javascript - typescript 方法覆盖不起作用

Question

我有一个基类 A 和一个覆盖 A 的 fetch 方法的扩展类 B。 但是，当我调用 B.fetch() 时，它会记录“基类”。为什么不调用B中定义的方法定义？

     class A {
        protected apiPath: string = ''
        static async fetch (id = null, Model = this, apiPath = this.apiPath): Promise<any> {
            console.log('basefetch')
            let url = apiPath
            if(id) url += '/' + id
            let { data } = await axios.get(url)
            let item: any = new Model(data)
            return item
        }
    }

    class B extends A {
        static async fetch(id = null, Model = this, apiPath = this.apiPath): Promise<Data> {
            console.log('childfetch')
            let {data} = await axios.get(apiPath)
            if (typeof data.challenge.participating == 'undefined') {
                data.challenge.participating = null
            }
            if (typeof data.challenge.progress == 'undefined') {
                data.challenge.progress = null
            }
            return new Model(data)
        }

class SomeOtherModule {
    async doSomething() {
        let b: B = await B.fetch()
    }
}

Answer 1

嗯，方法签名是不同的。在 A 中，您返回 Promise<any>但在 B 中你返回 Promise<Data> .

可以在this snippet中看到您正在尝试做的事情肯定会奏效:

class A {
  static fetch (): string {
    return 'basefetch';
  }
}

class B extends A {
  static fetch(): any {
    return 'childfetch';
  }
}

let aVal: string = A.fetch();
console.log(`aVal: ${aVal}`);

let bVal: string = B.fetch();
console.log(`bVal: ${bVal}`);