Union type generates anyOf instead oneOf

[ddontsov-b2broker]

export const example = z.union([
  z.string().min(10),
  z.string().min(5)],
);

will generate schema like this in zod@4.1.13

"exampleField": {
  "anyOf":[
    {"type":"string","minLength":10},
    {"type":"string","minLength":5}
  ]
}

but in zod v3 it generates:

"exampleField": {
  "oneOf":[
    {"type":"string","minLength":10},
    {"type":"string","minLength":5}
  ]
}

I think it should use oneOf instead of anyOf

[dosubot]

Hi @ddontsov-b2broker! I'm Dosu and I’m helping the zod team.

Zod v4 intentionally generates anyOf for union types in JSON Schema, while Zod v3 used oneOf. This change is based on the semantics of Zod unions: Zod's .union() accepts any value that matches any of the schemas, not exactly one, which aligns with how anyOf works in JSON Schema. In contrast, oneOf requires exactly one schema to match, which is only guaranteed for discriminated unions (created with z.discriminatedUnion), so Zod v4 uses oneOf for those cases only. This is an intentional design decision, not a bug or regression (maintainer comment, code reference).

If you need oneOf instead of anyOf for unions, you can use the override option in z.toJSONSchema() to post-process the schema. For example:

const jsonSchema = z.toJSONSchema(example, {
  override(ctx) {
    if (ctx.jsonSchema.anyOf) {
      ctx.jsonSchema.oneOf = ctx.jsonSchema.anyOf;
      delete ctx.jsonSchema.anyOf;
    }
  }
});

(example from issue #4519)

Let me know if this answers your question, or if you need help with a specific use case! If you're satisfied, feel free to close the issue.

To reply, just mention @dosu.

---

Share context across your team and agents. Try Dosu.