Suppose there is an indexed type:
type X = {
a: 'A',
b: 'B'
}
Is there a way to derive the following type from it:
type V = 'A' | 'B'
without using an explicit method like this:
type V = X['a'] | X['b']
What I am looking for is something similar to using keyof (for obtaining a keys union type), but for values.
To access the values of the keys in object X, you can use a type query like this:
type Value = X[keyof X]
When using a type query, it will give you a union of all possible field types, so X['a'] | X['b'] is equivalent to X['a' | 'b']. This is due to the fact that keyof returns a union of string literal types representing all the keys in the object.
While this question may have already been addressed, if you want to refine your collection to specific values rather than basic types and you're dealing with unchangeable state, here's a method you can use:
const Y = {
x: 'X',
y: 'Y'
// set properties as constant, otherwise string type assumed
} as const
type YValues = typeof Y[keyof typeof Y]
// "X" | "Y"
On the other hand, if you're handling both modifiable and unchangeable versions, establishing a distinct readonly type could come in handy:
const Y = {
x: 'X',
y: 'Y'
}
type Y = Readonly<typeof Y>
type YValues = keyof Y
// "X" | "Y"
If you're looking to handle a union of types rather than just one type, you can achieve it like this:
type allSubTypesOfUnionType<T> = T extends any ? T[keyof T] : never
For example, let's say you have a union of 2 types:
type car = {
make: string
model: string
}
type bike = {
gears: number
isMountainBike: boolean
}
type vehicles = car | bike
You can then define a type that extracts all the sub-types as a union:
type possibleTypesOfVehicles = allSubTypesOfUnionType<vehicles>
// types included are: string | number | boolean
This method was inspired by this answer which focused on keys rather than sub-types.
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