Does zod support recursive together with discriminated unions?

[jzimmek]

Hi,

i am trying to implement the following type in zod:

type filterNot = { type: "not"; condition: filterTypes };
type filterAnd = { type: "and"; condition: Array<filterTypes> };

type filterTypes =
  | filterNot
  | filterAnd

this is a simplified version of the real type.

i tried the following:

import z from "zod";

const base_filterNot = z.object({
  type: z.literal("not"),
});

const filterNot: z.ZodDiscriminatedUnionOption<"type"> = base_filterNot.extend({
  condition: z.lazy(() => filterTypes),
});

const base_filterAnd = z.object({
  type: z.literal("and"),
});

const filterAnd: z.ZodDiscriminatedUnionOption<"type"> = base_filterAnd.extend({
  condition: z.lazy(() => filterTypes),
});

const filterTypes = z.discriminatedUnion("type", [filterNot, filterAnd]);

but this is wrong, because

type the_filterTypes = z.infer<typeof filterTypes>;

is inferred as

type the_filterTypes = {
    [x: string]: any;
    type?: unknown;
}

instead of the discriminated union type.

i already searched for similar questions, but i only found questions about "recursive" types or only about the discriminatedUnion, not both together. according to docs there needs to be a type-hint provided for "recursive" but i do not know how this is supposed to be provided if the "type-hint" itself needs to contains the "condition" field, which is supposed to be of type of the final schema.

any help would be really appreciated.