Type inference in with predefined types

[ColemanDunn]

Hello! I am wanting to get type inference with my own types I have already defined. Using the example on the readme as a reference:

import { email, minLength, object, type Output, parse, string } from 'valibot'; // 1.2 kB

// Create login schema with email and password
const LoginSchema = object({
  email: string([email()]),
  password: string([minLength(8)]),
});

// Infer output TypeScript type of login schema
type LoginData = Output<typeof LoginSchema>; // { email: string; password: string }

// Returns data as { email: string; password: string }
parse(LoginSchema, { email: 'jane@example.com', password: '12345678' });

I clearly now have the type LoginData to use in functions and whatnot, but what if I have defined my own types ahead of time and want those to be the inferred type that could also help me define my schema? For example, if I have the type:

type MyLoginType = {email: string, password: string}

I want to know if I could do something along the lines of (this is just pseudo-code idea):

const LoginSchema = object< MyLoginType >({
  email: string([email()]),
  password: string([minLength(8)]),
});

Then I could get the following:

const loginData = parse(LoginSchema, SomeObject);

And this would result in loginData being of type MyLoginType.

Taking it a step further, it could help with the definition of my schema like I mentioned before like so:

const LoginSchema = object< MyLoginType >({
  email: string([email()]),
  password: number(), // <------------Some TypeScript error here
});

Is anything like this currently possible?
Thanks!

[fabian-hiller]

Thank you for creating this discussion. I would write it like this:

import * as v from 'valibot';

type LoginData = {
  email: string;
  password: string;
};

const LoginSchema = v.object({
  email: v.string([v.email()]),
  password: v.string([v.minLength(8)]),
}) satisfies v.BaseSchema<LoginData>;

[ColemanDunn]

Thank you!