我不喜欢上面的答案，因为它们都不能正确地处理字符串/数字的混合，可以在TypeScript enum中作为值。

下面的函数遵循TypeScript枚举的语义，给出键到值的正确映射。从那里，获取一个对象数组或者仅仅是键或者仅仅是值是很简单的。

/**
 * Converts the given enum to a map of the keys to the values.
 * @param enumeration The enum to convert to a map.
 */
function enumToMap(enumeration: any): Map<string, string | number> {
  const map = new Map<string, string | number>();
  for (let key in enumeration) {
      //TypeScript does not allow enum keys to be numeric
      if (!isNaN(Number(key))) continue;

      const val = enumeration[key] as string | number;

      //TypeScript does not allow enum value to be null or undefined
      if (val !== undefined && val !== null)
          map.set(key, val);
  }

  return map;
}

使用示例:

enum Dog {
    Rover = 1,
    Lassie = "Collie",
    Fido = 3,
    Cody = "Mutt",
}

let map = enumToMap(Dog); //Map of keys to values

let objs = Array.from(map.entries()).map(m => ({id: m[1], name: m[0]})); //Objects as asked for in OP
let entries = Array.from(map.entries()); //Array of each entry
let keys = Array.from(map.keys()); //An array of keys
let values = Array.from(map.values()); //An array of values

我还要指出OP是逆向思考枚举的。枚举中的“键”在技术上是在左边，而值在右边。TypeScript允许你尽可能多地重复RHS上的值。

只有一句话:

Object.entries(GoalProgressMeasurements).map(([key, value]) => ({id: key, value: value}))

 this worked for me :

    export enum FeedBackType {
    FEEDBACK1= 'FEEDBACK1',
    FEEDBACK2= 'FEEDBACK2',
    FEEDBACK3= 'FEEDBACK3',
    }

----------------------------------------------------------------- 
    export function getTypeFeedBackList() {
    let feedbackList: FeedBackType[] = [];
    Object.keys(FeedBackType).map((key) => {
    let strEnum = key as unknown as FeedBackType;
    feedbackList.push(strEnum);
    });
    return feedbackList;
    }
---------------------------------------------------------------- 
declare this :

    public feedbackList: FeedBackType[] = [];

and after call your function in  :

    ngOnInit(): void {
    this.feedbackList = getTypeFeedBackList();
    console.log(this.feedbackList); 
    }

Happy coding ;) 

我不建议在需要枚举条目列表的情况下使用TS enum。

在运行时，Enum是作为对象实现的，但它只在这种情况下工作:

enum X {
  Z = 'z',
  F = 'f'
};

console.log(Object.values(X))
console.log(Object.keys(X))
>>>
[LOG]: ["z", "f"] 
[LOG]: ["Z", "F"] 

在这种情况下，它与陷阱一起工作(TS让你通过它的数值访问值):

enum X {
  Z,
  F
};

console.log(Object.values(X))
console.log(Object.keys(X))
>>>
[LOG]: ["Z", "F", 0, 1] 
[LOG]: ["0", "1", "Z", "F"] 

因此，根据Enum定义，任何你写在Enum上循环的函数都将工作/失败。这是…不好的。

我的结论是:Enum不是被设计用来作为对象的。如果你需要访问键和值集合，请使用const而不是enum:

const Enumed = {
    X: 1,
    Y: 2
}

Typescript将控制对象键的存在，你将能够使用object。钥匙等，以安全一致的方式。

class EnumHelpers {

    static getNamesAndValues<T extends number>(e: any) {
        return EnumHelpers.getNames(e).map(n => ({ name: n, value: e[n] as T }));
    }

    static getNames(e: any) {
        return EnumHelpers.getObjValues(e).filter(v => typeof v === 'string') as string[];
    }

    static getValues<T extends number>(e: any) {
        return EnumHelpers.getObjValues(e).filter(v => typeof v === 'number') as T[];
    }

    static getSelectList<T extends number, U>(e: any, stringConverter: (arg: U) => string) {
        const selectList = new Map<T, string>();
        this.getValues(e).forEach(val => selectList.set(val as T, stringConverter(val as unknown as U)));
        return selectList;
    }

    static getSelectListAsArray<T extends number, U>(e: any, stringConverter: (arg: U) => string) {
        return Array.from(this.getSelectList(e, stringConverter), value => ({ value: value[0] as T, presentation: value[1] }));
    }

    private static getObjValues(e: any): (number | string)[] {
        return Object.keys(e).map(k => e[k]);
    }
}

export function enumKeys(E: any): string[] {
    return Object.keys(E).filter(k => isNaN(Number(k)));
}

export function enumValues(E: any): string[] | number[] {
    return enumKeys(E).map(k => E[k as any]);
}

两者都适用:

enum TestA {
    RED = "red",
    BLUE = "blue"
}

enum TestB {
    ONE = 1,
    TWO = 2
}