如果你想在html中获取名字

对于这个枚举

enum CanadianProvinces {
    AB,
    BC,
    MB,
    NB,
    NL,
    NT,
    NS,
    NU,
    ON,
    PE,
    QC,
    SK,
    YT
}

将组件中的变量赋值为

canadianProvinces = CanadianProvinces 

然后，在你的html

{{canadianProvinces[0]}}

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]

简单地说

如果你的枚举如下:

export enum Colors1 {
  Red = 1,
  Green = 2,
  Blue = 3
}

要获得特定的文本和值:

console.log(Colors1.Red); // 1 
console.log(Colors1[Colors1.Red]); // Red

获取值和文本列表:

public getTextAndValues(e: { [s: number]: string }) {
  for (const enumMember in e) {
    if (parseInt(enumMember, 10) >= 0) {
      console.log(e[enumMember]) // Value, such as 1,2,3
      console.log(parseInt(enumMember, 10)) // Text, such as Red,Green,Blue
    }
  }
}
this.getTextAndValues(Colors1)

如果你的枚举如下:

export enum Colors2 {
  Red = "Red",
  Green = "Green",
  Blue = "Blue"
}

要获得特定的文本和值:

console.log(Colors2.Red); // Red
console.log(Colors2["Red"]); // Red

获取值和文本列表:

public getTextAndValues(e: { [s: string]: string }) {
  for (const enumMember in e) {
    console.log(e[enumMember]);// Value, such as Red,Green,Blue
    console.log(enumMember); //  Text, such as Red,Green,Blue
  }
}
this.getTextAndValues(Colors2)

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]
    }))

这里有很多答案，尽管这是一个7年前的问题，但我还是查了一下，我猜会有更多的答案出现在这里。这是我的解决方案，它比其他解决方案简单一点，它处理数字/文本/混合值枚举，都是一样的。

enum funky {
    yum , tum='tum', gum = 'jump', plum = 4
}

const list1 = Object.keys(funky)
  .filter(k => (Number(k).toString() === Number.NaN.toString()));
console.log(JSON.stringify(list1)); // ["yum","tum","gum","plum"]" 

 // for the numeric enum vals (like yum = 0, plum = 4), typescript adds val = key implicitly (0 = yum, 4 = plum)
 // hence we need to filter out such numeric keys (0 or 4)