这里发现的另一个有趣的解决方案是使用ES6 Map:

export enum Type {
  low,
  mid,
  high
}

export const TypeLabel = new Map<number, string>([
  [Type.low, 'Low Season'],
  [Type.mid, 'Mid Season'],
  [Type.high, 'High Season']
]);

USE

console.log(TypeLabel.get(Type.low)); // Low Season


TypeLabel.forEach((label, value) => {
  console.log(label, value);
});

// Low Season 0
// Mid Season 1
// High Season 2    

基于上面的一些回答，我提出了这个类型安全的函数签名:

export function getStringValuesFromEnum<T>(myEnum: T): (keyof T)[] {
  return Object.keys(myEnum).filter(k => typeof (myEnum as any)[k] === 'number') as any;
}

用法:

enum myEnum { entry1, entry2 };
const stringVals = getStringValuesFromEnum(myEnum);

stringVals的类型是'entry1' | 'entry2'

看看它的实际应用

Typescript游乐场示例

enum TransactionStatus {
  SUBMITTED = 'submitted',
  APPROVED = 'approved',
  PAID = 'paid',
  CANCELLED = 'cancelled',
  DECLINED = 'declined',
  PROCESSING = 'processing',
}


let set1 = Object.entries(TransactionStatus).filter(([,value]) => value === TransactionStatus.SUBMITTED || value === TransactionStatus.CANCELLED).map(([key,]) => {
    return key
})


let set2 = Object.entries(TransactionStatus).filter(([,value]) => value === TransactionStatus.PAID || value === TransactionStatus.APPROVED).map(([key,]) => {
    return key
})

let allKeys = Object.keys(TransactionStatus)



console.log({set1,set2,allKeys})

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]
] 

这里发现的另一个有趣的解决方案是使用ES6 Map:

export enum Type {
  low,
  mid,
  high
}

export const TypeLabel = new Map<number, string>([
  [Type.low, 'Low Season'],
  [Type.mid, 'Mid Season'],
  [Type.high, 'High Season']
]);

USE

console.log(TypeLabel.get(Type.low)); // Low Season


TypeLabel.forEach((label, value) => {
  console.log(label, value);
});

// Low Season 0
// Mid Season 1
// High Season 2    

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]
    }))