Example:

type User = {
    id: number
    name: string
}

const user: { fields: (keyof User)[] } = {
    fields: ["name"]
}

type result = Pick<User, (typeof user.fields)[number]>

Result contains all User fields, but I need only "name"

It is possible if move fields outside, but I want to keep type safety.

Example:

type User = {
    id: number
    name: string
}

const user: { fields: (keyof User)[] } = {
    fields: ["name"]
}

type result = Pick<User, (typeof user.fields)[number]>

Result contains all User fields, but I need only "name"

It is possible if move fields outside, but I want to keep type safety.

• typescript

Share Improve this question Follow edited Mar 6 at 14:52 jonrsharpe 122k3030 gold badges268268 silver badges476476 bronze badges asked Mar 6 at 14:51 Pavel PetrovPavel Petrov 1133 bronze badges 1

• You explicitly said that user.fields is an array of any keyof User. You want a narrower type, e.g. tsplay.dev/m34ybW. – jonrsharpe Commented Mar 6 at 14:53

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2 Answers 2

Sorted by: Reset to default -1

Annotating a variable means assigning the annotation's type to it, so it would key keyof User. To preserve an exact narrow type use satisfies keyword:

Playground

const user = {
    fields: ["name"]
} satisfies { fields: (keyof User)[] } ;

Yes, you can achieve this using as const to preserve the literal type of fields:

const user = { fields: ["name"] as const };