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[]
)。
最佳答案
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"
])
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"
])
关于javascript - Typescript:使用 .map 时如何保留长度信息,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/71107948/