非常非常简单的Enum string (TypeScript 2.4)

import * from '../mylib'

export enum MESSAGES {
    ERROR_CHART_UNKNOWN,
    ERROR_2
}

export class Messages {
    public static get(id : MESSAGES){
        let message = ""
        switch (id) {
            case MESSAGES.ERROR_CHART_UNKNOWN :
                message = "The chart does not exist."
                break;
            case MESSAGES.ERROR_2 :
                message = "example."
                break;
        }
        return message
    }
}

function log(messageName:MESSAGES){
    console.log(Messages.get(messageName))
}

最近在使用TypeScript 1.0.1时遇到了这个问题，并以这种方式解决了:

enum IEvents {
        /** A click on a product or product link for one or more products. */
        CLICK,
        /** A view of product details. */
        DETAIL,
        /** Adding one or more products to a shopping cart. */
        ADD,
        /** Remove one or more products from a shopping cart. */
        REMOVE,
        /** Initiating the checkout process for one or more products. */
        CHECKOUT,
        /** Sending the option value for a given checkout step. */
        CHECKOUT_OPTION,
        /** The sale of one or more products. */
        PURCHASE,
        /** The refund of one or more products. */
        REFUND,
        /** A click on an internal promotion. */
        PROMO_CLICK
}

var Events = [
        'click',
        'detail',
        'add',
        'remove',
        'checkout',
        'checkout_option',
        'purchase',
        'refund',
        'promo_click'
];

function stuff(event: IEvents):boolean {
        // event can now be only IEvents constants
        Events[event]; // event is actually a number that matches the index of the array
}
// stuff('click') won't work, it needs to be called using stuff(IEvents.CLICK)

我认为你应该试试这个，在这种情况下，变量的值不会改变，它的工作方式很像枚举，使用类也可以，唯一的缺点是你可以错误地改变静态变量的值这是我们不希望在枚举中发生的。

namespace portal {

export namespace storageNames {

    export const appRegistration = 'appRegistration';
    export const accessToken = 'access_token';

  }
}

在最新版本(1.0RC)的TypeScript中，你可以像这样使用枚举:

enum States {
    New,
    Active,
    Disabled
} 

// this will show message '0' which is number representation of enum member
alert(States.Active); 

// this will show message 'Disabled' as string representation of enum member
alert(States[States.Disabled]);

更新1

要从string value中获取enum成员的number值，你可以使用这个:

var str = "Active";
// this will show message '1'
alert(States[str]);

更新2

在最新的TypeScript 2.4中，引入了字符串enum，如下所示:

enum ActionType {
    AddUser = "ADD_USER",
    DeleteUser = "DELETE_USER",
    RenameUser = "RENAME_USER",

    // Aliases
    RemoveUser = DeleteUser,
}

有关TypeScript 2.4的更多信息，请阅读MSDN上的博客。

TypeScript 2 + 4。

你现在可以直接将字符串值分配给enum成员:

enum Season {
    Winter = "winter",
    Spring = "spring",
    Summer = "summer",
    Fall = "fall"
}

更多信息见#15486。

打印稿1.8 +

在TypeScript 1.8+中，你可以创建一个字符串文字类型来定义类型，并为值列表创建一个同名的对象。它模仿字符串enum的预期行为。

这里有一个例子:

type MyStringEnum = "member1" | "member2";

const MyStringEnum = {
    Member1: "member1" as MyStringEnum,
    Member2: "member2" as MyStringEnum
};

它将像字符串enum一样工作:

// implicit typing example
let myVariable = MyStringEnum.Member1; // ok
myVariable = "member2";                // ok
myVariable = "some other value";       // error, desired

// explict typing example
let myExplicitlyTypedVariable: MyStringEnum;
myExplicitlyTypedVariable = MyStringEnum.Member1; // ok
myExplicitlyTypedVariable = "member2";            // ok
myExplicitlyTypedVariable = "some other value";   // error, desired

确保输入对象中的所有字符串!如果不这样做，那么在上面的第一个例子中，变量将不会隐式地类型化为MyStringEnum。

使用typescript@next中提供的自定义转换器(https://github.com/Microsoft/TypeScript/pull/13940)，您可以从字符串字面类型创建字符串值的枚举对象。

请查看我的npm包，ts-transformer-enumerate。

使用示例:

// The signature of `enumerate` here is `function enumerate<T extends string>(): { [K in T]: K };`
import { enumerate } from 'ts-transformer-enumerate';

type Colors = 'green' | 'yellow' | 'red';
const Colors = enumerate<Colors>();

console.log(Colors.green); // 'green'
console.log(Colors.yellow); // 'yellow'
console.log(Colors.red); // 'red'