Using a unit type instead of a variant?

[jawnothin]

Is there a way to make use of a unit type instead of having to re-type it out as a picklist?

type Choices = 'a' | 'b' | 'c'

v.picklist<Choices>()

instead of having to do

v.picklist(['a', 'b', 'c'])

[fabian-hiller]

No, this would require a compiler to translate your TypeScript type into run-time code. TypeScript types do not exist at runtime and are removed before the transpiled JavaScript code is executed. However, you can retrieve the TypeScript type from the runtime code. Here is an example:

import * as v from 'valibot';

const options = ['a', 'b', 'c'] as const;
type Choices = typeof options[number];
const Schema = v.picklist(options);

[jawnothin]

Right I figured that would be the case.

The problems lies in the package that generates this list in the first place, as it doesn't generate the array of strings, but just the unit type instead.

Thanks for the reply.

[fabian-hiller]

If your main concern is a mismatch between the generated type and the schema, you could use the satisfies keyword:

import * as v from 'valibot';

type Choices = 'a' | 'b' | 'c';
const Schema = v.picklist(['a', 'b', 'c']) satisfies v.GenericSchema<Choices>;