Typescript: Strong-typing (and autocomplete) for a value based on a sibling object's keys

Question

Say I have an object that looks like this:

const configuration: Config = {
  options: {
    'Option 1': 'some value here',
    'Option 2': 'some other value here'
  },
  defaultOption: 'Option 1'
}

How can I write the type Config such that:

• defaultOption can only be one of the keys in the .options object. ie. 'Option 1' or 'Option 2' in the case above.
• .options can have any number of key-value pairs.
• The user doesn't have to specify the .options keys upfront. So, the user shouldn't have to specify const configuration: Config<'Option 1' | 'Option 2'> = .... It's ok to use generics, but the user shouldn't have to specify them.

Answer 1

In order to do that and validate, you need to infer your config object from function argument:

type Configuration<
  Options
  >
  = {
    options: Options,
    defaultOption: keyof Options
  }

const config = <
  Options extends Record<string, string>,
  Config extends Configuration<Options>,
  Default extends keyof Config['options'],
  >(config: { options: Options, defaultOptions: Default }) => config

const result = config({
  options: {
    'Option 1': 'some value here',
    'Option 2': 'some other value here'
  },
  defaultOptions: 'Option 1'
}) // ok

const result2 = config({
  options: {
    'Option 1': 'some value here',
    'Option 2': 'some other value here'
  },
  defaultOptions: 'invalid property'
}) // error

Playground