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,但是让typescript推断出键并且没有那么多的开销来制作和维护所有键的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,但它不想像这样保持一组键:

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密钥集？

Answer 1

TypeScript >=4.9.0

TypeScript4.9.0添加了satisfies关键字，可用于在推断键时约束对象的值。

type ElementValue = {
  nodes: number;
  symmetric?: boolean;
};

const elements = {
  square: { nodes: 4, symmetric: true },
  triangle: { nodes: 3 },
} satisfies Record<string, ElementValue>;

type Elements = typeof elements;
type ElementType = keyof Elements;

function isSymmetric(elementType: ElementType): boolean {
  const element = elements[elementType];
  return 'symmetric' in element && element.symmetric;
}

isSymmetric('asdf'); // doesn't work

打字稿 <4.9.0

中间函数可用于在推断键时约束对象的值。

type ElementValue = {
  nodes: number;
  symmetric?: boolean;
};

function typedElements<T extends Record<string, ElementValue>>(o: T) {
  return o;
}

const elements = typedElements({
  square: { nodes: 4, symmetric: true },
  triangle: { nodes: 3 },
});

type Elements = typeof elements;
type ElementType = keyof Elements;

function isSymmetric(elementType: ElementType): boolean {
  const element = elements[elementType];
  return 'symmetric' in element && element.symmetric;
}

isSymmetric('asdf'); // doesn't work

Answer 2

所以你想要一些推断密钥但限制值类型的东西,并使用多余的属性检查来禁止额外的属性.我认为获得该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;

这个辅助函数从映射类型推断出类型.现在您可以像这样使用它:Tet

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

该类型所产生的elementsTyped意志(一旦你修复错误)已经推断出密钥square,triangle以及line与价值ElementType.

希望对你有用.祝好运!