How to find only files which have the same name as a parent directory

[onufriievkyrylo]

For example I have a simple directory 'routes':
-routes
--chanels
---helper.js
---chanels.js
--users
---helper.js
---users.js
--medias
---index.js
How can I get something like this
/routes/chanels/chanels.js
/routes/users/users.js

[isaacs]

There's no way to do backreferences or capture groups in glob patterns, so this is a little bit tricky.

With custom Ignore function options (planned, not implemented), you'd be able to kind of do this with a clever ignore function, but otherwise, I think you're stuck doing some programming.

import type { Path } from 'path-scurry'
import { glob } from 'glob'
const findFilesMatchingParentDir = async (extension: string) => {
  const results:string[] = []
  for (const dir of await glob(`**/`, { withFileTypes: true })) {
    const file = await dir.child(dir.name + '.' + extension).lstat()
    if (file) results.push(file.relative())
  }
  return results
}

Or if you don't know the extension:

import { Glob } from 'glob'
const findFilesMatchingParentDir = async () => {
  const results:string[] = []
  const g = new Glob(`**/`, { withFileTypes: true })
  for await (const dir of g) {
    const files = await glob(dir.name + '.*', { ...g, withFileTypes: false, cwd: dir.fullpath() })
    results.push(...files)
  }
  return results
}

[isaacs]

If we went the way suggested in #261, then it would be something like this:

import { Glob } from 'glob'
const findFilesMatchingParentDir = async (extension: string) => {
  return await glob(`**`, {
    ignore: {
      ignore: (p) => !!p.parent && (extension ? !p.isNamed(p.parent.name + '.' + extension) : p.name.startsWith(p.parent.name + '.')),
      ignoreChildren: () => false,
    }
  })
}