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"
}
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
我看不正确的答案看累了,就自己做了。
这个有测试。 适用于所有类型的枚举。 正确地输入。
type EnumKeys<Enum> = Exclude<keyof Enum, number>
const enumObject = <Enum extends Record<string, number | string>>(e: Enum) => {
const copy = {...e} as { [K in EnumKeys<Enum>]: Enum[K] };
Object.values(e).forEach(value => typeof value === 'number' && delete copy[value]);
return copy;
};
const enumKeys = <Enum extends Record<string, number | string>>(e: Enum) => {
return Object.keys(enumObject(e)) as EnumKeys<Enum>[];
};
const enumValues = <Enum extends Record<string, number | string>>(e: Enum) => {
return [...new Set(Object.values(enumObject(e)))] as Enum[EnumKeys<Enum>][];
};
enum Test1 { A = "C", B = "D"}
enum Test2 { A, B }
enum Test3 { A = 0, B = "C" }
enum Test4 { A = "0", B = "C" }
enum Test5 { undefined = "A" }
enum Test6 { A = "undefined" }
enum Test7 { A, B = "A" }
enum Test8 { A = "A", B = "A" }
enum Test9 { A = "B", B = "A" }
console.log(enumObject(Test1)); // {A: "C", B: "D"}
console.log(enumObject(Test2)); // {A: 0, B: 1}
console.log(enumObject(Test3)); // {A: 0, B: "C"}
console.log(enumObject(Test4)); // {A: "0", B: "C"}
console.log(enumObject(Test5)); // {undefined: "A"}
console.log(enumObject(Test6)); // {A: "undefined"}
console.log(enumObject(Test7)); // {A: 0,B: "A"}
console.log(enumObject(Test8)); // {A: "A", B: "A"}
console.log(enumObject(Test9)); // {A: "B", B: "A"}
console.log(enumKeys(Test1)); // ["A", "B"]
console.log(enumKeys(Test2)); // ["A", "B"]
console.log(enumKeys(Test3)); // ["A", "B"]
console.log(enumKeys(Test4)); // ["A", "B"]
console.log(enumKeys(Test5)); // ["undefined"]
console.log(enumKeys(Test6)); // ["A"]
console.log(enumKeys(Test7)); // ["A", "B"]
console.log(enumKeys(Test8)); // ["A", "B"]
console.log(enumKeys(Test9)); // ["A", "B"]
console.log(enumValues(Test1)); // ["C", "D"]
console.log(enumValues(Test2)); // [0, 1]
console.log(enumValues(Test3)); // [0, "C"]
console.log(enumValues(Test4)); // ["0", "C"]
console.log(enumValues(Test5)); // ["A"]
console.log(enumValues(Test6)); // ["undefined"]
console.log(enumValues(Test7)); // [0, "A"]
console.log(enumValues(Test8)); // ["A"]
console.log(enumValues(Test9)); // ["B", "A"]
在线版本。
其他回答
这里的答案似乎都不能在严格模式下使用string-enum。
考虑enum为:
enum AnimalEnum {
dog = "dog", cat = "cat", mouse = "mouse"
}
使用AnimalEnum["dog"]访问可能会导致如下错误:
元素隐式具有“any”类型,因为类型“any”的表达式不能用于索引类型“typeof AnimalEnum”.ts(7053)。
这种情况下的正确解,写为:
AnimalEnum["dog" as keyof typeof AnimalEnum]
具有数字enum:
enum MyNumericEnum {
First = 1,
Second = 2
}
你需要先把它转换成数组:
const values = Object.values(MyNumericEnum);
// ['First', 'Second', 1, 2]
如您所见,它同时包含键和值。钥匙先放。
之后,你可以检索它的键:
values.slice(0, values.length / 2);
// ['First', 'Second']
和值:
values.slice(values.length / 2);
// [1, 2]
对于字符串enum,你可以使用Object.keys(MyStringEnum)来分别获取key和Object.values(MyStringEnum)来分别获取值。
尽管提取混合枚举的键和值有点挑战性。
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
}
}
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]
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"]