foo = function()
Utils.print"hello"
end
}
end
package.preload['foo.utils'] = FooUtils
package.preload['foo'] = Foo
--- main application
require "foo".foo()
In this example assume that FooUtils and Foo just example of compiled modules.
E.g. it can be like FooUtils = loadstring('path/to/utils.lua) and it can be done
even in separate Lua state and then used in any other.
It is important to remember that Foo module have no idea about how host application does lookup of foo.utils.
So there no standart way to provide original file path or relevant paths.
So if you write some module wich relays on relative paths then this module
may be not working in some environments.
So I just suggest use full namespace like require 'foo.utils' instead of require 'utils'
So in lua, i want to import a module.
I want to have my "polygon" lib in a subfolder, so i reference it like this
local polygon = require('polygon.polygon')
however, it needs another module called 'delaunay', it cannot find it as it checks the main folder
Is there anyway short of editing my library, to get this to work? (some kind of ability to add search paths?)
Thanks
To know where to look for modules, Lua's require uses the variables package.path (.lua) and package.cpath (.so/.dll). You can change them in your program to look in the directory you have it in. For consistency's sake, you can look at their contents to know which OS-specific separator to use. For example:
local sep = package.path:find("\\") and "\\" or "/"
package.path = package.path .. ";." .. sep .. "polygon" .. sep .. "?.lua"
This would include ./polygon/?.lua into the search path, and a call to require "delaunay" would therefore have the require function look for ./polygon/delaunay.lua in addition to existing paths. Bear in mind that in require strings, . denotes a separator as far as file searching is concerned, so calling require "polygon.delaunay" in this scenario will mean searching for ./polygon/polygon/delaunay.lua.
From what I understand of your question, changing the package.path variable to include the path to where your delaunay library is stored would solve your issue, although to give a specific solution more information about your project and directory structure is required.
All what loaded into package.loaded can be load with require() because require() looks first in package.loaded...
-- Lua 5.3 ( lua -i )
> package.loaded.code=load(code.dump)()
> test=require('code')
> test
function: 0x565cb820
So you can use load() or loadfile('/path/to/your_code.lua') to do this. Another nice feature of this method is to load dumped code...
> package.loaded.shell=loadfile('shell.bin')
> shell=require('shell')
> shell('cat shell.bin')
uaS�
�
xV(w#F#�d�#��F�#G���d#��F�#G���d#&�typestringoexecute
/bin/bash>
In any way, it appears that you will have to deal with the subfolders explicitly.
As in, either the module polygon will have to import delaunay as polygon.delaunay.
Or module names will have to be appended to package.path so that lua could search through subfolders for filenames:
package.path=package.path..";./polygon/?.lua"
More info is here.
It is pointed out in comments, you'd probably want to ensure that path concatenation happens only once.
Also, one should be wary of name shadowing.
Finally, while we are at stirring up the past, a pretty nifty trick to solve the issue was proposed here five years before the question.
I would like to separate init.lua script used in Hammerspoon to enhance the readability and maintainance.
So it looks like the following:
init.lua
AppWatcher.lua
WiFiWatcher.lua
KeyRemap.lua
And then from within init.lua I would read these files and make the watcher activate.
However, it seems that there is no such function defined (maybe I may be missing it, though). Is it possible to separate the logic like that in Hammerspoon?
Yes, you can do this using require.
If you put your Lua files in ~/.hammerspoon/, you can then load them using require('modulename'). For example, if you have the following modules:
~/.hammerspoon/AppWatcher.lua
~/.hammerspoon/WiFiWatcher.lua
~/.hammerspoon/KeyRemap.lua
Then you can load them from ~/.hammerspoon/init.lua like this:
local AppWatcher = require('AppWatcher')
local WiFiWatcher = require('WiFiWatcher')
local KeyRemap = require('KeyRemap')
You can load any Lua modules, as long as they appear in package.path. To see the directories you can use, take a look at HammerSpoon's package.path setup file. This references the default Lua package.path, which is defined in luaconf.h.
If you want to put your Lua modules in a directory not included in package.path, you can do it by adding them to the LUA_PATH_5_3 or LUA_PATH environment variables.
I need to call the require on a lua file that will not always be in the same place. I was trying to call require on the full path name but that doesn't seem to be working either. I even tried replacing one of my working normal requires with a correct full path name to the same file
example changing
require "foo"
to
require "C:\Users\Me\MyLuaProject\foo"
but when i switched it to the full path name it could no longer find it. So I am wondering if you can even call require on a full path and if not how would i achieve the same result differently?
If you just need to load a file, use dofile, which takes a path:
dofile("C:\\Users\\Me\\MyLuaProject\\foo")
Add the directory containing the file to package.path:
package.path = package.path .. ";C:\\Users\\Me\\MyLuaProject"
require "foo"
You can also add it to the LUA_PATH environment variable, but this is probably less easy to modify on the fly.
A common pattern for modules is to have abc.lua and abc/xyz.lua; to require files in a subdirectory like that, use the following:
require "abc"
require "abc.xyz"