在数组angular 5中添加两种不同类型的对象

Question

I have two different arrays, heroes: Hero[] and monsters: Monster[]. They share a common field, named totalInitiative. These two arrays need to be put into the same array and sorted on their totalInitiative.

The goal I'm trying to achieve is something like this:

Array[hero1, hero2, hero3, monster1, monster2] 

I created a superclass called Participant:

import {Participant} from './participant';

export class Hero extends Participant{
id: number;
name: string;
player: string;
hitPoints: number;
armor: number;
initModif: number;
imageUrl: string;
totalInitiave: number;
}

import {Participant} from './participant';

export class Monster extends Participant{
id:number;
name: string;
hitPoints: number;
armor: number;
initModif: number;
imageUrl: string;
}

export class Participant{

}

I did not add the common fields in Participant because I have a Hero and a Monster component where I need those common properties to add a new Hero/Monster.

Now I need to adjust my Encounter model so it consists of a Participant[] which holds the Hero[] and Monster[]

import {Hero} from './hero';
import {Monster} from './monster';
import {Participant} from './participant';

export class Encounter {
id: number;
name: string;
participants: Participant[ Hero[] Monster[]]; //Doesn't work
}

I'm not even sure that this is the correct way?

Answer 1

The type of your array needs to be a union of Hero and Monster :

const participants: (Hero | Monster)[] = [];

Here is a cut down example with a rudimentary sort...

class Hero {
  constructor(public initiative: number) { };
}

class Monster {
  constructor(public initiative: number) { };
  evil = 5;
}

const heroes = [
  new Hero(5),
  new Hero(3)
];

const monsters = [
  new Monster(2),
  new Monster(7)
];

const participants: (Hero | Monster)[] = heroes.concat(monsters);

const sorted = participants.sort((a, b) => a.initiative - b.initiative);

console.log(sorted);

Answer 2

Hero和Monster也是Participant ，因此您可以简单地说participants是一组Participant元素。

participants: Participant[] = [];