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Hello, I'm trying to remove the Brand type from an inferred zod type (the branding can happen anywhere), do anyone know how to do this? const schema = z
.object({
name: z.string(),
age: z.number().min(1).int().brand('age'),
})
.brand("schema");
type SchemaWithBrand = z.infer<typeof schema>;
type SchemaWithoutBrand = ??<SchemaWithBrand> |
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Replies: 5 comments 3 replies
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Is this what you are looking for? const schema = z.object( {
name: z.string(),
age: z.number().min( 1 ).int(),
} ).brand( 'schema' )
type SchemaWithBrand = z.infer<typeof schema>
// type SchemaWithBrand = {
// name: string
// age: number
// } & z.BRAND<'schema'>
type SchemaWithoutBrand = z.infer<typeof schema._def.type>
// type SchemaWithoutBrand = {
// name: string
// age: number
// }If you found my answer satisfactory, please consider supporting me. Even a small amount is greatly appreciated. Thanks friend! 🙏 |
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thank you, but i need a way to get the schema without the brand without re using |
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after a little bit of digging, I found this: const schema = z.object( {
name: z.string(),
age: z.number().min( 1 ).int(),
} ).brand( 'schema' )
type SchemaWithBrand = z.infer<typeof schema>
// type SchemaWithBrand = {
// name: string
// age: number
// } & z.BRAND<'schema'>
type SchemaWithoutBrand = Omit<SchemaWithBrand, typeof z.BRAND>
// type SchemaWithoutBrand = {
// name: string
// age: number
// }If you found my answer satisfactory, please consider supporting me. Even a small amount is greatly appreciated. Thanks friend! 🙏 |
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thank you for your time, unfortunately I forgot to say that I need this to be able to unbrand not only object as the first level, but anywhere in the type :/ |
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Hey @afoures, The following should deeply to unbrand anywhere in the type. Slight disadvantage is that, In case you're passing a branded primitive such as Although if your goal is to infer such as // Our Unbranding Type.
type RemoveBranding<T> = T extends z.ZodType<any, any, infer U> ? z.ZodType<U> : T;
// Usage
let Unbranded: RemoveBranding<typeof mySchema>; |
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Hello! To remove the brand type from an inferred zod type, you can use the Omit utility type provided by TypeScript. Here's an example of how you can use it in your code: The Omit type takes two types of arguments - the first argument is the original type, and the second argument is a union of keys to be removed from the original type. In this case, we want to remove the '__brand' key from the SchemaWithBrand type, which is the key used by zod to store the branding information. So, by using Omit<SchemaWithBrand, '__brand'>, we create a new type that has all the properties of SchemaWithBrand, except for the __brand key. I hope this helps! |
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So in the end here is my current solution for this: import { UnionToIntersection } from 'type-fest';
type Primitive =
| string
| number
| boolean
| undefined
| null
| bigint
| Function
| Symbol
| Date
| never
| void;
type IterateOnTuple<T extends [...any[]]> = T extends [
infer Head,
...infer Tail
]
? [Unbrand<Head>, ...IterateOnTuple<Tail>]
: [];
type RemoveBrand<T> = T extends z.BRAND<infer Brand>
? T extends (
| z.BRAND<Brand>
| UnionToIntersection<{ [K in Brand]: z.BRAND<K> }[Brand]>
) &
infer X
? RemoveBrand<X>
: never
: T;
type Unbrand<T> = T extends Primitive
? RemoveBrand<T>
: T extends Promise<infer E>
? Promise<Unbrand<E>>
: T extends [any, ...any[]]
? IterateOnTuple<RemoveBrand<T>>
: T extends Array<infer E>
? Array<Unbrand<E>>
: T extends Set<infer E>
? Set<Unbrand<E>>
: T extends Map<infer E, infer F>
? Map<Unbrand<E>, Unbrand<F>>
: {
[k in Exclude<keyof T, keyof z.BRAND<any>>]: Unbrand<T[k]>;
}; |
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This almost worked for me, except that it removes the optional behavior. I ended up with the following in a import type { Primitive, UnionToIntersection } from 'type-fest';
import type { BRAND } from 'zod';
type IterateOnTuple<T extends [...unknown[]]> = T extends [infer Head, ...infer Tail]
? [Unbrand<Head>, ...IterateOnTuple<Tail>]
: [];
type RemoveBrand<T> =
T extends BRAND<infer Brand>
? T extends (BRAND<Brand> | UnionToIntersection<{ [K in Brand]: BRAND<K> }[Brand]>) & infer X
? RemoveBrand<X>
: never
: T;
/**
* Recursively removes the brand from T.
*
* @see https://github.com/colinhacks/zod/discussions/1994#discussioncomment-6068940
*/
type Unbrand<T> = T extends Primitive
? RemoveBrand<T>
: T extends Promise<infer E>
? Promise<Unbrand<E>>
: T extends [unknown, ...unknown[]]
? IterateOnTuple<RemoveBrand<T>>
: T extends (infer E)[]
? Unbrand<E>[]
: T extends Set<infer E>
? Set<Unbrand<E>>
: T extends Map<infer E, infer F>
? Map<Unbrand<E>, Unbrand<F>>
: { [K in keyof T]: Unbrand<T[K]> };
export default Unbrand; |
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So in the end here is my current solution for this: