require-ing support code in a lua filter

[pragdave]

I'm writing a lua filter that I've split into a number of lua source files.

How can I require these files from the main filter script? (I've tried the ... trick but that doesn't work... )

Cheers

Dave

[mcanouil]

I don't know what the ... trick is supposed to be, but to import Lua script inside Lua script, Lua uses require("lua-script-name-without-extension").
See http://www.lua.org/pil/8.1.html.

PS: could you change the category from "general" to "Q&A" (using the desktop version of the website, not mobile view or app)?

[dragonstyle]

As mcanouil said, using require is advised:

See more about require here: https://www.lua.org/pil/8.1.html

You can either use require to import the file and anything global will become accessible, or more preferable, the 'required' file can return a table of function, properties and so on which you can assign to a local variable- for example:

local importedStuff = require '<filename-without-extension>'

[pragdave]

I understand "require". My problem is how to specify a path that is relative to the file doing the requiring. For example, I might have

| index.qmd
| _quarto.yml
| _ extensions
    | filters
     _| my_filter.lua
      | lib_file1.lua
      | lib_file2.lua

In my_filter.lua, I want to be able to do something like:

lib1 = require("./lib_file1")
lib2 = require("./lib_file2")

I can't, though, because if I run quarto from the top level, because the working directory it the top level.

OK, you say: make the require something like `require("_extensions/filter/..."). But then, how, can I distribute this filter?

TIA

Dave

[mcanouil]

A working reproducible example would make things a lot easier to understand thus easier to properly help you.

[pragdave]

I've worked around it. Thanks all.