这对于基于键值的enum更有效:

enum yourEnum {
  ["First Key"] = "firstWordValue",
  ["Second Key"] = "secondWordValue"
}

Object.keys(yourEnum)[Object.values(yourEnum).findIndex(x => x === yourValue)]
// Result for passing values as yourValue
// FirstKey
// SecondKey

如果你有enum

enum Diet {
  KETO = "Ketogenic",
  ATKINS = "Atkins",
  PALEO = "Paleo",
  DGAF = "Whatever"
}

然后你可以得到如下的键和值:

Object.keys(Diet).forEach((d: Diet) => {
  console.log(d); // KETO
  console.log(Diet[d]) // Ketogenic
});

根据TypeScript文档，我们可以通过Enum和静态函数来实现这一点。

使用静态函数获取Enum名称

enum myEnum { 
    entry1, 
    entry2 
}

namespace myEnum {
    export function GetmyEnumName(m: myEnum) {
      return myEnum[m];
    }
}


now we can call it like below
myEnum.GetmyEnumName(myEnum.entry1);
// result entry1 

要阅读更多关于Enum的静态函数，请点击下面的链接 https://basarat.gitbooks.io/typescript/docs/enums.html

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
  }
}

你可以这样做，我认为这是最短、最干净、最快的:

Object.entries(test).filter(([key]) => (!~~key && key !== "0"))

给定以下混合类型枚举定义:

enum testEnum {
  Critical = "critical",
  Major = 3,
  Normal = "2",
  Minor = "minor",
  Info = "info",
  Debug = 0
};

它将会变成以下内容:

var testEnum = { 关键:“至关重要的”, 主要:3, 正常:“2”, 小:“小”, 信息:“信息”, 调试:0, [0]:“关键”, [1]: 3, [2]:“2”, [3]:“小”, [4]:“信息”, [5]: 0 ｝ 函数safeEnumEntries(test) { return Object.entries(test).filter(([key]) => (!~~key && key !== "0"); }; console.log (safeEnumEntries (testEnum));

执行函数后，你只会得到好的条目:

[
  ["Critical", "critical"],
  ["Major", 3],
  ["Normal", "2"],
  ["Minor", "minor"],
  ["Info", "info"],
  ["Debug", 0]
] 

我通过搜索“TypeScript iterate over enum keys”找到了这个问题。所以我只想给出对我来说有用的解。也许对别人也有帮助。

我的情况如下:我想在每个枚举键上迭代，然后过滤一些键，然后访问一些对象，其中键作为枚举的计算值。这就是没有TS误差的方法。

    enum MyEnum = { ONE = 'ONE', TWO = 'TWO' }
    const LABELS = {
       [MyEnum.ONE]: 'Label one',
       [MyEnum.TWO]: 'Label two'
    }


    // to declare type is important - otherwise TS complains on LABELS[type]
    // also, if replace Object.values with Object.keys - 
    // - TS blames wrong types here: "string[] is not assignable to MyEnum[]"
    const allKeys: Array<MyEnum> = Object.values(MyEnum)

    const allowedKeys = allKeys.filter(
      (type) => type !== MyEnum.ONE
    )

    const allowedLabels = allowedKeys.map((type) => ({
      label: LABELS[type]
    }))