TypeScript泛型'not assignable'错误：自定义对象结构赋值失败

你的代码报错的核心原因是**Winner类型的定义逻辑错误**。

错误原因拆解 #

1. 首先，GameNames是一个字符串联合类型："cash3" | "cash4"。
2. 当你写type Winner<GameNames> = { [Property in keyof GameNames]: Drawing[] }时，keyof GameNames对于字符串联合类型来说，结果是string | number | symbol（所有字符串字面量的父类型都是string）。
3. 这导致Winner<GameNames>变成了一个带有索引签名的类型：{ [x: string]: Drawing[]; }，要求对象的所有字符串键对应的属性都必须是Drawing[]。但你创建的{ cash3: Drawing[]; cash4: Drawing[]; }只包含两个特定键，TypeScript认为它不符合“任意字符串键都要有Drawing[]类型值”的索引签名要求，所以抛出错误。

正确解决方案 #

你需要让Winner类型生成一个以GameNames联合类型中的每个值为键的对象类型，有两种直接可行的方式：

方案1：使用Record工具类型

type Drawing = {
    date: string,
    draw_time: string,
    winning_numbers: string
}

// 用内置Record工具类型定义目标结构
type Winner = Record<GameNames, Drawing[]>

const gameNames = {
    name3: 'cash3',
    name4: 'cash4',
} as const

type GameNames = (typeof gameNames)[keyof typeof gameNames]

const { name3, name4 } = gameNames

const winners: Winner = {
    [name3]: [] as Drawing[],
    [name4]: [] as Drawing[],
}

方案2：修正自定义映射类型

如果你想自己实现映射逻辑，需要确保泛型约束指向联合类型的成员，而非联合类型的keyof结果：

type Drawing = {
    date: string,
    draw_time: string,
    winning_numbers: string
}

// 调整泛型约束，直接映射联合类型的每个成员
type Winner<K extends string | number | symbol> = {
    [Property in K]: Drawing[]
}

const gameNames = {
    name3: 'cash3',
    name4: 'cash4',
} as const

type GameNames = (typeof gameNames)[keyof typeof gameNames]

const { name3, name4 } = gameNames

const winners: Winner<GameNames> = {
    [name3]: [] as Drawing[],
    [name4]: [] as Drawing[],
}

两种方案都能让TypeScript正确识别winners对象的类型，消除报错。

内容的提问来源于stack exchange，提问作者Mike S.