Omit an object property in typescript-CodePudding

I've got the following code

export interface UserModel {
  id: string;
  email: string;
  password: string;
  firstName: string;
  lastName: string;
}

This model defines the DB model. I also have the interface that is used as a response from the API.

export type UserGetResponse = Omit<UserModel, "id" | "password">;

I don't want to respond with the id and password properties. Later when fetching the data from DB I do something like this

User.scan().then((users: UserModel[]) => users.map((user: UserModel) => user as UserGetResponse))

but the id and password are not removed from the response. Is there any better way to do this. Should I create a class with a constructor which gets UserModel object and assigns only properties I need?

CodePudding user response：

You can do this with a blacklist approach. Something like:

User
  .scan()
  .then((users: UserModel[]) => users.map((user: UserModel) => {
    ...user,
    id: undefined,
    password: undefined,
  } as UserGetResponse))

Or better yet you can do it with a whitelist approach. Something like:

User
  .scan()
  .then((users: UserModel[]) => users.map((user: UserModel) => {
    firstName: user.firstName,
    lastName: user.lastName,
    email: user.email,
  } as UserGetResponse))

The white list it better in my opinion as it has no risk of including more properties automatically if/when the data structure gets extended.