如何根据键的另一个属性为键分配条件值答案

【问题标题】：How to assign a conditional value to a key based on another of its properties如何根据键的另一个属性为键分配条件值
【发布时间】：2021-11-23 06:31:05
【问题描述】：

我想知道如何为依赖于同一对象内另一个键的键分配值。在这种情况下，我想说如果动物有主人，它不是食物，但如果它没有主人，它就是食物。因此，在我的示例中，对象 animal 应评估为 true，而 animal2 应评估为 false，但它们不会。

> I have tried isFood: animal.owner ? false: true 
> I have tried isFood: animal.hasOwnProperty("owner") ? false: true
> I have tried isFood: "owner" in animal ? false: true
> I have tried isFood: this.owner ? false: true

interface personInterface {
    name: string,
    age: number,
    isMember: boolean
};
interface animalInterface<T> {
    owner?: T,
    sound: string,
    species: string,
    isFood: boolean
}

let person: personInterface;

person = {
    name: "justin",
    age: 30,
    isMember: true
}

let animal: animalInterface<personInterface>;
let animal2: animalInterface<personInterface>;

animal = {
    owner: {...person},
    sound: "Woof",
    species: "Dog",
    isFood: this.owner ? false: true
}

animal2 = {
    sound: "Woof",
    species: "Dog",
    isFood: this.owner ? false: true 
}

【问题讨论】：

标签： javascript typescript javascript-objects

【解决方案1】：

this 可能因使用位置、函数内部或全局而异。为了引用当前对象并仍将键作为属性访问，您可以使用get 访问器，如下所示。您可以阅读有关get 访问器here 的更多信息。

animal = {
    owner: { ...person },
    sound: "Woof",
    species: "Dog",
    get isFood() {
        // Here this refers to current object
        return this.owner ? false : true
    }
}

animal2 = {
    sound: "Woof",
    species: "Dog",
    get isFood() {
        return this.owner ? false : true
    }
}

console.log(animal.isFood);
console.log(animal2.isFood);

【讨论】：

【解决方案2】：

为了实现它，你需要使用discriminated unions：

interface PersonInterface {
    name: string,
    age: number,
    isMember: boolean
};

interface AnimalInterfaceBase<T> {
    owner?: T,
    sound: string,
    species: string,
    isFood: boolean
}

type WithOwner<T> = {
    owner: T,
    sound: string,
    species: string,
    isFood: false
}

type WithoutOwner = {
    sound: string,
    species: string,
    isFood: true
}


type AnimalInterface<T> = WithOwner<T> | WithoutOwner


let person: PersonInterface;

person = {
    name: "justin",
    age: 30,
    isMember: true
}

let animal: AnimalInterface<PersonInterface>;
let animal2: AnimalInterface<PersonInterface>;

animal = {
    owner: person,
    sound: "Woof",
    species: "Dog",
    isFood: false
}

animal2 = {
    owner: person, // expected error
    sound: "Woof",
    species: "Dog",
    isFood: true
}

Playground

附：接口名称大写是惯例。

【讨论】：

【解决方案3】：

在这种情况下，您可以使用标记的联合类型。基本上，您创建两个接口并通过属性识别它们，在您的情况下为isFood：

interface BaseAnimalInterface {
    sound: string;
    species: string;
    isFood: boolean;
}

interface AnimalInterface extends BaseAnimalInterface {
    isFood: true;
}

interface PetInterface<T = PersonInterface> extends BaseAnimalInterface {
    owner: T;
    isFood: false;
}

type AnimalType = AnimalInterface | PetInterface;

然后，如果您在 if/switch 语句中检查该属性，您将获得正确的类型：

function playSound(animal: AnimalType): void {
    if (animal.isFood) {
        // Casted automatically to `AnimalInterface`
    } else {
        // Casted automatically to `PetInterface`
    }
}

Full example.

【讨论】：