Javascript 如何为Mongoose创建多个过滤

我有一个在线游戏store 和代码，可以通过标签和流派过滤数据.但我还想添加按平台和系统的过滤.例如，系统是PC，平台是Steam.我该如何实现这一点？

Schema个

const mongoose = require("mongoose");
const slug = require('mongoose-slug-updater');
const options = {
    separator: "-",
    lang: "en",
    truncate: 120
}
mongoose.plugin(slug, options);
const GamesSchema = new mongoose.Schema({
    title: String,
    slug: { type: String, slug: ["title"], slugPaddingSize: 2, unique: true },
    last_url: String,
    description: String,
    preview: {
        src: String,
        alt: String
    },
    stock_price: String,
    discount: Number,
    total_price: String,
    system: Array,
    platform: String,
    info: {
        Genre: Array
    },
    tags: Array,
    config: Object,
    images: Array,
    keys: Array,
    content: String
})
module.exports = mongoose.model("Games", GamesSchema);

api controller

async get(req, res) {
        try {
            const page = parseInt(req.query.page) - 1 || 0;
            const limit = parseInt(req.query.limit) || 5;
            const search = req.query.search || '';
            let sort = req.query.sort || "top";
            let tags = req.query.tags || 'all';
            const tagsOpt = [];
            Games.find().then(async games => {
                games.forEach(game => {
                    game.tags.forEach(tag => {
                        if (!tagsOpt.includes(tag)) {
                            tagsOpt.push(tag)
                        }
                    })
                })

                tags === 'all' ? (tags = [...tagsOpt]) : (tags = req.query.tags.split(','));
                req.query.sort ? (sort = req.query.sort.split(',')) : (sort = [sort]);

                let sortBy = {};
                if (sort[1]) {
                    sortBy[sort[0] = sort[1]];
                } else {
                    sortBy[sort[0] = "top"]
                }
                const gamesData = await Games.find({ title: { $regex: search, $options: 'i' } })
                    .where('tags')
                    .in([...tags])
                    .sort(sortBy)
                    .skip(page * limit)
                    .limit(limit);

                const total = await Games.countDocuments({
                    tags: { $in: [...tags] },
                    title: { $regex: search, $options: 'i' },
                });
                const response = {
                    error: false,
                    total,
                    page: page + 1,
                    limit,
                    tags: tagsOpt,
                    gamesData
                };
                res.status(200).json(response);
            })
        } catch (error) {
            console.log(error);
            res.status(400).json({ message: 'Ошибка получения' })
        }
    }

我能够将流派和标签结合在一起并根据它们进行过滤，现在我真的需要将平台和系统都放入一组标签中吗？