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

其他回答

假设您坚持使用规则,只生成带有数值的枚举,您可以使用这段代码。这正确地处理了名称恰好是有效数字的情况

enum Color {
    Red,
    Green,
    Blue,
    "10" // wat
}

var names: string[] = [];
for(var n in Color) {
    if(typeof Color[n] === 'number') names.push(n);
}
console.log(names); // ['Red', 'Green', 'Blue', '10']

从TypeScript 2.4开始,枚举可以包含字符串初始化器https://www.typescriptlang.org/docs/handbook/release-notes/typescript-2-4.html

这允许你这样写:

 enum Order {
      ONE = "First",
      TWO = "Second"
 }

console.log(`One is ${Order.ONE.toString()}`);

得到这样的输出:

一个是First

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

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[];