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Asked:2022-09-28 19:23:15 +0800 CST2022-09-28 19:23:15 +0800 CST

TypeScript 中的 Object.entries

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如何转换这种类型：

{
    a: number
    b: string
    c: number
}

在那里面：

[['a', number], ['b', string], ['c', number]]

那些。效果是 1 in 1 as from Object.entries，只是代替 sv-in 他们的类型的值。

javascript

1 个回答

  1. Best Answer

    通过稍微修改将类型转换为的代码uniontuple- 您可以获得预期的结果。但是，由于union类型没有排序，因此元素的顺序tuple可能会发生变化：

    type TuplePrepend<Tuple extends readonly unknown[], NewElement> =
    [NewElement, ...Tuple]

type Consumer<Value> = (value: Value) => void;

type IntersectionFromUnion<Union> = 
    (Union extends unknown ? Consumer<Union> : never) extends (Consumer<infer ResultIntersection>)
    ? ResultIntersection
    : never;

type OverloadedConsumerFromUnion<Union> = IntersectionFromUnion<Union extends unknown ? Consumer<Union> : never>;

type UnionLast<Union> = OverloadedConsumerFromUnion<Union> extends ((a: infer A) => void) ? A : never;

type UnionExcludingLast<Union> = Exclude<Union, UnionLast<Union>>;

type TupleFromUnionRec<RemainingUnion, CurrentTuple extends readonly unknown[], TObj extends Record<any, any>> =
    [RemainingUnion] extends [never]
    ? CurrentTuple
    : TupleFromUnionRec< // рекурсивный тип
        UnionExcludingLast<RemainingUnion>, 
        TuplePrepend<
          CurrentTuple, 
          [UnionLast<RemainingUnion>, TObj[UnionLast<RemainingUnion>]] // добавляем тип ключа с типом значения в конечный tuple
        >, 
        TObj>;

export type TupleFromUnion<TObj> = TupleFromUnionRec<keyof TObj, [], TObj>;

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