extending a ZodType

[reharik]

Hi, I have the following types Address and Location. Location extends Address. I'm trying to get zod schemas that are typed by the TS types. I have to resort to the solution below because ZodType<Address> does not have .extend or .merge on it. My question is, is this the proper way of doing it? Perhaps the way I"m "typing" the schema is incorrect. I tried ZodObject but the generic takes a ZodRawShape.
Any help on "typing" and or extending would be greatly appreciated.
Thanks

import { z } from 'zod';

export interface Address {
  id?: string;
  addressText: string;
  line1: string;
  line2?: string;
  city: string;
  state: string;
  postalCode: string;
  country: string;
}

export interface Location extends Address {
  latitude: number;
  longitude: number;
  businessName?: string;
}

const addressSchema_ = z.object({
  id: z.string().optional(),
  addressText: z.string(),
  line1: z.string(),
  line2: z.string().optional(),
  city: z.string(),
  state: z.string(),
  postalCode: z.string(),
  country: z.string(),
});

export const addressSchema: z.ZodType<Address> = addressSchema_;
export const locationSchema: z.ZodType<Location> = addressSchema_.extend({
  latitude: z.number(),
  longitude: z.number(),
  businessName: z.string().optional(),
});

[JacobWeisenburger]

Is this what you are looking for?

interface Address {
    id?: string
    addressText: string
    line1: string
    line2?: string
    city: string
    state: string
    postalCode: string
    country: string
}

interface Location extends Address {
    latitude: number
    longitude: number
    businessName?: string
}

const addressSchema = z.object( {
    id: z.string().optional(),
    addressText: z.string(),
    line1: z.string(),
    line2: z.string().optional(),
    city: z.string(),
    state: z.string(),
    postalCode: z.string(),
    country: z.string(),
} ) satisfies z.ZodType<Address>

const locationSchema = addressSchema.extend( {
    latitude: z.number(),
    longitude: z.number(),
    businessName: z.string().optional(),
} ) satisfies z.ZodType<Location>

[JacobWeisenburger]

it's the best way I can think of

[reharik]

@JacobWeisenburger
Hi, this does seem to satisfy TS, but where does the satisfies keyword come from? I can't find it anywhere and it seems to break prettier :)

[JacobWeisenburger]

https://www.typescriptlang.org/docs/handbook/release-notes/typescript-4-9.html#the-satisfies-operator

[reharik]

Interesting I guess it hasn't made it into the docs yet. Or perhaps my docs are set to the wrong version. Whatever.
Thanks!

[alexanderhorner]

The problem with this is that it won't infer the type anymore.

interface Child {
  name: string
}
const ChildSchema = z.object({
  name: z.string()
}) satisfies z.ZodType<Child>;

interface Parent {
  child: Child
}

const ParentSchema = z.object({
  child: ChildSchema
}) satisfies z.ZodType<Parent>;

type Test = z.infer<typeof ParentSchema>

Result:

type Test = {
    child: {
        name: string;
    };
}

---

Not using satisfies:

interface Child {
  name: string
}
const ChildSchema: z.ZodType<Child> = z.object({
  name: z.string()
})

interface Parent {
  child: Child
}

const ParentSchema: z.ZodType<Parent> = z.object({
  child: ChildSchema
})

type Test = z.infer<typeof ParentSchema>
type Test2 = Test['child']

Result:

type Test = Parent
type Test2 = Child