如何正确覆盖打字稿中的方法？-解网

问：

为什么方法中是空的？this.fullNameshow()

class Person {
  protected name: string = "";
  constructor(name: string) {
    this.makeSir(name);
  }

  makeSir(name: string) {
    this.name = "sir" + name;  
  }
}

class Man extends Person {
  protected fullName = "";
  constructor(name: string) {
    super(name);
  }

  makeSir(name: string) {
    super.makeSir(name);
    this.fullName = "***" + name;
    console.log(this.fullName);//[LOG]: "***john" 
  }

  show() {
    console.log(this.fullName);//[LOG]: "" 
  }
}


const man = new Man("john");
man.show();

我应该如何解决这个问题？

TypeScript 继承这个

class Person  {
    
protected name: String ="";
    
    constructor(name: string){
        this.makeSir(name);
    }
    
    makeSir(name: string){
        this.name="sir"+name ;
    }
    
    test(){
        console.log("PERSON");
    }
}

class Man extends Person {
    public name: string ="";
    
    constructor(name:string){
        super(name);
        this.makeSir(name);
    }
    
    makeSir(name:string){
        this.name="####"+name;
    }
    
    test(){
        console.log("Man ")
    }
    show(){
        console.log("#####"+this.name);
    }
}

const n = new Man("Man ")
n.show();
n.test();


const m = new Person("Person");
m.test();

你只调用构造函数，而构造函数在 tern 中只调用超级构造函数。super

如果您看到测试被正确覆盖

输出：

#########Man 
Man 
PERSON

2赞 y2bd 3/12/2021 #2

super()在构造像 your 这样的派生类时，总是首先调用的。这包括在您执行的任何变量初始化之前，例如 .因此，当 u 被调用并设置为一个好的值时，紧接着你的空字符串赋值就会启动并清除其值。Manprotected fullName = ''Man.makeSirfullName

避免这种延迟覆盖的一种方法是不为设置初始值：fullName

// give it a type because we don't give it an initial value
// use `!` syntax to hint to TypeScript that this will be initialized during the constructor
// even though TS cannot figure that out itself
protected fullName!: string;

现在，由于您从未设置为“初始”值，因此您永远不会覆盖调用完成的工作。fullNamesuper()

如何正确覆盖打字稿中的方法？

How to properly override a method in typescript?

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