TypeScript：向枚举添加更多数据

Question

在 TypeScript 中，是否可以像在 Java 中那样向枚举常量添加更多内容（属性、方法等）？

演示添加字段、方法和构造函数的 Java 示例：

public enum Planet {
    MERCURY (3.303e+23, 2.4397e6),
    VENUS   (4.869e+24, 6.0518e6),
    EARTH   (5.976e+24, 6.37814e6),
    MARS    (6.421e+23, 3.3972e6),
    JUPITER (1.9e+27,   7.1492e7),
    SATURN  (5.688e+26, 6.0268e7),
    URANUS  (8.686e+25, 2.5559e7),
    NEPTUNE (1.024e+26, 2.4746e7);

    private final double mass;   // in kilograms
    private final double radius; // in meters

    Planet(double mass, double radius) {
        this.mass = mass;
        this.radius = radius;
    }
    private double mass() { return mass; }
    private double radius() { return radius; }

    // universal gravitational constant  (m3 kg-1 s-2)
    public static final double G = 6.67300E-11;

    double surfaceGravity() {
        return G * mass / (radius * radius);
    }
    double surfaceWeight(double otherMass) {
        return otherMass * surfaceGravity();
    }
    // ...
}

Answer 1

不使用枚举，但您可以使用类和一些静态成员获得相同的结果：

class Planet {
    public static MERCURY = new Planet(3.303e+23, 2.4397e6);
    public static VENUS = new Planet(4.869e+24, 6.0518e6);
    public static EARTH = new Planet(5.976e+24, 6.37814e6);
    public static MARS = new Planet(6.421e+23, 3.3972e6);
    public static JUPITER = new Planet(1.9e+27, 7.1492e7);
    public static SATURN = new Planet(5.688e+26, 6.0268e7);
    public static URANUS = new Planet(8.686e+25, 2.5559e7);
    public static NEPTUNE = new Planet(1.024e+26, 2.4746e7);

    private mass: number;
    private radius: number;

    private constructor(mass: number, radius: number) {
        this.mass = mass;
        this.radius = radius;
    }

    public static G = 6.67300E-11;

    public surfaceGravity(): number {
        return Planet.G * this.mass / (this.radius * this.radius);
    }

    public surfaceWeight(otherMass: number) {
        return otherMass * this.surfaceGravity();
    }
}

console.log(Planet.MERCURY.surfaceGravity());

(code in playground)

在java中为枚举中的每个项目创建一个静态实例，这意味着这确实做了同样的事情，只是java有更好的语法来定义枚举。

---

编辑

这是一个具有等效 Planet.values() 的版本，java 将生成：

class Planet {
    public static MERCURY = new Planet(3.303e+23, 2.4397e6);
    public static VENUS = new Planet(4.869e+24, 6.0518e6);
    ...

    private static VALUES: Planet[] = [];

    private mass: number;
    private radius: number;

    private constructor(mass: number, radius: number) {
        this.mass = mass;
        this.radius = radius;

        Planet.VALUES.push(this);
    }

    public static values() {
        return Planet.VALUES;
    }

    ...
}

---

第二次编辑

这是一种实现valueOf的方法：

   public static valueOf(name: string): Planet | null {
        const names = Object.keys(this);
        for (let i = 0; i < names.length; i++) {
            if (this[names[i]] instanceof Planet && name.toLowerCase() === names[i].toLowerCase()) {
                return this[names[i]];
            }
        }

        return null;
    }

Answer 2

很遗憾，这是不可能的。您可以只分配属性名称

请参阅此参考以了解 TypeScript 规范 https://github.com/microsoft/TypeScript/blob/30cb20434a6b117e007a4959b2a7c16489f86069/doc/spec-ARCHIVED.md#a7-enums

Answer 3

您可以使用 const 获得“相同的东西”。

显然，这个解决方案并不完全像枚举，它只是对 const 对象的键的不同使用，而不是实际的枚举值。

export const OrderStateEnum = {
    WAITING_SCORING: { key: 1, value: "WAITING for SCORING" },
    WAITING_ACCEPTANCE: { key: 2, value: "Waiting for acceptance" },
    ORDER_ACCEPTED: { key: 3, value: "Order accepted" },
 } as const

export type OrderStateEnum = keyof typeof OrderStateEnum

export function valueOf(value: string): OrderStateEnum{
    for(var key in OrderStateEnum) {
        if(value.toLowerCase()=== OrderStateEnum[key].value.toLowerCase()){
            return OrderStateEnum[key];
        }
    }
}

接收假枚举的所有值：

  states = OrderStateEnum;

要接收某个值的假枚举：

  OrderStateEnum.WAITING_SCORING.value;

下面是 *ngFor 的用法

<select formControlName="statoOrdine" [(ngModel)]="stato">
    <option *ngFor="let item of states | keyvalue" [value]="item.value.key"> 
      {{item.value.value}}</option>
</select>