JSDoc - 如何记录超类中返回其子类实例的静态方法？

Question

所以我有一个像这样的基类：

class BaseAPIClass {
    /**
     * Async factory
     * Does some async operation, then instantiates the inheriting class.
     *
     * @param {String} id
     *
     * @returns {Promise<???>}
     */
    static async create(id) {
        const data = await fetchFromNetwork(id)
        return new this.prototype.constructor(data)
    }
}

还有一个像这样的子类：

class RadAPIClass extends BaseAPIClass {
    /**
     * Output a compliment.
     * @param {name} [name='my dude']
     */
    complimentJorts(name = 'my dude') {
        const possibleCompliments = ['gnarly', 'tubular', 'bodacious', 'radical', 'sick']
        const i = Math.floor(Math.random() * possibleCompliments.length)
        const compliment = possibleCompliments[i]
        console.log(`Those are some ${compliment} jorts, ${name}!`)
    }
}

我的问题是：

如何以正确告诉读者和智能感知

BaseAPIClass.create()

返回

RadAPIClass.create()

的方式记录

{Promise<RadAPIClass>}

的返回类型？

async function main()
{
    const rad = await RadAPIClass.create()
    rad.complimentJorts() // I want VSCode to be able to autocomplete this method for me.
}

Answer 1

年纪大了，更聪明了。答案是通用的

@this

。

class BaseAPIClass {
    /**
     * Async factory
     * Does some async operation, then instantiates the inheriting class.
     *
     * @template {{new(...args: any): any}} T
     * @this {T}
     *
     * @param {String} id
     *
     * @returns {Promise<InstanceType<T>>}
     */
    static async create(id) {
        const data = await fetchFromNetwork(id)
        return new this.prototype.constructor(data)
    }
}

由于

T

，泛型

@this {T}

是从子类的类型推断出来的，因此

Promise<InstanceType<T>>

的返回类型（记住

T

是

typeof BaseAPIClass

或

typeof RadAPIClass

，而不是实例类型）解析为正确的班级。