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 types?

type C = Something<T1, T2> // { y: number } 

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.

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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 };