我如何使require()指向文件的直接路径 [英] How do I make require() take a direct path to a file

问题描述

所以我有以下代码，问题是当我遍历数组中的每个文件并尝试要求文件路径时，它给我一个找不到模块的错误.

so I have the following code, and the problem is that when I loop through each file in my array and try to require the file path, it gives me an error of the module isn't found.

local Commands = {}

function getCommands()
    local readdir = fs.readdir
    local readdirRecursive = require('luvit-walk').readdirRecursive
    readdirRecursive('./Desktop/Discord/ArtifexBot/Discordia/resources/commands/', function(k, files) 
        for i,v in pairs(files) do
            if v:match(".lua") and not v:match("commands.lua") then
                local cmd = v:match("([^/]-)%..-$")
                fs.readlink(v,function(err,thing)
                    print(err,thing)
                end)
                Commands[cmd] = require(v)
            end
        end
    end)
end
getCommands()

递归函数有效，并且文件只是路径的字符串.但是经过研究，require()需要相对路径，而不是直接路径.所以我认为我需要对fs进行一些操作，以使文件路径成为相对路径?我在任何地方都找不到答案.

The recursive function works, and the files are just strings of the path. But after research, require() needs a relative path, not a direct path. So I think I need to do something with fs to make the file path a relative path instead? I couldn't find the answer anywhere.

谢谢！

推荐答案

require根本没有路径.标准加载程序只是根据您的算法，按照一系列模式使用您给它提供的字符串.

require doesn't take a path at all. The standard loaders simply use the string you give it in a sequence of patterns, in accord with its algorithm.

您想要的是在磁盘上加载并执行给定的Lua脚本.那不是拼写的require；拼写为dofile.

What you want is to load and execute a given Lua script on disk. That's not spelled require; that's spelled dofile.

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