重载方法的自动完成参数值的类型推断

我正在尝试为通用方法的参数添加一些自动完成支持

 subscribeToTelemetry<T extends keyof TypeMap>(name: T): Stream<TypeMap[T]>;
 subscribeToTelemetry<T extends NonNullable<{}>>(name: EntityName<T>): Stream<T>; 

简而言之，如果没有提供类型，subscribeToTelemetry则应根据名称和一些可用的注入/自动生成类型推断返回类型。此外，应使用可用的注入/自动生成类型提出可能的名称建议。例如，使用以下类型：

type TypeMap = {
    ranger1: RangerTelemetry,
    ranger2: RangerTelemetry,
    mavic2: DroneTelemetry
};

以及以下代码

// should suggest: ['mavic2' | 'ranger2' | 'ranger1'] since no type T is provided 
const myStream1 = r.subscribeToTelemetry('ranger1'); // myStream1: Stream<RangerTelemetry

建议值应为 ['mavic2' | 'ranger2' | 'ranger1']，因为未提供类型 T，并且 myStream1 类型将是Stream<RangerTelemetry>

相反，如果我们提供一个类型subscribeToTelemetry ，并考虑可用的注入/自动生成类型

const entityNames = {
    RangerTelemetry: ['ranger1', 'ranger2', 'ranger3'],
    DroneTelemetry: ['mavic1', 'mavic2', 'mavic3'],
} as const;

和其他一些类型的技巧（参见操场链接），使用以下代码

// should suggest only: ['mavic1' | 'mavic2' | 'mavic3'] since type T is provided as DroneTelemetry,
const myStream3 = r.subscribeToTelemetry<DroneTelemetry>('mavic1');

我们应该建议：['mavic1' | 'mavic2' | 'mavic3']，因为提供了类型DroneTelemetry，但显示的却是 ['mavic1' | 'mavic2' | 'mavic3', 'ranger1', 'ranger2', 'ranger3']。的类型myStream3将是Stream<DroneTelemetry>

我有多个重载方法的签名实现变体，但 subscribeToTelemetry在提供类型时无法提供正确的建议，即使我指定 DroneTelemetry为类型，我总是会得到游骑兵的建议。

class Registry {
    subscribeToTelemetry<T extends keyof TypeMap>(name: T): Stream<TypeMap[T]>;
    subscribeToTelemetry<T extends NonNullable<{}>>(name: EntityName<T>): Stream<T>;
    subscribeToTelemetry<T extends NonNullable<{}>>(name: EntityName<T>): Stream<T> {
        return new Stream<T>(name); 
    }
}

如果有人有任何建议，我将不胜感激

温贝托

完整代码如下：

type DroneTelemetry = { kind: 'DroneTelemetry', batteryLevel: number, velocity: number, location: {latitude: number, longitude: number}}
type RangerTelemetry = { kind: 'RangerTelemetry', velocity: number, location: {latitude: number, longitude: number}}


////// the entity names and type mapping will be autogenerated
const entityNames = {
    RangerTelemetry: ['ranger1', 'ranger2', 'ranger3'],
    DroneTelemetry: ['mavic1', 'mavic2', 'mavic3'],
} as const;

type TypeMap = {
    ranger1: RangerTelemetry,
    ranger2: RangerTelemetry,
    mavic2: DroneTelemetry
};
////////

// Define the types for the keys and values of the mapping
type EntityNames = typeof entityNames;
type EntityKeys = keyof EntityNames;
type EntityValues<T extends EntityKeys> = EntityNames[T][number];

// Define a mapping from type aliases to their corresponding keys
type TypeToKey<T> = T extends { kind: infer K } ? K : never;

type ExtractEntitiesNames<T> = TypeToKey<T> extends keyof EntityNames ? EntityValues<TypeToKey<T>> : never;
type EntityName<T> = ExtractEntitiesNames<T> |  (string  & {});

/////////////////////////////////////////////////////////////////////

class Stream<T> {
    name: string
    constructor(name: string){
        this.name = name
    }
    getLastValue() {
        return {} as T // dummy impl
    }
}

class Registry {
    subscribeToTelemetry<T extends keyof TypeMap>(name: T): Stream<TypeMap[T]>;
    subscribeToTelemetry<T extends NonNullable<{}>>(name: EntityName<T>): Stream<T>;
    subscribeToTelemetry<T extends NonNullable<{}>>(name: EntityName<T>): Stream<T> {
        return new Stream<T>(name); 
    }
}

const r = new Registry();
// should suggest: ['mavic2' | 'ranger2' | 'ranger1'] since no type T is provided // OK
const myStream1 = r.subscribeToTelemetry('ranger1'); // myStream1 type will be Stream<RangerTelemetry>

// should suggest: ['mavic2' | 'ranger2' | 'ranger1'] since no type T is provided // OK
const myStream2 = r.subscribeToTelemetry('ranger2'); // myStream2 type will be Stream<RangerTelemetry>

 // should suggest only: ['mavic1' | 'mavic2' | 'mavic3'] since type T is provided as DroneTelemetry,
 // but ['mavic1' | 'mavic2' | 'mavic3', 'ranger1', 'ranger2', 'ranger3'] is shown instead.                // WRONG
const myStream3 = r.subscribeToTelemetry<DroneTelemetry>('mavic1'); // myStream3 type will be Stream<RangerTelemetry>

这是一个包含完整代码的ts-playground