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
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。
其他回答
在当前的TypeScript版本1.8.9中,我使用类型化enum:
export enum Option {
OPTION1 = <any>'this is option 1',
OPTION2 = <any>'this is option 2'
}
与结果在这个Javascript对象:
Option = {
"OPTION1": "this is option 1",
"OPTION2": "this is option 2",
"this is option 1": "OPTION1",
"this is option 2": "OPTION2"
}
所以我必须通过键和值查询,只返回值:
let optionNames: Array<any> = [];
for (let enumValue in Option) {
let optionNameLength = optionNames.length;
if (optionNameLength === 0) {
this.optionNames.push([enumValue, Option[enumValue]]);
} else {
if (this.optionNames[optionNameLength - 1][1] !== enumValue) {
this.optionNames.push([enumValue, Option[enumValue]]);
}
}
}
我在数组中收到选项键:
optionNames = [ "OPTION1", "OPTION2" ];
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
}
}
如果你有enum
enum Diet {
KETO = "Ketogenic",
ATKINS = "Atkins",
PALEO = "Paleo",
DGAF = "Whatever"
}
然后你可以得到如下的键和值:
Object.keys(Diet).forEach((d: Diet) => {
console.log(d); // KETO
console.log(Diet[d]) // Ketogenic
});
唯一的解决方案,适用于我在所有情况下(即使值是字符串)是以下:
var enumToString = function(enumType, enumValue) {
for (var enumMember in enumType) {
if (enumType[enumMember]==enumValue) return enumMember
}
}
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