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
从TypeScript 2.4开始,枚举可以包含字符串初始化器https://www.typescriptlang.org/docs/handbook/release-notes/typescript-2-4.html
这允许你这样写:
enum Order {
ONE = "First",
TWO = "Second"
}
console.log(`One is ${Order.ONE.toString()}`);
得到这样的输出:
一个是First
其他回答
简单地说
如果你的枚举如下:
export enum Colors1 {
Red = 1,
Green = 2,
Blue = 3
}
要获得特定的文本和值:
console.log(Colors1.Red); // 1
console.log(Colors1[Colors1.Red]); // Red
获取值和文本列表:
public getTextAndValues(e: { [s: number]: string }) {
for (const enumMember in e) {
if (parseInt(enumMember, 10) >= 0) {
console.log(e[enumMember]) // Value, such as 1,2,3
console.log(parseInt(enumMember, 10)) // Text, such as Red,Green,Blue
}
}
}
this.getTextAndValues(Colors1)
如果你的枚举如下:
export enum Colors2 {
Red = "Red",
Green = "Green",
Blue = "Blue"
}
要获得特定的文本和值:
console.log(Colors2.Red); // Red
console.log(Colors2["Red"]); // Red
获取值和文本列表:
public getTextAndValues(e: { [s: string]: string }) {
for (const enumMember in e) {
console.log(e[enumMember]);// Value, such as Red,Green,Blue
console.log(enumMember); // Text, such as Red,Green,Blue
}
}
this.getTextAndValues(Colors2)
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" ];
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]);
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]
}))
假设您有一个枚举
export enum SCROLL_LABEL_OFFSET {
SMALL = 48,
REGULAR = 60,
LARGE = 112
}
你想要创建一个基于枚举的类型而不仅仅是复制和粘贴。 你可以像这样使用枚举来创建你的类型:
export type ScrollLabelOffset = keyof typeof SCROLL_LABEL_OFFSET;
结果你会收到一个可能值为'SMALL' | 'REGULAR' | 'LARGE'的类型