使用TypeScript超級（）

Question

我想在TypeScript中擴展一個類。我在編譯時收到這個錯誤：'提供的參數不匹配調用目標的任何簽名'。我已經嘗試在超級調用中引用artist.name屬性作爲超級（名稱），但不起作用。

您可能會有任何想法和解釋將不勝感激。謝謝 - 亞歷克斯。

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); 

Answer 1

超級調用必須提供基類的所有參數。構造函數不是繼承的。評論出藝術家，因爲我猜這樣做時不需要。

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，如果你想要的藝術參數來填充基本屬性，但在這種情況下，我想有是不是真的需要使用公共藝術作爲參數的屬性會被繼承和那隻存儲重複數據。

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}`); 
    } 
}