typescript - 如何使 Typescript 推断对象的键但定义其值的类型？

Question

我想定义一个对象的类型，但让 typescript 推断键并且没有那么多的开销来创建和维护所有键的 UnionType。

键入一个对象将允许所有字符串作为键:

const elementsTyped: { 
    [key: string]: { nodes: number, symmetric?: boolean }
} = {
    square: { nodes: 4, symmetric: true },
    triangle: { nodes: 3 }
}

function isSymmetric(elementType: keyof typeof elementsTyped): boolean {
    return elementsTyped[elementType].symmetric;
}
isSymmetric('asdf'); // works but shouldn't

推断整个对象将显示错误并允许所有类型的值:

const elementsInferred = {
    square: { nodes: 4, symmetric: true },
    triangle: { nodes: 3 },
    line: { nodes: 2, notSymmetric: false /* don't want that to be possible */ }
}

function isSymmetric(elementType: keyof typeof elementsInferred): boolean {
    return elementsInferred[elementType].symmetric; 
    // Property 'symmetric' does not exist on type '{ nodes: number; }'.
}

我得到的最接近的是这个，但它不想维护这样的键集:

type ElementTypes = 'square' | 'triangle'; // don't want to maintain that :(
const elementsTyped: { 
    [key in ElementTypes]: { nodes: number, symmetric?: boolean }
} = {
    square: { nodes: 4, symmetric: true },
    triangle: { nodes: 3 },
    lines: { nodes: 2, notSymmetric: false } // 'lines' does not exist in type ...
    // if I add lines to the ElementTypes as expected => 'notSymmetric' does not exist in type { nodes: number, symmetric?: boolean }
}

function isSymmetric(elementType: keyof typeof elementsTyped): boolean {
    return elementsTyped[elementType].symmetric;
}
isSymmetric('asdf'); // Error: Argument of type '"asdf"' is not assignable to parameter of type '"square" | "triangle"'.

有没有更好的方法来定义对象而不维护键集？

Answer 1

所以你想要一些可以推断键但限制值类型并使用 excess property checking 的东西禁止额外的属性。我认为获得这种行为的最简单方法是引入辅助函数:

// Let's give a name to this type
interface ElementType {
  nodes: number,
  symmetric?: boolean
}

// helper function which infers keys and restricts values to ElementType
const asElementTypes = <T>(et: { [K in keyof T]: ElementType }) => et;

这个辅助函数 infers et 的映射类型中的类型 T。现在你可以像这样使用它:

const elementsTyped = asElementTypes({
  square: { nodes: 4, symmetric: true },
  triangle: { nodes: 3 },
  line: { nodes: 2, notSymmetric: false /* error where you want it */} 
});

生成的 elementsTyped 的类型(一旦您修复了错误)将具有推断键 square、triangle 和 line ，值为 ElementType。

希望对你有用。祝你好运!