Add a way to generate a glob from a file name

[andreaskundig]

I need to generate a valid glob that uniquely matches a filename.

For a file called ba[ra].foo the glob needs to escape the brackets like this: bar\[ra\].foo, in order to avoid matching files called baa.foo or bar.foo. There are enough other characters and expressions that need escaping that the task does not seem trivial.

Maybe the solution to this is as simple as a regex that you could add to the documentation.

[andreaskundig]

Here's my solution, with some tests, but I don't know if this catches everything

export function convertPathToGlob(filePath) {
  if (!filePath) {
    throw new Error(`cannot create glob for file path "${filePath}"`)
  }
  return filePath.replace(/([[\]{}*!()])/g, '\\$1')
}

test('convertPathToGlob', () => {
  const testConversion = (path, glob, shouldNotMatch) => {
    expect(convertPathToGlob(path)).toBe(glob)
    expect(micromatch.isMatch(path, glob)).toBe(true)
    expect(micromatch.isMatch(path, convertPathToGlob(path))).toBe(true)
    expect(micromatch.isMatch(shouldNotMatch, glob)).toBe(false)
    expect(micromatch.isMatch(shouldNotMatch, path)).toBe(true)
  }
  testConversion('ba[r].foo', 'ba\\[r\\].foo', 'bar.foo')
  testConversion('ba[r-t].foo', 'ba\\[r-t\\].foo', 'bas.foo')
  testConversion('ba*.foo', 'ba\\*.foo', 'bar.foo')
  testConversion('!baa.foo', '\\!baa.foo', 'bar.foo')
  testConversion('*.!(a)', '\\*.\\!\\(a\\)', 'a.b')
  testConversion('foo/{a,b,c}/bar', 'foo/\\{a,b,c\\}/bar', 'foo/a/bar')
  testConversion('ba?(a).foo', 'ba?\\(a\\).foo', 'baa.foo')
  testConversion('ba*(a).foo', 'ba\\*\\(a\\).foo', 'baa.foo')
  testConversion('ba@(a).foo', 'ba@\\(a\\).foo', 'baa.foo')
  testConversion('ba+(a).foo', 'ba+\\(a\\).foo', 'baa.foo')
  testConversion('ba[[:alpha:]]', 'ba\\[\\[:alpha:\\]\\]', 'baa')
  testConversion(
    '{f,b}*/{1..3}/{b,q}*',
    '\\{f,b\\}\\*/\\{1..3\\}/\\{b,q\\}\\*',
    'foo/2/qux'
  )
})