How to get all subdirectories and return their paths?

February 19, 2023

Assume there is the following folder structure:

/apps
|- /group1
|  |- /project1
|  |- /project2
|- /group2
|  |- /project3
|  |- /project4

I need to get the path of all projects – which are all two level deep (group folder and then project folder).

So I started with this:

import { readdirSync } from 'fs'

const directoriesInDirectory = readdirSync([rootPath, 'apps'].join('/'), {
  withFileTypes: true
})
.filter((item) => item.isDirectory())
.map((item) => item.name)

This gives me all groups ['group1', 'group2'].

But I need to get this:

[
    'apps/group1/project1',
    'apps/group1/project2',
    'apps/group2/project3',
    'apps/group2/project4'
]

>Solution :

You need to perform readdirSync multiple times, once for the app directory, and the rest for the nested directories. Also, instead of using sync, if you adapt the non-blocking nature, you can use promises instead:

import { readdir } from 'fs/promises'

const startPath = [rootPath, 'apps'].join('/')

const parentDirs = (
  await readdir(startPath, {
    withFileTypes: true,
  })
)
  .filter((item) => item.isDirectory())
  .map((item) => item.name)

const projectPaths = (
  await Promise.all(
    parentDirs.map(async (dir) => {
      const projects = (
        await readdir(startPath + '/' + dir, {
          withFileTypes: true,
        })
      )
        .filter((item) => item.isDirectory())
        .map((item) => item.name)
      return { dir, projects }
    })
  )
)
  .map(({ dir, projects }) =>
    projects.map((project) => [startPath, dir, projects].join('/'))
  )
  .flat()

console.log(projectPaths) // ['apps/group1/project1', 'apps/group1/project2', ...]