如何将 object（类实例属性）传播到父级的超级构造函数？

Question

如何将 class 属性传播到父级的超级构造函数？

ActionLog 是一个基础 class，它在 ActionRequestAccess 内部的方法中实例化

ActionRequestAccess.ts

export class ActionRequestAccess extends ActionLog {
  constructor(public actionLog: ActionLog, public customerId: string) {
    super(
      actionLog.id, // Want to get rid of these assignments <<< and switch to something like: ...actionLog
      actionLog.type,
      actionLog.date,
      actionLog.address,
      actionLog.location
    );
  }

  static override fromMap(map: any) {
    if (!map) {
      return null;
    }
    const baseMap = super.fromMap(map);
    if (!baseMap) {
      return null;
    }
    return new ActionRequestAccess(baseMap, map['customerId'] ?? null);
  }
}

动作日志.ts

import { ActionType } from '../../enums';

export class ActionLog {
  constructor(
    public id: string,
    public type: ActionType,
    public date: Date,
    public address: string,
    public location: string
  ) {}

  static fromMap(map: any) {
    if (!map) {
      return null;
    }
    return new ActionLog(
      map['id'] ?? null,
      map['type'] ?? null,
      map['date'] ?? null,
      map['address'] ?? null,
      map['location'] ?? null
    );
  }
}

Answer 1

您可以在ActionRequestAccess内的 object 上使用Object.values 。

 constructor(public actionLog: ActionLog, public customerId: string) {
    super(...Object.values(actionLog));
  }

但这意味着您的 object 的值的顺序传递给 function 数学，所以这种模式有点不安全。 这将起作用

const myObject = new ActionRequestAccess({
    id: 1,
    type: 'myType',
    date: 'myDate',
    adress: 'myAdress',
    location: 'myLocation',
});

但这会切换 id 和 type 的 position 并且不起作用。

const myObject = new ActionRequestAccess({
    type: 'myType',
    id: 1,
    date: 'myDate',
    adress: 'myAdress',
    location: 'myLocation',
});

如果您可以将ActionLog的构造函数的预期值更改为 object 而不是参数，那么这可能会更容易和更安全。