Typescript how to create type with common properties of two types?

Question

There are 2 types

type A = {
  x: number
  y: number
}

type B = {
  y: number
  z: number
}

How to get type with common properties of that

Answer 1

Common Properties

Use static keyof operator:

type Ka = keyof A // 'x' | 'y'
type Kb = keyof B // 'y' | 'z'
type Kc = Ka & Kb // 'y'

And define a Mapped Type with properties in Kc:

type C = {
  [K in keyof A & keyof B]: A[K] | B[K]
}

This defines a new type where each key will be present both in A and B.

Each value associated to this key will have type A[K] | B[K], in case A[K] and B[K] are different.

---

Common Properties with Same Types only

Use Conditional Type to map key to value only if type same in A and B:

type MappedC = {
  [K in keyof A & keyof B]:
    A[K] extends B[K] // Basic check for simplicity here.
    ? K // Value becomes same as key
    : never // Or `never` if check did not pass
}

From this object, get union of all values, by accessing all keys:

// `never` will not appear in the union
type Kc = MappedC[keyof A & keyof B]

Finally:

type C = {
  [K in Kc]: A[K]
}

Answer 2

Based on @kube's answer, you could use generics to create a reusable type:

type Common<A, B> = {
    [P in keyof A & keyof B]: A[P] | B[P];
}

This allows you to create intersections on the fly:

const c: Common<T1, T2> = { y: 123 };

Answer 3

Extract common props with same types only

Based on @kube's answer,

type Common<A, B> = Pick<
  A,
  {
    [K in keyof A & keyof B]: A[K] extends B[K]
      ? B[K] extends A[K]
        ? K
        : never
      : never;
  }[keyof A & keyof B]
>;