Implement a type that adds a new field to the interface. The type takes the three arguments. The output should be an object with the new field.
For example
type Test = { id: '1' }
type Result = AppendToObject<Test, 'value', 4> // expected to be { id: '1', value: 4 }
/* _____________ Your Code Here _____________ */
type Merge<T extends Record<PropertyKey, any>> = {
[Key in keyof T]: T[Key]
}
type AppendToObject<T extends Record<PropertyKey, any>, U extends string | number | symbol, V extends any> = Merge<T & {
[Key in U]: V
}>
/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'
type test1 = {
key: 'cat'
value: 'green'
}
type testExpect1 = {
key: 'cat'
value: 'green'
home: boolean
}
type test2 = {
key: 'dog' | undefined
value: 'white'
sun: true
}
type testExpect2 = {
key: 'dog' | undefined
value: 'white'
sun: true
home: 1
}
type test3 = {
key: 'cow'
value: 'yellow'
sun: false
}
type testExpect3 = {
key: 'cow'
value: 'yellow'
sun: false
isMotherRussia: false | undefined
}
type cases = [
Expect<Equal<AppendToObject<test1, 'home', boolean>, testExpect1>>,
Expect<Equal<AppendToObject<test2, 'home', 1>, testExpect2>>,
Expect<Equal<AppendToObject<test3, 'isMotherRussia', false | undefined>, testExpect3>>,
]
标签:Typescript,_____________,sun,value,98,Medium,Expect,key,type From: https://www.cnblogs.com/Answer1215/p/16884174.html