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
我通过搜索“TypeScript iterate over enum keys”找到了这个问题。所以我只想给出对我来说有用的解。也许对别人也有帮助。
我的情况如下:我想在每个枚举键上迭代,然后过滤一些键,然后访问一些对象,其中键作为枚举的计算值。这就是没有TS误差的方法。
enum MyEnum = { ONE = 'ONE', TWO = 'TWO' }
const LABELS = {
[MyEnum.ONE]: 'Label one',
[MyEnum.TWO]: 'Label two'
}
// to declare type is important - otherwise TS complains on LABELS[type]
// also, if replace Object.values with Object.keys -
// - TS blames wrong types here: "string[] is not assignable to MyEnum[]"
const allKeys: Array<MyEnum> = Object.values(MyEnum)
const allowedKeys = allKeys.filter(
(type) => type !== MyEnum.ONE
)
const allowedLabels = allowedKeys.map((type) => ({
label: LABELS[type]
}))
其他回答
你可以用下面的方法从Enum中获取一个名称数组:
const enumNames: string[] = Object.keys(YourEnum).filter(key => isNaN(Number(key)));
让ts-enum-util (github, npm)为您工作,并提供许多额外的类型安全实用程序。适用于字符串和数字enum,正确忽略数字enum的数字索引反向查找条目:
字符串枚举:
import {$enum} from "ts-enum-util";
enum Option {
OPTION1 = 'this is option 1',
OPTION2 = 'this is option 2'
}
// type: ("OPTION1" | "OPTION2")[]
// value: ["OPTION1", "OPTION2"]
const keys= $enum(Option).getKeys();
// type: Option[]
// value: ["this is option 1", "this is option 2"]
const values = $enum(Option).getValues();
数字枚举:
enum Option {
OPTION1,
OPTION2
}
// type: ("OPTION1" | "OPTION2")[]
// value: ["OPTION1", "OPTION2"]
const keys= $enum(Option).getKeys();
// type: Option[]
// value: [0, 1]
const values = $enum(Option).getValues();
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
}
}
虽然答案已经提供了,但几乎没有人指向文档
下面是一个片段
enum Enum {
A
}
let nameOfA = Enum[Enum.A]; // "A"
请记住,string enum成员根本不会生成反向映射。
这里有很多答案,尽管这是一个7年前的问题,但我还是查了一下,我猜会有更多的答案出现在这里。这是我的解决方案,它比其他解决方案简单一点,它处理数字/文本/混合值枚举,都是一样的。
enum funky {
yum , tum='tum', gum = 'jump', plum = 4
}
const list1 = Object.keys(funky)
.filter(k => (Number(k).toString() === Number.NaN.toString()));
console.log(JSON.stringify(list1)); // ["yum","tum","gum","plum"]"
// for the numeric enum vals (like yum = 0, plum = 4), typescript adds val = key implicitly (0 = yum, 4 = plum)
// hence we need to filter out such numeric keys (0 or 4)