jsDoc `@discriminator kind` tag generates `z.union()` instead of `z.discriminatedUnion()`

[alexkonst]

Bug description

Hi. Thanks for the helpful tool.

When we using the jsdoc @discriminator tag to generate a schema, we get an incorrect result that differs from the expected one (z.union() instead of z.discriminatedUnion()).

According to the zod library's author (issue comment), there are no plans to fix the issue with union types working with discriminatedUnion. This behavior can be corrected for generated schema by explicitly using the .extend(Scheme.shape) method or an equivalent spread syntax (see example below).

Input

export type Foo = { a: string }
export type Bar = { b: number }

/**
 * @discriminator kind
 */
export type Item =
  | { kind: 'foo' } & Foo
  | { kind: 'bar' } & Bar

Expected output

// Generated by ts-to-zod
import { z } from 'zod'

export const fooSchema = z.object({ a: z.string() })
export const barSchema = z.object({ b: z.number() })

export const itemSchema = z.discriminatedUnion('kind', [
  z.object({ kind: z.literal('foo'), ...fooSchema.shape }),
  z.object({ kind: z.literal('bar'), ...barSchema.shape }),
])

Actual output

// Generated by ts-to-zod
import { z } from 'zod'

export const fooSchema = z.object({ a: z.string() })
export const barSchema = z.object({ b: z.number() })

export const itemSchema = z.union([
  z.object({ kind: z.literal('foo') }).and(fooSchema),
  z.object({ kind: z.literal('bar') }).and(barSchema),
])

Versions

• Typescript: v5.8.3
• Zod: v4.3.5