NodeJS fs and require looking in different places

Question

I am trying to loop through a directory and require every file in it. My code is working, but I was wondering why I had to modify the path in fs functions. (Code below stripped of useless info)

Folder structure:

project
|-- bin
|   `-- start
|-- modules
|   |-- file1.js
|   `-- file2.js
`-- package.json

/bin/start:

#!/usr/bin/env node

// Require dependencies
var fs = require('fs')

// Get modules
// NOTE: fs is looking in the project folder, but this file is in the bin folder
fs.readdir('./modules', function (err, files) {
  // Handle error
  if (err) {
    console.log(err)
    process.exit(1)
  }

  // Loop through files
  files.forEach(function (file, index) {
    // Get info about file
    fs.stat('./modules/' + file, function (err, stat) {
      // Handle error
      if (err) {
        console.log(err)
        process.exit(1)
      }

      // If it is a file
      if (stat.isFile()) {
        // NOTE: require is looking in the bin folder, and this file is in the bin folder
        require('../modules/' + file)
      }
    })
  })
})

package.json:

{
  "name": "modular-chat",
  "version": "1.0.0",
  "description": "A simple modular example of a chat application",
  "scripts": {
    "start": "node ./bin/start"
  },
  "author": "JoshyRobot",
  "license": "MIT"
}

Answer 1

require is a Node function. It's just that require uses __dirname as the basename when resolving relative paths rather than the current working directory which is also the result of process.cwd().

You simply need to join __dirname to your relative paths before passing them to fs functions. Using using path.join(__dirname, '../modules') and path.join(__dirname, '../modules', file). Use these in your fs.readdir and fs.stat calls:

fs.readdir(path.join(__dirname, '../modules'), function (err, files) {

fs.stat(path.join(__dirname, '../modules', file), function (err, stat) {

This makes both your fs calls and the require align so that the right files are loaded.

---

It is also not that hard to do the reverse and make require load any path:

Rather than require(path.join('../modules/', file)):

require(path.join(process.cwd(), '../modules/', file))

This works because require doesn't change absolute paths as it does with relative paths by prepending __dirname.