如何从具有 super

Question

我想做这样的事情但是出错了。 我在官方文档中找不到任何帮助，我认为这就像调用父方法一样简单。

“错误类型错误：childA.getChildOfType 不是函数”

export class Parent
{
    constructor(public type: string)
    {}

    public getChildOfType = (type: string) =>
    {
        if(this[type] && this[type].length > 0) return this[type];
        return null;
    }
}


export class ChildA extends Parent
{
    Plants = ['Amaryllis', 'African Violet', 'Bird Of Paradise'];
    
    constructor(public type: string)
    {
        super(type);
    }
}

export class ChildB extends Parent
{
    Animals = ['Monkey', 'Tiger', 'Mouse'];
    
    constructor(public type: string)
    {
        super(type);
    }
}

const childA = new ChildA('Plants');
const childB = new ChildB('Animals');

console.log(childA.getChildOfType('Plants'));
console.log(childA.getChildOfType('Animals'));

Answer 1

尝试类似：

Parent.prototype.getChildOfType = (type: string) =>
    {
        if(this[type] && this[type].length > 0) return this[type];
        return null;
    }