Question

我有一个库存类。在那个类中，我有2个函数，其中包含一系列项目。我想抓住两个阵列中的所有项目，然后将它们组合成一个单个数组，然后我用它来过滤掉项目。

class Inventory {

    private _lions = [];
    private _wolves = [];
    private _allAnimals: Array<any>  = [];

    getAllLions(): void {
        const lions: Array<ILion> = [
            { id: 1, name: 'Joffrey', gender: Gender.male, age: 20, price: 220, species: Species.lion, vertrabrates: true, warmBlood: true, hair: 'Golden', runningSpeed: 30, makeSound() { } },
            { id: 2, name: 'Tommen', gender: Gender.male, age: 18, price: 230, species: Species.lion, vertrabrates: true, warmBlood: true, hair: 'Golden', runningSpeed: 30, makeSound() { } },
            { id: 3, name: 'Marcella', gender: Gender.female, age: 24, price: 180, species: Species.lion, vertrabrates: true, warmBlood: true, hair: 'Golden', runningSpeed: 30, makeSound() { } },
        ];

        for (let lion of lions) {
            this._lions.push(lion);
        }        
    }

    getAllWolves(): void {
        const wolves: Array<IWolf> = [
            { id: 1, name: 'Jon', gender: Gender.male, price: 130, species: Species.wolf, age: 13, vertrabrates: true, warmBlood: true, hair: 'Grey', runningSpeed: 30, makeSound() { } },
            { id: 2, name: 'Robb', gender: Gender.male, price: 80, species: Species.wolf, age: 18, vertrabrates: true, warmBlood: true, hair: 'Black', runningSpeed: 30, makeSound() { } },
            { id: 3, name: 'Sansa', gender: Gender.female, price: 10, species: Species.wolf, age: 35, vertrabrates: true, warmBlood: true, hair: 'Grey', runningSpeed: 30, makeSound() { } },
            { id: 4, name: 'Arya', gender: Gender.female, price: 190, species: Species.wolf, age: 8, vertrabrates: true, warmBlood: true, hair: 'White', runningSpeed: 30, makeSound() { } },
        ];

        for (let wolf of wolves) {
            this._wolves.push(wolf);
        }
    }

    getAllAnimals(): void {
        this._allAnimals = this._lions.concat(this._wolves);
    };

    findByTemplate(template: any): Array<any>  {
        return this._allAnimals.filter(animal => {
            return Object.keys(template).every(propertyName => animal[propertyName] === template[propertyName]);
        });
    }
}

对狮子和狼的数组进行循环操作，但我宁愿在该字段中推送整个数组。但后来我在数组中推出一个数组，导致过滤函数出现问题。

是否可以将lion数组推入_lion字段，而无需在数组中插入数组？

Answer 1

使用spread运算符，您可以使用push执行此操作，例如：

this._wolves.push(...wolves);

Answer 2

似乎：

this._wolves = this._wolves.concat(wolves);

是一个很好的解决方案，使用concat可以抓取数组中的所有对象，将它们与另一个数组合并。但是如果你不这样做，它只会返回当前数组中的对象。

将对象推入数组

2 个答案: