Javascript 使用typescriptsuper（）

我正在尝试用TypeScript扩展一个类。我在编译时不断收到此错误：“提供的参数与调用目标的任何签名都不匹配。”我已尝试将super调用中的artist.name属性作为super（name）引用，但不起作用

如果您有任何想法和解释，我们将不胜感激。谢谢，亚历克斯

class Artist {
  constructor(
    public name: string,
    public age: number,
    public style: string,
    public location: string
  ){
    console.log(`instantiated ${name}, whom is ${age} old, from ${location}, and heavily regarded in the ${style} community`);
  }
}

class StreetArtist extends Artist {
  constructor(
    public medium: string,
    public famous: boolean,
    public arrested: boolean,
    public art: Artist
  ){
    super();
    console.log(`instantiated ${this.name}. Are they famous? ${famous}. Are they locked up? ${arrested}`);
  }
}

interface Human {
  name: string,
  age: number
}

function getArtist(artist: Human){
  console.log(artist.name)
}

let Banksy = new Artist(
  "Banksy",
   40,
  "Politcal Graffitti",
  "England / Wolrd"
)

getArtist(Banksy);

---

超级调用必须提供基类的所有参数。构造函数不是继承的。注释了艺术家，因为我想这样做时不需要它

class StreetArtist extends Artist {
  constructor(
    name: string,
    age: number,
    style: string,
    location: string,
    public medium: string,
    public famous: boolean,
    public arrested: boolean,
    /*public art: Artist*/
  ){
    super(name, age, style, location);
    console.log(`instantiated ${this.name}. Are they famous? ${famous}. Are they locked up? ${arrested}`);
  }
}

或者，如果您打算使用art参数填充基本属性，但在这种情况下，我想实际上不需要使用public on art参数，因为这些属性将被继承，并且只存储重复的数据

class StreetArtist extends Artist {
  constructor(
    public medium: string,
    public famous: boolean,
    public arrested: boolean,
    /*public */art: Artist
  ){
    super(art.name, art.age, art.style, art.location);
    console.log(`instantiated ${this.name}. Are they famous? ${famous}. Are they locked up? ${arrested}`);
  }
}

---

如果将pubic添加到每个构造函数参数中，此参数将在child和Parents中分配。我确实打算用art:Artister填充基类。第二个解决方案可以无缝地工作。非常感谢。很高兴能帮忙@mortezaT你是对的，这意味着前四个论点没有公开。我不知道如果你把Streetarist投给艺术家，然后取一个名字，会发生什么，例如，他们会是一样的吗？它隐藏了基本产权？值得注意的是第二种解决方案。您基本上是使用相同的信息创建两个艺术家对象。来自c#world，这对我来说似乎有点不合适，但它是有效的：）对于typescript，也许我会在art参数中使用一个接口，并使用一个对象初始值设定项作为接口后面的

{name:'aoue'，…}

。@CodyBugsteina据我所知。看起来有人要求它或多或少像Java规范一样隐式。