如何在柯里化函数中推断泛型？

Question

我正在尝试输入一个用于查询集合的 API。我希望它像这样使用：

interface Car {
  model: string
  doors: 2 | 4 | 5
  designers: string[]
}

const result = query({
  model: [ contains('tesla') ],
  doors: [ is(3) ],
  designers: [ includes('Franz') ],
  take: 5,
});

所以这是我不完美的方法：

type Query<Entity extends EntityType> = (config: QueryConfig<Entity>) => Promise<Entity[]>
type QueryConfig<Entity extends EntityType> =
  { [K in FilterScalarMembers<Entity>]?: ScalarBuilderFn<Entity[K]>[] }
  & { [K in FilterCollectionMembers<Entity>]?: CollectionBuilderFn<Entity[K]>[] }
  & { take: number };

type FilterCollectionMembers<Entity extends EntityType> =
  { [K in keyof Entity]: Entity[K] extends CollectionType ? K : never }[keyof Entity]

type FilterScalarMembers<Entity extends EntityType> =
  { [k in keyof Entity]: Entity[k] extends (string | number) ? k : never }[keyof Entity]

type ScalarBuilderFn<T extends string | number> = (builder: ScalarQueryBuilder<T>) => ScalarQueryBuilder<T>
type CollectionBuilderFn<T extends CollectionType> = (builder: CollectionQueryBuilder<T>) => CollectionQueryBuilder<T>
type ScalarQueryBuilder<T extends string | number> = {}
type CollectionQueryBuilder<T extends CollectionType> = {}
type EntityType = Record<string, any>
type CollectionType = ((string | number)[]) | Set<string | number>

const is = <T extends string | number>(arg: T): ScalarBuilderFn<T> => function() {} as unknown as ScalarBuilderFn<T>;

问题在这里变得很明显：doors: [ is(3) ] 因为类型没有缩小到2 | 4 | 5，TypeScript 不会给出关于3 参数的错误。我们可以对is 函数做些什么来帮助它正确推断类型吗？

Answer 1

您可以使用indexed types：

const result = query({
    doors: [is<Car['doors']>(3)]
});

您可以将is 传递给Car['doors'] 的通用参数，这将告诉is 它的参数必须可分配给Car['doors']，即2 | 4 | 5。

我不确定您是否需要所有这些类型 - 除非它们执行此处未显示的其他操作。你所需要的只是is 有一个泛型类型参数，它的参数需要匹配

const is = <T>(val: T) => { } // rest of the implementation