如何在子构造函数中使用 ES6 super 调用父方法？

Question

情况：

我正在扩展 Node.js (v. 8.4.0) 带有附加属性（时间戳、id）的错误对象，然后扩展此对象以获得更精细的错误处理。

class MyError extends Error {
  constructor (msg) {
    super(msg);
    this.id = uuid();
    this.timestamp = Date.now();
    // I reckon this can be replaced by this.init(this) ?
    this.name = this.constructor.name;
    Error.captureStackTrace && Error.captureStackTrace(this, this.constructor);
  }

  init (self) {
    self.name = self.constructor.name;
    Error.captureStackTrace && Error.captureStackTrace(self, self.constructor);
  }
}

我希望不会在子错误中重复Error.captureStackTrace 和this.name 调用。所以我创建了一个 init 函数，我在孩子中使用这样的函数：

class GranularError extends MyError {
  constructor (msg) {
    super(msg);
    this.type = "error";
    this.status = 500;
    this.code = "internalServerError";
    super.init(this);
  }
}

GranularError 然后将再次扩展以获取 MoreGranularError 等。这就是为什么我想让它保持 DRY。

问题：

当 GranularError 或 MoreGranularError 被抛出时，它会失败并显示一个

TypeError: (intermediate value).init is not a function

我主要阅读了以下资料，但我无法将它们应用于问题。任何帮助表示赞赏。

Call parent function which is being overridden by child during constructor chain in JavaScript(ES6)

Parent constructor call overridden functions before all child constructors are finished

http://2ality.com/2015/02/es6-classes-final.html#referring_to_super-properties_in_methods

Answer 1

我不知道你得到这个错误是什么，但没有必要创建一个init 函数。 this.name 和 Error.captureStack 的内容也将在子实例中起作用，因为 this 指的是子实例。

换句话说，您正在尝试解决一个不存在的问题。

class MyError extends Error {
  constructor (msg) {
    super(msg);
    this.id = Math.random();
    this.timestamp = Date.now();
    this.name = this.constructor.name;
    Error.captureStackTrace && Error.captureStackTrace(this, this.constructor);
  }
}
class GranularError extends MyError {
  constructor (msg) {
    super(msg);
    this.type = "error";
    this.status = 500;
    this.code = "internalServerError";
  }
}

console.dir(new GranularError("this is the error message"));