如何在NodeJs中创建消耗文件相对路径的功能（内部模块）

Question

What I am trying to do is creating function like NodeJS require . You can do require("./your-file") and require understand that the file ./your-file is sibling of the calling module, without specifying the full path of the file.

My current problem is getting the directory of current executing function ( __dirname of the executing function)

I tried things below:

• Using module.parent.fileName failed when the calling method wrapped with other function.
• By reading from V8 stack trace almost there but failed when run inside test runner (Jest)

It should be an easy solution for this, maybe I'm over complicate it?

Answer 1

May be you should use:

var path = require('path'); 
//_dirname will give you the current position

using path package you can achieve this. find Documentation here

Answer 2

Both module.parent.fileName and V8 stack trace works fine , but it will failed if you export the module in the index.js

example:

//src/my-lib.js

function myFunction(){
   const path = module.parent.fileName
   //do stuff
}

then you export the function above in the index.js

//src/index.js
const lib = require("./my-lib")

exports.myFunction = lib.myFunction

in the client code if you import myFunction from index.js the path detected will always /src/index.js