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
在TypeScript中,一个enum在javascript中被编译成一个map(从键中获取值):
enum MyEnum {
entry0,
entry1,
}
console.log(MyEnum['entry0']); // 0
console.log(MyEnum['entry1']); // 1
它还创建了一个反向映射(从值中获取键):
console.log(MyEnum[0]); // 'entry0'
console.log(MyEnum[0]); // 'entry1'
所以你可以通过以下方式访问一个条目的名称:
console.log(MyEnum[MyEnum.entry0]); // 'entry0'
console.log(MyEnum[MyEnum.entry1]); // 'entry1'
但是,string enum在设计上没有反向映射(参见注释和pull request),因为这可能导致映射对象中的键和值之间的冲突。
enum MyEnum {
entry0 = 'value0',
entry1 = 'value1',
}
console.log(MyEnum['value0']); // undefined
console.log(MyEnum['value1']); // undefined
如果你想强制你的字符串enum编译一个反向映射(你必须确保所有的键和值都是不同的),你可以使用这个技巧:
enum MyEnum {
entry0 = <any>'value0',
entry1 = <any>'value1',
}
console.log(MyEnum['value0']); // 'entry0'
console.log(MyEnum['value1']); // 'entry1'
console.log(MyEnum[MyEnum.entry0]); // 'entry0'
console.log(MyEnum[MyEnum.entry1]); // 'entry1'
其他回答
当我遇到同样的问题时,你可以使用我写的enum-values包:
Git: enum-values
var names = EnumValues.getNames(myEnum);
要获得枚举值的列表,您必须使用:
enum AnimalEnum {
DOG = "dog",
CAT = "cat",
MOUSE = "mouse"
}
Object.values(AnimalEnum);
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
});
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
}
}
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]
]