迭代枚举 TypeScript

Question

我正在尝试迭代这个枚举。然而，我面临着从枚举中提取值的问题。如果我给它一个像这样的硬编码字符串，我可以提取该值。

export enum Values {
  ACE = 'A',
  TWO = '2',
  THREE = '3',
  FOUR = '4',
  FIVE = '5',
  SIX = '6',
  SEVEN = '7',
  EIGHT = '8',
  NINE = '9',
  TEN = '10',
  JACK = 'J',
  QUEEN = 'Q',
  KING = 'K',
}

console.log(Values['TWO']);

但是一旦我给它来自枚举的多个字符串值，如下所示。

export enum Values {
  ACE = 'A',
  TWO = '2',
  THREE = '3',
  FOUR = '4',
  FIVE = '5',
  SIX = '6',
  SEVEN = '7',
  EIGHT = '8',
  NINE = '9',
  TEN = '10',
  JACK = 'J',
  QUEEN = 'Q',
  KING = 'K',
}

for (let value in Values) {
  console.log(Values[value]);
}

它给出以下错误：

error TS7053: Element implicitly has an 'any' type because expression of type 'string' can't be used to index type 'typeof Values'.
  No index signature with a parameter of type 'string' was found on type 'typeof Values'.

25   console.log(Values[value]);

对于 TypeScript 来说是全新的，所以可能会遗漏一些东西。

为任何能给我一些指导的人干杯。

Answer 1

for-in使用循环迭代时循环变量的类型enum Values是隐式的any，因此会出现编译错误。

for (let value in Values) {
  console.log(Values[value]);
}

该变量value的类型为any，不能用于索引enum Values。

在此处打开 TypeScript 问题。

如果您想迭代 的键值对，enum您可以借助以下方法来实现Object.entries：

enum Values {
  ACE = 'A',
  TWO = '2',
  THREE = '3',
  FOUR = '4',
  FIVE = '5',
  SIX = '6',
  SEVEN = '7',
  EIGHT = '8',
  NINE = '9',
  TEN = '10',
  JACK = 'J',
  QUEEN = 'Q',
  KING = 'K',
}

//Both key and value
Object.entries(Values).forEach(([key, value]) => {
    console.log(key, value)
});

//keys
Object.keys(Values).forEach(key => {
    console.log(key);
});

//values
Object.values(Values).forEach(value => {
    console.log(value);
});