require() all files in a folder?

[Henness 0]

I have a file structure for a program I'm working on that looks like this:

Main.lua
Addons folder
    -API.lua
    -Addon A.lua
    -Addon B.lua
    -Addon Z.lua

My problem is all the information I can find about listing files in a directory is inconsistent, and I don't quite under stand file path format in lua. Some people say its easier with lua filesystem but is that even available in open computers? The way I have been thinking about doing it would be something like this:

for dir in io.popen([[dir "PATHHERESOMEHOW" /b]]):lines() do require(dir) end

First will something like this even work is io.popen even available to use in open computers? Seconded what whould I have to put in the path area to make it work? And Third what does the modifiers after the path mean I saw some use /b and some use /a /b?

I have another question that isn't really related to the above. If I require() the A B and Z files in the API file would the A B and Z files be usable in the Main file if I requre() the API file?

 

Thanks in advance,

Henness

[Fingercomp 37]

OpenOS has the filesystem API. There's the page about it on the wiki. You'll see the filesystem.list function there. It's exactly what you need.

local fs = require("filesystem")

for node in fs.list("/path") do
  print(node)
end

This code prints all files and directories in /path.

You should also check if the node is a file or a directory. This can be done using the filesystem.isDirectory function.

local fs = require("filesystem")

for file in fs.list("/path") do
  if not fs.isDirectory(file) then
    print(file)
  end
end

You shouldn't use require to load your addon scripts. It just won't work. Use dofile instead.

local fs = require("filesystem")

for file in fs.list("/path") do
  if fs.isDirectory(file) do
    dofile(fs.concat("/path", file))
  end
end

dir "PATH" /b is a Windows command. It won't work in OpenOS.

[Henness 0]

Thank you this is exactly what I was looking for!