是否有一种基于字典值自动生成类型的方法?例如,给定这个常量值字典:
export const SHIPMETHODS: OptionsMap<UpsShipMethodOption> = {
'UPS': 'GND',
'UPS AIR2': '2DA',
'UPS 3DAY SELECT': '3DS'
}有没有一种方法可以自动生成这样的类型,而不必复制所有的东西?
export type ShipMethodOptionMap = {
'UPS': 'GND',
'UPS AIR2': '2DA',
'UPS 3DAY SELECT': '3DS'
}不幸的是,keyof似乎不起作用,因为它是一个对象,而不是一个类型,因此它不起作用:
export type UpsShipMethodOption = keyof SHIPMETHODS // Doesn't work我还尝试将映射类型与预先声明的键联合类型一起使用,但它也不起作用:
/*
Type 'KeyOptions' is not assignable to type 'string | number | symbol'.
Type 'KeyOptions' is not assignable to type 'symbol'.ts(2322)
*/
export type OptionsMap<KeyOptions> = {
readonly [Property in KeyOptions]: string;
}
export const SHIPMETHODS: OptionsMap<UpsShipMethodOption> = {
'UPS': 'GND',
'UPS AIR2': '2DA',
'UPS 3DAY SELECT': '3DS'
}
export type UpsShipMethodOption = 'UPS' | 'UPS AIR2' | 'UPS 3DAY SELECT'我有一堆这样的字典,其中大多数都很大(30+键)。将字典复制到任何地方都会使维护变得很痛苦,以防值发生变化,因此对此提供任何帮助都是很好的!
发布于 2022-02-07 19:21:57
我想你只是想要typeof接线员。它只获取值的类型。
const SHIPMETHODS = {
'UPS': 'GND',
'UPS AIR2': '2DA',
'UPS 3DAY SELECT': '3DS'
} as const // as const is important.
type ShipMethodOptionMap = typeof SHIPMETHODS
/*
type ShipMethodOptionMap = {
readonly UPS: "GND";
readonly 'UPS AIR2': "2DA";
readonly 'UPS 3DAY SELECT': "3DS";
}
*/
type UpsShipMethodOption = keyof ShipMethodOptionMap
// type UpsShipMethodOption = "UPS" | "UPS AIR2" | "UPS 3DAY SELECT"只需注意在你的目标上,它允许类型记录推断字符串文字,这样它们就可以成为类型的一部分。
https://stackoverflow.com/questions/71024171
复制相似问题