type PropParser<F extends object, C extends object> = { // keys in both get unioned [K in keyof F | keyof C]: K extends keyof F ? K extends keyof C ? F[K] | C[K] : F[K] : K extends keyof C ? C[K] : never } type MergeAll<XS extends object[], C extends object={}> = XS extends [infer F extends object, ...infer R extends object[]] ? MergeAll<R, PropParser<F, C>> : C
Solution by mxniliuc #37540
// 你的答案
type MergeAll<XS extends any[],B extends object = {}> = XS extends [infer R,...infer RR] ? R extends R ? MergeAll<RR, {
[p in (keyof B | keyof R)]: p extends keyof B? p extends keyof R ? B[p]|R[p] : B[p] : p extends keyof R ? R[p]:never
}> :never : B
Solution by yuadh #36835
// your answers
type Merge<A extends any = {}, B extends any = {}> = {
[key in (keyof A | keyof B)]: key extends keyof A ? key extends keyof B ? A[key] | B[key] : A[key] : key extends keyof B ? B[key] : never
}
type MergeAll<T extends any[], Result extends any = {}> =
T extends [infer R, ...infer Rest]
? MergeAll<Rest, Merge<R, Result>>
: Result
Solution by duanlvxin #36654
// your answers
type Merge<A,B> = {
[T in keyof A | keyof B] : T extends keyof A & keyof B
? A[T] | B[T] : T extends keyof A ? A[T] : T extends keyof B ? B[T] : never
}
type MergeAll<T, P = {}> =
T extends [infer H, ...infer R] ? MergeAll<R,Merge<P,H>> : P
Solution by Mohmn #35344
type MergeAll<XS extends object[], U = XS[number], keys extends PropertyKey = U extends U ? keyof U : never> = {
[k in keys]: U extends U ? U[k & keyof U] : never
}
Solution by wendao-liu #35149
type MergeAll<XS> = XS extends [infer F, ...infer R]
? MergeAll<R> extends infer U
? Omit<
{
[P in keyof F & keyof U]: F[P] | U[P];
} & Omit<F, keyof F & keyof U> &
Omit<U, keyof F & keyof U>,
never
>
: never
: {};
Solution by yukicountry #34405
// U extends U是为了把联合类型展开
// XS = [{ a: 1 }, { c: 2 }]
// U = { a: 1 } | { c: 2 }
// Keys = a | b
type MergeAll<XS extends object[], U = XS[number], Keys extends PropertyKey = U extends U ? keyof U : never> = {
[K in Keys]: U extends U ? U[K & keyof U] : never
}
// 方法二
type Merge<T, P> = {
[K in keyof T | keyof P]: K extends keyof T ? (K extends keyof P ? T[K] | P[K] : T[K]) : (K extends keyof P ? P[K] : never)
}
type MergeAll<XS extends any[], U extends object = {}> = XS extends [infer R, ...infer rest] ? MergeAll<rest, Merge<U, R>> : U
Solution by ouzexi #34166
type MergeAll<List, Result = {}> =
List extends [infer A, ...infer Rest]
? MergeAll<Rest, MergeTwo<Result, A>>
: Result
type MergeTwo<A, B> = {
[Key in (keyof A | keyof B)]:
(Key extends (keyof B) ? B[Key] : never) | (Key extends (keyof A) ? A[Key] : never)
}
Solution by drylint #34137
type MergeAll<T, I extends keyof T = keyof T & `${number}`>
= {[P in I extends I ? keyof T[I] : I]: I extends I ? P extends keyof T[I] ? T[I][P] : never : I}
Solution by teamchong #33000
type MergeAll<XS, Acc extends object = NonNullable<unknown>> = XS extends [infer F, ...infer R]
? MergeAll<R, {
[P in keyof F | keyof Acc]: P extends keyof Acc
? Acc[P] | F[keyof F & P]
: F[keyof F & P]
}>
: Acc
Solution by keyurparalkar #32844
type MergeObject<M, N, O extends PropertyKey = keyof M | keyof N> = {
[P in O]: M[P & keyof M] | N[P & keyof N];
};
type MergeAll<
T extends any[],
M extends Record<PropertyKey, any> = {}
> = T extends [infer L, ...infer R] ? MergeAll<R, MergeObject<L, M>> : M;
Solution by Vampirelee #32585
type MergeAll<XS, M = {}> = XS extends [infer X, ...infer R]
? MergeAll<
R,
{
[K in keyof M | keyof X]: K extends keyof X
? K extends keyof M
? M[K] | X[K]
: X[K]
: M[K & keyof M];
}
>
: M;
Solution by vangie #32202
type MergeAll<T extends any[], U extends object = {}> = T extends [
infer F,
...infer R
]
? F extends object
? MergeAll<
R,
{
[K in keyof F | keyof U]: K extends keyof U
? K extends keyof F
? F[K] | U[K]
: U[K]
: K extends keyof F
? F[K]
: never;
}
>
: MergeAll<R, U>
: U;
Solution by moonpoet #31047
type Merge<T, U> = {[K in keyof T | keyof U]: (K extends keyof T ? T[K] : never) | (K extends keyof U ? U[K] : never)};
type MergeAll<XS extends any[]> =
