我想迭代一个TypeScript枚举对象,并获得每个枚举符号名称,例如: enum myEnum {entry1, entry2}

for (var entry in myEnum) { 
    // use entry's name here, e.g., "entry1"
}

当前回答

我看不正确的答案看累了,就自己做了。

这个有测试。 适用于所有类型的枚举。 正确地输入。

type EnumKeys<Enum> = Exclude<keyof Enum, number>

const enumObject = <Enum extends Record<string, number | string>>(e: Enum) => {
    const copy = {...e} as { [K in EnumKeys<Enum>]: Enum[K] };
    Object.values(e).forEach(value => typeof value === 'number' && delete copy[value]);
    return copy;
};

const enumKeys = <Enum extends Record<string, number | string>>(e: Enum) => {
    return Object.keys(enumObject(e)) as EnumKeys<Enum>[];
};

const enumValues = <Enum extends Record<string, number | string>>(e: Enum) => {
    return [...new Set(Object.values(enumObject(e)))] as Enum[EnumKeys<Enum>][];
};

enum Test1 { A = "C", B = "D"}
enum Test2 { A, B }
enum Test3 { A = 0, B = "C" }
enum Test4 { A = "0", B = "C" }
enum Test5 { undefined = "A" }
enum Test6 { A = "undefined" }
enum Test7 { A, B = "A" }
enum Test8 { A = "A", B = "A" }
enum Test9 { A = "B", B = "A" }

console.log(enumObject(Test1)); // {A: "C", B: "D"}
console.log(enumObject(Test2)); // {A: 0, B: 1}
console.log(enumObject(Test3)); // {A: 0, B: "C"}
console.log(enumObject(Test4)); // {A: "0", B: "C"}
console.log(enumObject(Test5)); // {undefined: "A"}
console.log(enumObject(Test6)); // {A: "undefined"}
console.log(enumObject(Test7)); // {A: 0,B: "A"}
console.log(enumObject(Test8)); // {A: "A", B: "A"}
console.log(enumObject(Test9)); // {A: "B", B: "A"}

console.log(enumKeys(Test1)); // ["A", "B"]
console.log(enumKeys(Test2)); // ["A", "B"]
console.log(enumKeys(Test3)); // ["A", "B"]
console.log(enumKeys(Test4)); // ["A", "B"]
console.log(enumKeys(Test5)); // ["undefined"]
console.log(enumKeys(Test6)); // ["A"]
console.log(enumKeys(Test7)); // ["A", "B"]
console.log(enumKeys(Test8)); // ["A", "B"]
console.log(enumKeys(Test9)); // ["A", "B"]

console.log(enumValues(Test1)); // ["C", "D"]
console.log(enumValues(Test2)); // [0, 1]
console.log(enumValues(Test3)); // [0, "C"]
console.log(enumValues(Test4)); // ["0", "C"]
console.log(enumValues(Test5)); // ["A"] 
console.log(enumValues(Test6)); // ["undefined"] 
console.log(enumValues(Test7)); // [0, "A"]
console.log(enumValues(Test8)); // ["A"]
console.log(enumValues(Test9)); // ["B", "A"]

在线版本。

其他回答

这里的答案似乎都不能在严格模式下使用string-enum。

考虑enum为:

enum AnimalEnum {
  dog = "dog", cat = "cat", mouse = "mouse"
}

使用AnimalEnum["dog"]访问可能会导致如下错误:

元素隐式具有“any”类型,因为类型“any”的表达式不能用于索引类型“typeof AnimalEnum”.ts(7053)。

这种情况下的正确解,写为:

AnimalEnum["dog" as keyof typeof AnimalEnum]

具有数字enum:

enum MyNumericEnum {
 First = 1,
 Second = 2
}

你需要先把它转换成数组:

const values = Object.values(MyNumericEnum);
// ['First', 'Second', 1, 2]

如您所见,它同时包含键和值。钥匙先放。

之后,你可以检索它的键:

values.slice(0, values.length / 2);
// ['First', 'Second']

和值:

values.slice(values.length / 2);
// [1, 2]

对于字符串enum,你可以使用Object.keys(MyStringEnum)来分别获取key和Object.values(MyStringEnum)来分别获取值。

尽管提取混合枚举的键和值有点挑战性。

使用当前版本的TypeScript,你可以使用这些函数将Enum映射到你选择的记录。注意,不能用这些函数定义字符串值,因为它们查找值为数字的键。

enum STATES {
  LOGIN,
  LOGOUT,
}

export const enumToRecordWithKeys = <E extends any>(enumeration: E): E => (
  Object.keys(enumeration)
    .filter(key => typeof enumeration[key] === 'number')
    .reduce((record, key) => ({...record, [key]: key }), {}) as E
);

export const enumToRecordWithValues = <E extends any>(enumeration: E): E => (
  Object.keys(enumeration)
    .filter(key => typeof enumeration[key] === 'number')
    .reduce((record, key) => ({...record, [key]: enumeration[key] }), {}) as E
);

const states = enumToRecordWithKeys(STATES)
const statesWithIndex = enumToRecordWithValues(STATES)

console.log(JSON.stringify({
  STATES,
  states,
  statesWithIndex,
}, null ,2));

// Console output:
{
  "STATES": {
    "0": "LOGIN",
    "1": "LOGOUT",
    "LOGIN": 0,
    "LOGOUT": 1
  },
  "states": {
    "LOGIN": "LOGIN",
    "LOGOUT": "LOGOUT"
  },
  "statesWithIndex": {
    "LOGIN": 0,
    "LOGOUT": 1
  }
}

老问题了,为什么不使用const对象映射呢?

不要这样做:

enum Foo {
    BAR = 60,
    EVERYTHING_IS_TERRIBLE = 80
}

console.log(Object.keys(Foo))
// -> ["60", "80", "BAR", "EVERYTHING_IS_TERRIBLE"]
console.log(Object.values(Foo))
// -> ["BAR", "EVERYTHING_IS_TERRIBLE", 60, 80]

这样做(注意as const强制转换):

const Foo = {
    BAR: 60,
    EVERYTHING_IS_TERRIBLE: 80
} as const

console.log(Object.keys(Foo))
// -> ["BAR", "EVERYTHING_IS_TERRIBLE"]
console.log(Object.values(Foo))
// -> [60, 80]

我希望这个问题仍然有意义。我使用这样的函数:

function enumKeys(target: Record<string, number|string>): string[] {
  const allKeys: string[] = Object.keys(target);
  const parsedKeys: string[] = [];

  for (const key of allKeys) {
    const needToIgnore: boolean
      = target[target[key]]?.toString() === key && !isNaN(parseInt(key));

    if (!needToIgnore) {
      parsedKeys.push(key);
    }
  }

  return parsedKeys;
}

function enumValues(target: Record<string, number|string>): Array<string|number> {
  const keys: string[] = enumKeys(target);
  const values: Array<string|number> = [];

  for (const key of keys) {
    values.push(target[key]);
  }

  return values;
}

例子:

enum HttpStatus {
  OK,
  INTERNAL_ERROR,
  FORBIDDEN = 'FORBIDDEN',
  NOT_FOUND = 404,
  BAD_GATEWAY = 'bad-gateway'
}


console.log(enumKeys(HttpStatus));
// > ["OK", "INTERNAL_ERROR", "FORBIDDEN", "NOT_FOUND", "BAD_GATEWAY"] 

console.log(enumValues(HttpStatus));
// > [0, 1, "FORBIDDEN", 404, "bad-gateway"]