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

在TypeScript中,一个enum在javascript中被编译成一个map(从键中获取值):

enum MyEnum {
  entry0,
  entry1,
}

console.log(MyEnum['entry0']); // 0
console.log(MyEnum['entry1']); // 1

它还创建了一个反向映射(从值中获取键):

console.log(MyEnum[0]); // 'entry0'
console.log(MyEnum[0]); // 'entry1'

所以你可以通过以下方式访问一个条目的名称:

console.log(MyEnum[MyEnum.entry0]); // 'entry0'
console.log(MyEnum[MyEnum.entry1]); // 'entry1'

但是,string enum在设计上没有反向映射(参见注释和pull request),因为这可能导致映射对象中的键和值之间的冲突。

enum MyEnum {
  entry0 = 'value0',
  entry1 = 'value1',
}

console.log(MyEnum['value0']); // undefined
console.log(MyEnum['value1']); // undefined

如果你想强制你的字符串enum编译一个反向映射(你必须确保所有的键和值都是不同的),你可以使用这个技巧:

enum MyEnum {
  entry0 = <any>'value0',
  entry1 = <any>'value1',
}

console.log(MyEnum['value0']); // 'entry0'
console.log(MyEnum['value1']); // 'entry1'
console.log(MyEnum[MyEnum.entry0]); // 'entry0'
console.log(MyEnum[MyEnum.entry1]); // 'entry1'

其他回答

当我遇到同样的问题时,你可以使用我写的enum-values包:

Git: enum-values

var names = EnumValues.getNames(myEnum);

要获得枚举值的列表,您必须使用:

enum AnimalEnum {
  DOG = "dog", 
  CAT = "cat", 
  MOUSE = "mouse"
}

Object.values(AnimalEnum);

如果你有enum

enum Diet {
  KETO = "Ketogenic",
  ATKINS = "Atkins",
  PALEO = "Paleo",
  DGAF = "Whatever"
}

然后你可以得到如下的键和值:

Object.keys(Diet).forEach((d: Diet) => {
  console.log(d); // KETO
  console.log(Diet[d]) // Ketogenic
});

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
  }
}

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]
]