Inheritance 如何访问Typescript中基类的属性?
有人建议使用这样的代码Inheritance 如何访问Typescript中基类的属性?,inheritance,properties,typescript,superclass,Inheritance,Properties,Typescript,Superclass,有人建议使用这样的代码 class A { // Setting this to private will cause class B to have a compile error public x: string = 'a'; } class B extends A { constructor(){super();} method():string { return super.x; } } var b:B = new B(); a
class A {
// Setting this to private will cause class B to have a compile error
public x: string = 'a';
}
class B extends A {
constructor(){super();}
method():string {
return super.x;
}
}
var b:B = new B();
alert(b.method());
它甚至获得了9票。但是当你把它贴在官方的TS操场上
它给你带来了错误
如何从B访问A的x属性?使用
this
而不是super
:
class A {
// Setting this to private will cause class B to have a compile error
public x: string = 'a';
}
class B extends A {
// constructor(){super();}
method():string {
return this.x;
}
}
var b:B = new B();
alert(b.method());
冠军!抱歉,没有足够的声誉+1