XS extends [infer O, ...infer X]
? Merge<O, MergeAll<X>>
: {}
Solution by milletlovemouse #31004
type FlattenKeys<T extends {}[], U extends {} = {}> = T extends [infer F, ...infer Rest extends {}[]]
? FlattenKeys<
Rest,
{
[key in keyof F | keyof U]: [];
}
>
: U;
type Unionize<T> = T extends [infer F, ...infer Rest]
? unknown extends F
? Unionize<Rest>
: F | Unionize<Rest>
: never;
type MergeAll<T extends {}[], R extends Record<PropertyKey, any> = FlattenKeys<T>> = T extends [
infer F extends Record<PropertyKey, any>,
...infer Rest extends {}[],
]
? MergeAll<
Rest,
{
[key in keyof R]: [F[key], ...R[key]];
}
>
: {
[key in keyof R]: Unionize<R[key]>;
};
Solution by leejaehyup #30928
type Merge<T,V> = {
[Key in keyof T | keyof V] : Key extends keyof T
? Key extends keyof V
? T[Key] | V[Key]
: T[Key]
: V[Key];
}
type MergeAll<XS,All = {}> = XS extends [infer First,...infer Rest] ? MergeAll<Rest,Merge<All,First>>: All;
Solution by idebbarh #30173
type MergeAll<T, S = {}> = T extends [infer F, ...infer Rest]
? MergeAll<Rest, Omit<{
[P in Exclude<keyof S, keyof F>]: S[P]
} & {
[P in Exclude<keyof F, keyof S>]: F[P]
} & {
[P in Extract<keyof S, keyof F>]: S[P] | F[P]
}, never>>
: S
Solution by hesoso #30069
type MergeAll<
XS extends object[],
U = XS[number],
Keys extends PropertyKey = U extends U ? keyof U : never
> = {
[K in Keys]: U extends U ? U[K & keyof U] : never
}
Solution by Alexsey #29394
type Merge<T, M> = { [k in (keyof T) | (keyof M)]: k extends keyof T ? k extends keyof M ? T[k] | M[k] : T[k] : M[k]}
type MergeAll<T, R = {}> = T extends [infer I, ...infer Rest] ? MergeAll<Rest, Merge<R, I>> : R;
Solution by MrRENGE #29284
// your answers
type Merge<F, S> = {
[K in keyof F | keyof S]: K extends keyof F ? K extends keyof S ? F[K] | S[K] : F[K] : K extends keyof S ? S[K] : never
}
type MergeAll<XS extends any[], Result extends object = {}> = XS extends [infer F, ...infer Rest] ? MergeAll<Rest, Merge<Result, F>> : Result;
Solution by suchangv #29277
type MergeAll<XS extends unknown[], Res = {}> = XS extends [
infer Item,
...infer Rest
]
? MergeAll<
Rest,
Omit<
Omit<Res, keyof Item> & {
[K in keyof Item]: K extends keyof Res ? Res[K] | Item[K] : Item[K];
},
never
>
>
: Res;
Solution by DoubleWoodLin #28919
递归遍历，进行合并，用一个空对象收集所有的结果
type MergeAll<XS, P = {}> = XS extends [infer F, ...infer Rest]
? MergeAll<Rest, Merge<P, F>>
: P
type Merge<F, S> = {
[P in keyof F | keyof S]:
P extends keyof S
? P extends keyof F
? S[P] | F[P]
: S[P]
: P extends keyof F
? F[P]
:never
}
Solution by linjunc #28601
type MergeObj<T extends object, U extends object> = { [P in (keyof T | keyof U)]:
P extends keyof T ?
T[P] | (P extends keyof U ? U[P] : never) :
P extends keyof U ? U[P] : never };
type MergeAll<XS extends object[]> = XS extends [] ?
{} :
XS extends [infer F extends object, infer S extends object, ...infer R extends object[]] ? MergeAll<[MergeObj<F, S>, ...R]> : XS[0];
Solution by E-uler #28122
type Merge<A, B> = {
[K in keyof A | keyof B]: K extends keyof A
? K extends keyof B
? A[K] | B[K]
: A[K]
: K extends keyof B
? B[K]
: never
}
type MergeAll<T, R extends Record<string, any> = {}> = T extends [infer F, ...infer Rest]
? MergeAll<Rest, Merge<R, F>>
: R
Solution by XkSuperCool #28013
type Merge<T,U> = {
[K in (keyof T) | (keyof U)]:
K extends keyof T?
K extends keyof U?
T[K] | U[K]
: T[K]
: K extends keyof U? U[K]:never
}
type MergeAll<XS extends any[],R = {}> =
XS extends [infer F,...infer Z]?
MergeAll<Z, Merge<R,F>>:R
Solution by jiangshanmeta #27988
type MergeAll<XS extends object[], Res = {}> =
XS extends [infer L, ...infer R extends object[]]
? MergeAll<R, Omit<Res, keyof L> & {
[p in keyof L]: p extends keyof Res ? L[p] | Res[p] : L[p]
}>
: Omit<Res, never>;
Solution by omittee #27962
type MergeAll<XS, Prev = {}> = XS extends [infer First, ...infer Rest]
? MergeAll<Rest, Merge<Prev, First>>
: Prev
type Merge<O1, O2> = {
[K in (keyof O1 | keyof O2)]: PropOr<O1, K> | PropOr<O2, K>
}
type PropOr<Obj, Prop, Default = never> = Prop extends keyof Obj
? Obj[Prop]
: Default
Solution by Sun79 #27956