javascript - Typescript:使用 .map 时如何保留长度信息

Question

我目前的“工作”方法是这样的:

const generateMainOrientations = <T extends readonly string[]>(
  mainOrientationsNames: T
): { [Index in keyof T]: Orientation } => {
  const temp: Orientation[] = mainOrientationsNames.map(
    mainOrientationName => ({
      name: mainOrientationName,
      getYear(date) {
        return date.getFullYear()
      },
      getRecordContent: getMainOrientationRecordContent
    })
  )

  return temp as unknown as { [Index in keyof T]: Orientation }
}

const mainOrientations = generateMainOrientations([
  "One",
  "Two",
  "Three"
] as const)

但是，我必须使用 as unknown as { [Index in keyof T]: Orientation } ，这并不理想，否则(即使从 temp 变量中删除类型断言)它也会抛出

Type '{ name: string; getYear(date: any): any; getRecordContent: (values: number[]) => string[]; }[]' is not assignable to type '{ [Index in keyof T]: Orientation; }'.ts(2322)

仍然，{ name: string; getYear(date: any): any; getRecordContent: (values: number[]) => string[]; }是 Orientation 的定义

这表明在使用 map 后任何长度信息都丢失了。

有没有更有机的方法来实现这一点，最好根本不必使用类型断言，或者至少不必使用 as unknown .目标是使 mainOrientations 成为 Orientation 的元组与传递给 generateMainOrientations 的参数长度相同, 所以 [Orientation, Orientation, Orientation]在这种情况下，(不是 Orientation[] )。

Playground

Answer 1

你需要重载你的函数:

interface Orientation {
  name: string,
  getYear(date: Date): number,
  getRecordContent(values: number[]): string[]
}

declare function getMainOrientationRecordContent(values: number[]): string[]

function generateMainOrientations<T extends string, Tuple extends T[]>(
  mainOrientationsNames: [...Tuple]
): { [Index in keyof Tuple]: Orientation }
function generateMainOrientations(
  mainOrientationsNames: string[]
) {
  return mainOrientationsNames.map(
    mainOrientationName => ({
      name: mainOrientationName,
      getYear: (date: Date) => date.getFullYear(),
      getRecordContent: getMainOrientationRecordContent
    })
  )
}

// [Orientation, Orientation, Orientation]
const mainOrientations = generateMainOrientations([
  "One",
  "Two",
  "Three"
])

Playground

请记住，一旦您使用了 Array.prototype.map ， typescript 就不会保留结果的长度。 Here你可以找到原因。

因此，您只有两个选择:重载和类型断言。

如果将 name 属性参数化，您可以做得更好:

interface Orientation<Name extends string> {
  name: Name,
  getYear(date: Date): number,
  getRecordContent(values: number[]): string[]
}

declare function getMainOrientationRecordContent(values: number[]): string[]

function generateMainOrientations<T extends string, Tuple extends T[]>(
  mainOrientationsNames: [...Tuple]
): { [Index in keyof Tuple]: Orientation<Tuple[Index] & string> }
function generateMainOrientations(
  mainOrientationsNames: string[]
) {
  return mainOrientationsNames.map<Orientation<string>>(
    name => ({
      name,
      getYear: (date) => date.getFullYear(),
      getRecordContent: getMainOrientationRecordContent
    })
  )
}

// [Orientation<"One">, Orientation<"Two">, Orientation<"Three">]
const mainOrientations = generateMainOrientations([
  "One",
  "Two",
  "Three"
])