proper mechanism to integrate with 3rd party types

[cdaringe]

problem

import { Bananas } from '@fruit/banana'
import * as z from 'zod'
const schema = z.object({
  apple: z.string(),
  banana: z.doSomethingWith3pType() // z.object({}) ??
})
const myType = z.infer<typeof schema> // { apple: string, banana: ???? }

what the best way to get my 3p schema in? assuming because types have been compiled away, getting my 3p schema may not be possible, but, is there a mechanism to put my type into the schema, and get it reflected back out?

[colinhacks]

I suppose I could implement a generic z.external(...) that accepts any object that conforms to a particular interface, probably this:

interface Schema<T> {
  parse: (data:unknown): T;
  check: (data:unknown): data is T;
}

You'd have to define a class that defines these methods and delegates to the appropriate methods on your third-party schemas. If you define a generic class, you should be able to get type inference to work as well (assuming the third party library has some type inference support).

What libraries are you trying to integrate with?

[cdaringe]

ATM im extending my schema with @types/puppeteer::Viewport. i have something like { url: string, viewport: Viewport } 🤔

[colinhacks]

Ah so you're only interested in incorporating static types in your schemas, not doing runtime validation, correct?

You should just construct your own types using TypeScript's functionality:

const BaseType = z.object({ url: z.string() });
type BaseType = z.infer<typeof Test>;

type FinalType = BaseType & { viewport: Viewport };
 => { url: string; viewport: Viewport }

[colinhacks]

@cdaringe Just thought of a better way to do this!

const Viewport: z.ZodType<Viewport> = z.any();

const FullSchema = z.object({ url: z.string(), viewport: Viewport });

This way you can still do validations with parse as you'd expect since z.any() allows any inputs, and the type inference works as expected. 👍

[cdaringe]

Doh! I figured I was missing something! Perfecto, thx!