How does typescript define multiple type arrays? And array item is one of them.
interface A {
m: string,
n: number
}
interface B {
x: boolean
y: number[]
}
type C = A | B
const arr: C[] = [
{m: '1', n: 1, x: true}
]
I want array item is A or B type, only match one of them, like this:
const arr: C[] = [
{m: '1', n: 1}
]
// or
const arr: C[] = [
{x: true, y: [1,2,3]}
]
In order to achieve it, you need to make this union more strict.
interface A {
m: string,
n: number
}
interface B {
x: boolean
y: number[]
}
type UnionKeys<T> = T extends T ? keyof T : never;
type StrictUnionHelper<T, TAll> =
T extends any
? T & Partial<Record<Exclude<UnionKeys<TAll>, keyof T>, never>> : never;
// credits goes to https://stackoverflow.com/questions/65805600/type-union-not-checking-for-excess-properties#answer-65805753
type StrictUnion<T> = StrictUnionHelper<T, T>
type C = StrictUnion<A | B>
const arr1: C[] = [
{ m: '1', n: 1, x: true } // expected error
]
const arr2: C[] = [
{ m: '1', n: 1 } // ok
]
const arr3: C[] = [
{ x: true, y: [1, 2, 3] }, // ok
{ m: '1', n: 2 }
]
Please see related examples in my blog
If both the property names are same in the interface, then it will make sure that you can not mix them two. It also usually not make sense to create a union of two interfaces that do not have anything in common.
interface A {
m: string,
n: number
}
interface B {
m: boolean
n: number[]
}
type C = A | B
// no error
const arr: C[] = [
{m: '1', n: 1}
]
// error (you cant mix both, it has to be one or the other)
const arr1: C[] = [
{m: '1', n: [1]}
]
