首页 > 解决方案 > Typescript:如何提取 const 对象的值类型并将它们用作新类型中的键?

问题描述

我想取一个常量对象,比如说{key1: 'value1', key2: value2'} as const,并将它的值转换为键,然后在一个新的对象中使用它 type {value1: number; value2: number}。我想在一个函数中执行此操作,并且希望正确键入该函数。

下面是我试图开始工作的代码。当我尝试将类型的变量强制Partial<Blah>转换为Blah. 有人可以解释为什么会发生这种情况,以及是否有办法做我想做的事情?谢谢!

type BaseObject = Record<string, string>;


type ValuesAsKeys<T extends BaseObject> = {
    [K in keyof T as T[K]]: number;
};

function useValuesAsKeys<T extends BaseObject>(arg: T): ValuesAsKeys<T> {
  const x: Partial<ValuesAsKeys<T>> = {};
  for (const value of Object.values(arg) as Array<T[keyof T]>) {
    x[value] = 1;
  }

  // This line doesn't work
  return x as ValuesAsKeys<T>;
}

// Example use case
const obj = {key1: 'value1', key2: 'value2'} as const;
const result = useValuesAsKeys(obj); // Type of result is {value1: number; value2: number}


// Everything below here works
// Only including this because I'm contrasting this code with the code above.

type KeysAsKeys<T extends BaseObject> = {
    [K in keyof T]: number;
};

function useKeysAsKeys<T extends BaseObject>(arg: T): KeysAsKeys<T> {
    const x: Partial<KeysAsKeys<T>> = {};
  for (const value of Object.values(arg) as Array<T[keyof T]>) {
    x[value] = 1;
  }
  return x as KeysAsKeys<T>;
}

标签: typescriptgenericstypespartial

解决方案


显式返回类型reduce几乎总是不起作用,因为initial每次迭代都会改变值。

为了使它工作,最好重载你的函数:

type BaseObject = Record<string, string>;


type ValuesAsKeys<T extends BaseObject> = {
  [K in keyof T as T[K]]: number;
};

type Values<T> = T[keyof T]

function useValuesAsKeys<T extends BaseObject>(arg: T): ValuesAsKeys<T>
function useValuesAsKeys<T extends BaseObject>(arg: T) {
  return (Object.values(arg) as Array<Values<T>>).reduce((acc, elem) => ({
    ...acc,
    [elem]: 1
  }), {})
}

// Example use case
const obj = { key1: 'value1', key2: 'value2' } as const;

const result = useValuesAsKeys(obj); // Type of result is {value1: number; value2: number}
result.value1 //ok

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