如何在Typescript中访问基类的属性？

Question

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();
alert(b.method());

它甚至得到了9票.但是当你将它粘贴在官方TS操场 http://www.typescriptlang.org/Playground/上时, 它会给你和错误.

如何从B访问A的x属性？

Answer 1

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());