Converts anyOf: type: object to Record<string, never> about openapi-typescript HOT 3 CLOSED

That’s correct. A { "type": "object" } has no properties (Record<string, never>), and you didn’t declare additionalProperties explicitly. And the | null is correct. See the docs on this behavior.

openapi-typescript will never generate a any type because that’s basically not typechecking at all. In order to achieve that behavior, you’ll have to either make your OpenAPI schema more explicit (recommended), or override the type with the desired type (i.e. OpenAPI schema didn’t declare a descriptive type, therefore openapi-typescript couldn’t, so I’ll have to override it manually in TypeScript).

from openapi-typescript.

That’s correct. A { "type": "object" } has no properties (Record<string, never>), and you didn’t declare additionalProperties explicitly. And the | null is correct. See the docs on this behavior.

openapi-typescript will never generate a any type because that’s basically not typechecking at all. In order to achieve that behavior, you’ll have to either make your OpenAPI schema more explicit (recommended), or override the type with the desired type (i.e. OpenAPI schema didn’t declare a descriptive type, therefore openapi-typescript couldn’t, so I’ll have to override it manually in TypeScript).

What if it's intended that I can send something with whatever keys and values I want to the backend (for example, it's an extra field that's saved to the DB as a JSON type)? Is there a good way to represent that the type of the record doesn't matter?

from openapi-typescript.

Yes! That’s additionalProperties. You can give that a type, and be as descriptive as possible. It will let any arbitrary keys return the provided shape.

You can also set additionalProperties: true to generate Record<string, unknown>, but it’s always best to just pass a full Schema Object to additionalProperties ($refs work too)

from openapi-typescript.