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
这对于基于键值的enum更有效:
enum yourEnum {
["First Key"] = "firstWordValue",
["Second Key"] = "secondWordValue"
}
Object.keys(yourEnum)[Object.values(yourEnum).findIndex(x => x === yourValue)]
// Result for passing values as yourValue
// FirstKey
// SecondKey
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
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]
]
如果它是你的枚举,你定义如下所示,名称和值是相同的,它会直接给你条目的名称。
enum myEnum {
entry1="entry1",
entry2="entry2"
}
for (var entry in myEnum) {
// use entry's name here, e.g., "entry1"
}
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" ];