Typescript 如何在curried函数中推断泛型？

我正在尝试键入用于查询集合的API。我希望它能像这样使用：

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

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

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>;

type Query=（config:QueryConfig）=>Promise
类型查询配置=
{[K in FilterScalarMembers]？：ScalarBuilderFn[]}
&{[K in FilterCollectionMembers]？：CollectionBuilderFn[]}
&{take:number}；
类型FilterCollectionMembers=
{[K-in-keyof-Entity]：Entity[K]扩展了CollectionType？K:never}[keyof-Entity]
类型过滤器CalArMembers=
{[k-in-keyof-Entity]：Entity[k]扩展（字符串|编号）？k:never}[keyof-Entity]
键入ScalarBuilderFn=（生成器：ScalarQueryBuilder）=>ScalarQueryBuilder
类型CollectionBuilderFn=（生成器：CollectionQueryBuilder）=>CollectionQueryBuilder
类型ScalarQueryBuilder={}
类型CollectionQueryBuilder={}
类型EntityType=记录
类型集合类型=（（字符串|编号）[]）|集
常量is=（arg:T）：ScalarBuilderFn=>function（）{}与ScalarBuilderFn一样未知；

问题在这里变得很明显：

doors:[is（3）]

由于类型没有缩小到

2 | 4 | 5

，TypeScript没有给出关于

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？

您可以使用：

---

正如我提到的，我希望从使用

is

的上下文中推断出

is

参数类型。手动指定泛型类型容易出错。

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

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