How to make type optional in typeof typescript

Question

I have a code User.container.:

const mapState = (state: AppStateType) => ({
  users: getUsers(state),
  pageSize: getPageSize(state),
  isFetching: getIsFetching(state)
})
const mapDispatch = {
  toggleFollowingInProgress,
}
const connector = connect(mapState, mapDispatch)
type PropsFromRedux = ConnectedProps<typeof connector>

export type PropsType = PropsFromRedux & {
  onPageChanged?: (page: number) => void
}

in file Users.tsx i make this:

import { PropsType } from './Users.container'
const Users = (props: PropsType) => {
...
}

The variable isFetching is not required for Users.tsx. But the compiler throws an error because i can not through ? make mapState isFetching optional.
What to do? I need to get rid of code duplication

Answer 1

You have a few options. You could make that property optional itself. But if you want isFetching to be required in the original context you could make another a separate type. I tend to create my types via type or interface rather than rely on the typeof operator.

That being said, Typescript created Utility Types.

TypeScript provides several utility types to facilitate common type transformations. These utilities are available globally.

Partial is a Utility Type for the purpose of:

Constructs a type with all properties of T set to optional. This utility will return a type that represents all subsets of a given type.

So you could write

const Users = (props: Partial<PropsType>) => {

Just keep in mind that will make all the properties optional, not just isFetching.