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

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

其他回答

我希望这个问题仍然有意义。我使用这样的函数:

function enumKeys(target: Record<string, number|string>): string[] {
  const allKeys: string[] = Object.keys(target);
  const parsedKeys: string[] = [];

  for (const key of allKeys) {
    const needToIgnore: boolean
      = target[target[key]]?.toString() === key && !isNaN(parseInt(key));

    if (!needToIgnore) {
      parsedKeys.push(key);
    }
  }

  return parsedKeys;
}

function enumValues(target: Record<string, number|string>): Array<string|number> {
  const keys: string[] = enumKeys(target);
  const values: Array<string|number> = [];

  for (const key of keys) {
    values.push(target[key]);
  }

  return values;
}

例子:

enum HttpStatus {
  OK,
  INTERNAL_ERROR,
  FORBIDDEN = 'FORBIDDEN',
  NOT_FOUND = 404,
  BAD_GATEWAY = 'bad-gateway'
}


console.log(enumKeys(HttpStatus));
// > ["OK", "INTERNAL_ERROR", "FORBIDDEN", "NOT_FOUND", "BAD_GATEWAY"] 

console.log(enumValues(HttpStatus));
// > [0, 1, "FORBIDDEN", 404, "bad-gateway"]

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
});

根据TypeScript文档,我们可以通过Enum和静态函数来实现这一点。

使用静态函数获取Enum名称

enum myEnum { 
    entry1, 
    entry2 
}

namespace myEnum {
    export function GetmyEnumName(m: myEnum) {
      return myEnum[m];
    }
}


now we can call it like below
myEnum.GetmyEnumName(myEnum.entry1);
// result entry1 

要阅读更多关于Enum的静态函数,请点击下面的链接 https://basarat.gitbooks.io/typescript/docs/enums.html

老问题了,为什么不使用const对象映射呢?

不要这样做:

enum Foo {
    BAR = 60,
    EVERYTHING_IS_TERRIBLE = 80
}

console.log(Object.keys(Foo))
// -> ["60", "80", "BAR", "EVERYTHING_IS_TERRIBLE"]
console.log(Object.values(Foo))
// -> ["BAR", "EVERYTHING_IS_TERRIBLE", 60, 80]

这样做(注意as const强制转换):

const Foo = {
    BAR: 60,
    EVERYTHING_IS_TERRIBLE: 80
} as const

console.log(Object.keys(Foo))
// -> ["BAR", "EVERYTHING_IS_TERRIBLE"]
console.log(Object.values(Foo))
// -> [60, 80]

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[];