typescript - 如何在 TypeScript 中合并两个枚举

Question

假设我有两个枚举，如下面在 Typescript 中描述的那样，那么我该如何合并它们

enum Mammals {
    Humans,
    Bats,
    Dolphins
}

enum Reptiles {
    Snakes,
    Alligators,
    Lizards
}

export default Mammals & Reptiles // For Illustration purpose, Consider both the Enums have been merged.

现在，当我在另一个文件中import exported value 时，我应该能够访问两个枚举中的值。

import animalTypes from "./animalTypes"

animalTypes.Humans //valid

animalTypes.Snakes // valid

如何在 TypeScript 中实现这样的功能？

Answer 1

合并问题:

• 相同的值 => 值被覆盖

• 相同的键 => 键被覆盖

• ❌ 具有相同值的枚举(=> 值被覆盖)

enum AA1 {
  aKey, // = 0
  bKey // = 1
}
enum BB1 {
  cKey, // = 0
  dKey // = 1
}

• ❌ 具有相同键的枚举(=> 键被覆盖)

enum AA2 {
  aKey = 1
}
enum BB2 {
  aKey = 2
}

• ✅ 好

enum AA3 {
  aKey, // = 0
  bKey // = 1
}
enum BB3 {
  cKey = 2,
  dKey // = 3
}

• ✅ 还不错

enum AA4 {
  aKey = 'Hello',
  bKey = 0,
  cKey // = 1
}
enum BB4 {
  dKey = 2,
  eKey = 'Hello',
  fKey = 'World'
}

注意:aKey = 'Hello' 和 eKey = 'Hello' 可以工作，因为带有字符串值的枚举没有这个值作为键

// For aKey = 'Hello', key is working
type aa4aKey = AA4.aKey; // = AA4.aKey
// value is not.
type aa4aValue = AA4.Hello; // ❌ Namespace 'AA4' has no exported member 'Hello'
type aa4aValue2 = AA4['Hello']; // ❌ Property 'Hello' does not exist on type 'AA4'

console.log(AA4); // { 0: 'bKey', 1: 'cKey', aKey: 'Hello', bKey: 0, cKey: 1 }
console.log(BB4); // { 2: 'dKey', dKey: 2, eKey: 'Hello', fKey: 'World' }

合并

• ❌ 使用联合类型

type AABB1 = AA4 | BB4; // = AA4 | BB4
type AABB1key = AABB1['aKey']; // = never
type AABB1key2 = AABB1.aKey; // ❌ 'AABB1' only refers to a type, but is being used as a namespace here. ts(2702)

• ❌ 使用交集类型

type AABB1 = AA4 & BB4; // = never
type AABB1key = AABB1['aKey']; // = never

• ✅ 使用 typeof 的交集类型

type AABB2 = (typeof AA4) & (typeof BB4); // = typeof AA4 & typeof BB4
type AABB2key = AABB2['aKey']; // = AA4.aKey

• ✅ 使用js复制

const aabb1 = { ...AA4, ...BB4 };
const aabb2 = Object.assign({}, AA4, BB4); // also work
// aabb1 = {
// 0: 'bKey',
// 1: 'cKey',
// 2: 'dKey',
// aKey: 'Hello',
// bKey: 0,
// cKey: 1,
// dKey: 2,
// eKey: 'Hello',
// fKey: 'World' }

• ✅ 使用 typeof 和 js 副本

const aabb = { ...AA4, ...BB4 };
type TypeofAABB = typeof aabb;
// type TypeofAABB = {
// [x: number]: string;
// dKey: BB4.dKey;
// eKey: BB4.eKey;
// fKey: BB4.fKey;
// aKey: AA4.aKey;
// bKey: AA4.bKey;
// cKey: AA4.cKey;
// };

提示:您可以对类型和值使用相同的名称

const merged = { ...AA4, ...BB4 };
type merged = typeof merged;

const aValue = merged.aKey;
type aType = merged['aKey'];

你的情况

如果你想合并你的 2 个枚举，你有 ~3 个选择:

1。使用字符串枚举

enum Mammals {
  Humans = 'Humans',
  Bats = 'Bats',
  Dolphins = 'Dolphins'
}

enum Reptiles {
  Snakes = 'Snakes',
  Alligators = 'Alligators',
  Lizards = 'Lizards'
}

export const Animals = { ...Mammals, ...Reptiles };
export type Animals = typeof Animals;

2。使用唯一编号

enum Mammals {
  Humans = 0,
  Bats,
  Dolphins
}

enum Reptiles {
  Snakes = 2,
  Alligators,
  Lizards
}

export const Animals = { ...Mammals, ...Reptiles };
export type Animals = typeof Animals;

3。使用嵌套枚举

enum Mammals {
  Humans,
  Bats,
  Dolphins
}

enum Reptiles {
  Snakes,
  Alligators,
  Lizards
}

export const Animals = { Mammals, Reptiles };
export type Animals = typeof Animals;

const bats = Animals.Mammals.Bats; // = 1
const alligators = Animals.Reptiles.Alligators; // = 1

注意:您还可以使用以下代码合并嵌套枚举。如果这样做，请注意不要有重复的值!

type Animal = {
  [K in keyof Animals]: {
    [K2 in keyof Animals[K]]: Animals[K][K2]
  }[keyof Animals[K]]
}[keyof Animals];

const animal: Animal = 0 as any;

switch (animal) {
  case Animals.Mammals.Bats:
  case Animals.Mammals.Dolphins:
  case Animals.Mammals.Humans:
  case Animals.Reptiles.Alligators:
  case Animals.Reptiles.Lizards:
  case Animals.Reptiles.Snakes:
    break;
  default: {
    const invalid: never = animal; // no error
  }
}