arktype•16mo ago

How would you define an object with optional keys, but require at least one of the keys to be filled

An example of a very basic schema:

const schema = type({ a: type.string.optional(), b: type.string.optional() });

const schema = type({ a: type.string.optional(), b: type.string.optional() });

const schema = type({ a: type.string.optional(), b: type.string.optional() });

const schema = type({ a: type.string.optional(), b: type.string.optional() });

I'd like this to accept values

{a: 'foo'}

{a: 'foo'}

{a: 'foo'}

{a: 'foo'},

{ b: 'foo' }

{ b: 'foo' }

{ b: 'foo' }

{ b: 'foo' }, and

{ a: 'foo', b: 'bar' }

{ a: 'foo', b: 'bar' }

{ a: 'foo', b: 'bar' }

{ a: 'foo', b: 'bar' }, but fail on

{}

{}

{}

{}. It would be a lovely bonus for the type to be inferred as

{ a: string, b?: string } | { a?: string, b: string }

{ a: string, b?: string } | { a?: string, b: string }

{ a: string, b?: string } | { a?: string, b: string }

{ a: string, b?: string } | { a?: string, b: string } as well. Is there a way to do this without explicitly specifying that inferred type as an arktype union using

.or

.or

.or

.or?

How would you define an object with optional keys, but require at least one of the keys to be filled

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