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

简单地说

如果你的枚举如下:

export enum Colors1 {
  Red = 1,
  Green = 2,
  Blue = 3
}

要获得特定的文本和值:

console.log(Colors1.Red); // 1 
console.log(Colors1[Colors1.Red]); // Red

获取值和文本列表:

public getTextAndValues(e: { [s: number]: string }) {
  for (const enumMember in e) {
    if (parseInt(enumMember, 10) >= 0) {
      console.log(e[enumMember]) // Value, such as 1,2,3
      console.log(parseInt(enumMember, 10)) // Text, such as Red,Green,Blue
    }
  }
}
this.getTextAndValues(Colors1)

如果你的枚举如下:

export enum Colors2 {
  Red = "Red",
  Green = "Green",
  Blue = "Blue"
}

要获得特定的文本和值:

console.log(Colors2.Red); // Red
console.log(Colors2["Red"]); // Red

获取值和文本列表:

public getTextAndValues(e: { [s: string]: string }) {
  for (const enumMember in e) {
    console.log(e[enumMember]);// Value, such as Red,Green,Blue
    console.log(enumMember); //  Text, such as Red,Green,Blue
  }
}
this.getTextAndValues(Colors2)

其他回答

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"]

我发现这个解决方案更优雅:

for (let val in myEnum ) {

 if ( isNaN( parseInt( val )) )
     console.log( val );
}

它显示:

bar 
foo

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]
] 

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