Generic index signature only for specific sub types

Question

Every of our database tables has an identifier (id) and version. Which means every concrete Entity extends from the Entity interface in order to have those two attributes.

interface Entity {
  id: number;
  version: number;
}

Here is a multi-level entity for example purposes.

interface Address extends Entity {
    street: string;
    hno: string;
    zip: string;
    city: string;
    country: string;
}

interface Department extends Entity {
    name: string;
}

interface User extends Entity {
    name: string;
    birthday: string;
    address: Address;
    department: Department;
}

Now I want a generic type which replaces all appearances of id and version (even the fields in the sub-objects) into optional. Which means from

User {
    dbid: number;
    version: number;
    name: string;
    birthday: string;
    address: {
        dbid: number;
        version: number;
        street: string;
        hno: string;
        zip: string;
        city: string;
        country: string;
    };
    department: {
        dbid: number;
        version: number;
        name: string;
    };
}

into

User {
    dbid?: number;
    version?: number;
    name: string;
    birthday: string;
    address: {
        dbid?: number;
        version?: number;
        street: string;
        hno: string;
        zip: string;
        city: string;
        country: string;
    };
    department: {
        dbid?: number;
        version?: number;
        name: string;
    };
}

As I said as a generic type, because I have a lot of different entities where I want to perform the same.

Something like

export type NewEntity<T extends Entity> = {
  // remove id and version fields
  [K in keyof Omit<T, 'dbid' | 'version'>]: T[K] extends Entity ? NewEntity<T[K]> : T[K];
};

but this removes only the two attributes but lacks the redefinition of

id?: number;
version?: number

Tobias S. · Accepted Answer · 2022-09-23T10:27:21.093

You can define a recursive generic type like this one:

type NewEntity<T extends Entity> = ({
  [K in keyof T as K extends keyof Entity ? never : K]: 
    T[K] extends Entity 
      ? NewEntity<T[K]> 
      : T[K]
} & {
  [K in Extract<keyof T, keyof Entity>]?: T[K]
}) extends infer O ? { [K in keyof O]: O[K] } : never

It creates an intersection of two mapped types.

The first one is including all properties which are not in Entity (id and version). If a property type T[K] extends Entity, we can call the type NewEntity recursively with it.

The second type includes the properties of T which are also in Entity. Those are made optional.

The extends infer O ? { [K in keyof O]: O[K] } : never is only for aesthetic reasons and makes the resulting type readable. (See this for more information)

Here is the result:

type NewUser = NewEntity<User>

// type NewUser = {
//     name: string;
//     birthday: string;
//     address: {
//         street: string;
//         hno: string;
//         zip: string;
//         city: string;
//         country: string;
//         id?: number | undefined;
//         version?: number | undefined;
//     };
//     department: {
//         name: string;
//         id?: number | undefined;
//         version?: number | undefined;
//     };
//     id?: number | undefined;
//     version?: number | undefined;
// }

Playground

But what if e.g. address.hno is optional in the base interface: hno?: string; In NewEntity it will be hno: string | undefined but not hno?: string | undefined. Is there a generic way for adopting the exact same type in this case? — user2622344, Sep 23 '22 at 08:02

Generic index signature only for specific sub types

1 Answers1