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

我写了一个EnumUtil类,它通过枚举值进行类型检查:

export class EnumUtils {
  /**
   * Returns the enum keys
   * @param enumObj enum object
   * @param enumType the enum type
   */
  static getEnumKeys(enumObj: any, enumType: EnumType): any[] {
    return EnumUtils.getEnumValues(enumObj, enumType).map(value => enumObj[value]);
  }

  /**
   * Returns the enum values
   * @param enumObj enum object
   * @param enumType the enum type
   */
  static getEnumValues(enumObj: any, enumType: EnumType): any[] {
    return Object.keys(enumObj).filter(key => typeof enumObj[key] === enumType);
  }
}

export enum EnumType {
  Number = 'number',
  String = 'string'
}

如何使用:

enum NumberValueEnum{
  A= 0,
  B= 1
}

enum StringValueEnum{
  A= 'A',
  B= 'B'
}

EnumUtils.getEnumKeys(NumberValueEnum, EnumType.Number);
EnumUtils.getEnumValues(NumberValueEnum, EnumType.Number);

EnumUtils.getEnumKeys(StringValueEnum, EnumType.String);
EnumUtils.getEnumValues(StringValueEnum, EnumType.String);

NumberValueEnum键的结果:["A", "B"]

NumberValueEnum值的结果:[0,1]

StringValueEnumkeys的结果:["A", "B"]

StringValueEnumvalues的结果:["A", "B"]

其他回答

从TypeScript 2.4开始,枚举不再包含键作为成员。来源TypeScript自述文件

需要注意的是,字符串初始化的枚举不能反向映射到原始枚举成员名。换句话说,你不能写Colors["RED"]来获得字符串"RED"。

我的解决方案:

export const getColourKey = (value: string ) => {
    let colourKey = '';
    for (const key in ColourEnum) {
        if (value === ColourEnum[key]) {
            colourKey = key;
            break;
        }
    }
    return colourKey;
};

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)
 

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})

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。

我卑微的2美分基于阅读一个了不起的评论从github TS讨论

const EnvironmentVariants = ['development', 'production', 'test'] as const 
type EPredefinedEnvironment = typeof EnvironmentVariants[number]

然后在编译时:

// TS2322: Type '"qaEnv"' is not assignable to type '"development" | "production" | "test"'.
const qaEnv: EPredefinedEnvironment = 'qa' 

在运行时:

function isPredefinedEnvironemt(env: string) {
  for (const predefined of EnvironmentVariants) {
    if (predefined === env) {
      return true
    }
  }
  return false
}

assert(isPredefinedEnvironemet('test'), true)
assert(isPredefinedEnvironemet('qa'), false)

注意,for(const index in environmentvariables){…}将遍历"0","1","2"集合