使用标准的相对路径方法需要LUA文件-问题:尝试调用全局"myfunc"(nil值) [英] Require LUA files using a standard relative path approach - problems with: attempt to call global 'myfunc' (a nil value)

问题描述

我需要调用另一个我的.lua文件中定义的LUA函数；从另一个.因此，我想要的是经典的C/C ++包含方法. 我尝试了以下方法:

I need to call a LUA function, defined in another my .lua file; from another. So, what I want is the classic C/C++ include approach. I tried with the following:

(file funcs.lua)
function myfunc(arg1, arg2)
 ..dosomething
end

和

(file main.lua)
package.path = package.path .. ";/path/to/libs/?.lua"
require "funcs"
myfunc(1, 2)

require工作正常，但是在执行时出现此错误:

The require works good, but at execution I get this error:

attempt to call global 'myfunc' (a nil value)

为什么? 预先感谢，

推荐答案

谢谢大家的评论；我正在OpenResty/Nginx下运行LUA.

Thank you all for the comments; I'm running LUA under OpenResty/Nginx.

我通过直接导出函数来解决问题，我不知道这是否是首选方法，但是我注意到许多新的LUA模块都使用了该方法. 例如，我将代码更改如下:

I solved by exporting directly the function(s), I don't know if this is the preferred method, but I noticed that is used by lots of newer LUA modules. For example, I changed the code as follows:

file (funcs.lua)
local A = {}
function A.myfunc(arg1, arg2)
 ..dosomething
end
return A

(file main.lua)
package.path = package.path .. ";/path/to/libs/?.lua"
funcs = require "funcs"
funcs.myfunc(1, 2)

这很好用，并且可以以一种OOP风格手动导出每个功能.

This works good and it's nice to have every function to be manually exported, in a sort of OOP-style.

这篇关于使用标准的相对路径方法需要LUA文件-问题:尝试调用全局"myfunc"(nil值)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！