Typescript: How to extract only the optional keys from a type? [duplicate]

Question

Let's say I have a type like this:

type User = {
  uid: string,
  displayName?: string,
  bestFriend?: string,
}

Is it possible to extract the optional properties from my User type using a mapped type of some kind? I'm looking for how to define the OptioanlProperties<T> type below.

type OptionalUserProperties = OptionalProperties<User>
// type OptionalUserProperties = "displayName" | "bestFriend"

My use-case is to compute an UpdateOf<User> type that permits specific "operation" values, like DeleteProperty to be assigned to keys that are optional in the base type.

export type UpdateOf<T> =
  // optional fields may be their own type, or the special DeleteProperty type.
  { [P in OptionalProperties<T>]?: T[P] | DeleteProperty } &
  // required fields may be changed, but cannot be deleted.
  { [P in Diff<keyof T, OptionalProperties<T>>]?: T[P] }

Karol Majewski · Accepted Answer · 2018-12-22 22:39:32Z

49

Yes:

type OptionalPropertyOf<T extends object> = Exclude<{
  [K in keyof T]: T extends Record<K, T[K]>
    ? never
    : K
}[keyof T], undefined>

answered Dec 22, 2018 at 22:39

Karol Majewski

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3 Comments

Just Jake Over a year ago

This is much more concise than the solution in the linked duplicate question. Thank you.

wcandillon Over a year ago

Indeed this solution feels better as well.

Qwerty Over a year ago

This is amazing what one can do with TypeScript! I needed this to improve my ModifyDeep to allow changing of property optionality of an existing type.

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