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
我的Enum是这样的:
export enum UserSorting {
SortByFullName = "Sort by FullName",
SortByLastname = "Sort by Lastame",
SortByEmail = "Sort by Email",
SortByRoleName = "Sort by Role",
SortByCreatedAt = "Sort by Creation date",
SortByCreatedBy = "Sort by Author",
SortByUpdatedAt = "Sort by Edit date",
SortByUpdatedBy = "Sort by Editor",
}
这样做会返回undefined:
UserSorting[UserSorting.SortByUpdatedAt]
为了解决这个问题,我选择了另一种使用管道的方法:
import { Pipe, PipeTransform } from '@angular/core';
@Pipe({
name: 'enumKey'
})
export class EnumKeyPipe implements PipeTransform {
transform(value, args: string[] = null): any {
let enumValue = args[0];
var keys = Object.keys(value);
var values = Object.values(value);
for (var i = 0; i < keys.length; i++) {
if (values[i] == enumValue) {
return keys[i];
}
}
return null;
}
}
要使用它:
return this.enumKeyPipe.transform(UserSorting, [UserSorting.SortByUpdatedAt]);
其他回答
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]
我发现这个解决方案更优雅:
for (let val in myEnum ) {
if ( isNaN( parseInt( val )) )
console.log( val );
}
它显示:
bar
foo
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
}
}
我写了一个helper函数来枚举一个枚举:
static getEnumValues<T extends number>(enumType: {}): T[] {
const values: T[] = [];
const keys = Object.keys(enumType);
for (const key of keys.slice(0, keys.length / 2)) {
values.push(<T>+key);
}
return values;
}
用法:
for (const enumValue of getEnumValues<myEnum>(myEnum)) {
// do the thing
}
该函数返回可以轻松枚举的内容,并将其转换为枚举类型。
基于上面的一些回答,我提出了这个类型安全的函数签名:
export function getStringValuesFromEnum<T>(myEnum: T): (keyof T)[] {
return Object.keys(myEnum).filter(k => typeof (myEnum as any)[k] === 'number') as any;
}
用法:
enum myEnum { entry1, entry2 };
const stringVals = getStringValuesFromEnum(myEnum);
stringVals的类型是'entry1' | 'entry2'
看看它的实际应用