[question] ZodType with ZodObject props

[domosedov]

Hello, how create zodtype with object props from external types?

// Generated type from API
type User = {
 id: number
 name: string
}

const UserSchema = z.ZodType<User> = z.object({
  id: z.number(),
  name: z.string()
})

UserSchema.pick  // not working
// Property 'pick' does not exist on type 'ZodType<User, ZodTypeDef, User>'.

[suhaotian]

type User = {
  id: number;
  name: string;
};

const UserSchema = z.object({
  id: z.number(),
  name: z.string(),
});

type USER = z.infer<typeof UserSchema>;

[nicolapalavecino]

Maybe you are looking to construct a validation object (z.object(...)) with a type. I believe that's not possible because of the compile time/run time limitations. I suppose you are looking something like this:

// Generated type from API
type User = {
  id: number;
  name: string;
};

const UserSchema: z.ZodObject<any, any, any, User> = z.object({
  id: z.number(),
  name: z.string(),
});

UserSchema.pick; // now it works

[domosedov]

Modern solution TS 4.9+

type User = {
    id: number
    name: string
    age: number
}

const UserSchema = z.object({
    id: z.number(),
    name: z.string(),
    age: z.number()
}) satisfies z.ZodType<User>