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。

其他回答

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)
 

如果你只搜索名称,然后迭代使用:

Object.keys(myEnum).map(key => myEnum[key]).filter(value => typeof value === 'string') as string[];

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})

他们在官方文件中提供了一个叫做“反向映射”的概念。它帮助了我:

https://www.typescriptlang.org/docs/handbook/enums.html#reverse-mappings

解决方法很简单:

enum Enum {
 A,
}

let a = Enum.A;
let nameOfA = Enum[a]; // "A"

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