如何从打字稿中的实例方法访问静态成员？

Question

我尝试从实例方法使用静态成员。我知道accessing static member from non-static function in typescript，但我不想硬代码的类，允许继承：像

typeof this.PREFIX 

Answer 1

class Logger { 
    protected static PREFIX = '[info]'; 

    public log(msg: string) { 
    console.log(Logger.PREFIX + ' ' + msg); // What to use instead of Logger` to get the expected result? 
    } 
} 

class Warner extends Logger { 
    protected static PREFIX = '[warn]'; 
} 

(new Logger).log('=> should be prefixed [info]'); 
(new Warner).log('=> should be prefixed [warn]'); 

我试过的东西你只需要ClassName.property：

class Logger { 
    protected static PREFIX = '[info]'; 
    public log(message: string): void { 
    alert(Logger.PREFIX + string); 
    } 
} 

class Warner extends Logger { 
    protected static PREFIX = '[warn]'; 
} 

---

更多

从：http://basarat.gitbooks.io/typescript/content/docs/classes.html

打字稿类支持由该类的所有实例共享静态属性。一个自然的地方放（和访问），他们是在类本身，这就是打字稿做：

class Something { 
    static instances = 0; 
    constructor() { 
     Something.instances++; 
    } 
} 

var s1 = new Something(); 
var s2 = new Something(); 
console.log(Someting.instances); // 2 

---

UPDATE

如果你想让它从特定实例的构造继承使用this.constructor。可悲的是你需要使用某些类型的断言。我正在使用如下所示的typeof Logger：

class Logger { 
    protected static PREFIX = '[info]'; 
    public log(message: string): void { 
    var logger = <typeof Logger>this.constructor; 
    alert(logger.PREFIX + message); 
    } 
} 

class Warner extends Logger { 
    protected static PREFIX = '[warn]'; 
}