Typescript ERROR(2322) typescript 縮小類型失敗

Question

我試圖添加一個條件運算符來指示 function greet 的返回類型，但收到以下錯誤消息：

輸入“你好？ ${T & string}`' 不可分配給類型 'T extends string: string。 細繩[]'？ 類型“string”不可分配給類型“T extends string: string”。 細繩[]'。

下面是截圖：

下面是代碼，大家可以復制到Typescript playground進行測試。

// Welcome to the TypeScript Playground, this is a website
// which gives you a chance to write, share and learn TypeScript.

// You could think of it in three ways:
//
//  - A location to learn TypeScript where nothing can break
//  - A place to experiment with TypeScript syntax, and share the URLs with others
//  - A sandbox to experiment with different compiler features of TypeScript

const anExampleVariable = "Hello World"
console.log(anExampleVariable)

// To learn more about the language, click above in "Examples" or "What's New".
// Otherwise, get started by removing these comments and the world is your playground.

function isString(val: any): val is string{
    return Object.prototype.toString.call(val) === '[object String]';
}

function greet<T extends string | string[]>(person: T) : T extends string ? string: string[]  {
    if( isString(person) ){
        return `hello! ${person}`;
    } else {
        return person.map(p=>`hello! ${p}`);
    }
}

let greetStr: string = greet('Jhon');
let greetStrAry: string[] = greet(['Jhon', 'Miley']);
console.log(greetStr);
console.log(greetStrAry);

Answer 1

您可以使用function 重載來解決說person可以是一個string或string[]並且返回類型應該相同的困難：

function greet(person: string): string;
function greet(person: string[]): string[];
function greet(person: string | string[]): string | string[] {
    if( isString(person) ){
        return `hello! ${person}`;
    } else {
        return person.map(p=>`hello! ${p}`);
    }
}

演示