简化MongoDB聚合-前端问题

Simplify MongoDB aggregation(简化MongoDB聚合)

本文介绍了简化MongoDB聚合的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我正在使用这种聚合按深度嵌套字段ObjectID对所有产品进行排序。

首先我填充CatalogProduct字段。

然后填充CatalogProduct内的类别。

按类别ID对所有数据排序(如果ID arr包含类别._id，则返回product)

按相反顺序排序，返回分页并按8限制分页。

然后获取没有分页和限制的所有已排序产品的总计数。

const sortedProducts = await StorageModel.aggregate([
                // Unite products arr and totalCount of sorted products
                {$facet: {
                    "sortedProducts": [
                        // populate catalogProduct ref by Id
                        { $lookup: {
                                from: "catalogs",
                                localField: "catalogProduct",
                                foreignField: "_id",
                                as: "catalogProduct"
                            } },
                        // deconstruct this Arr, because we get only one Object
                        { $unwind: "$catalogProduct" },
                        // populate category ref by Id inside catalogProduct object
                        { $lookup: {
                                from: "categories",
                                localField: "catalogProduct.category",
                                foreignField: "_id",
                                as: "catalogProduct.category"
                            } },
                        // deconstruct this Arr, because we get only one Object
                        { $unwind: "$catalogProduct.category" },
                        // returns product, if ids arr includes a catalogProduct.category._id
                        { $match: {
                                "catalogProduct.category._id": { $in: ids }
                            } },
                        // sort in reverse order
                        { $sort: { _id: -1 } },
                        // returns only *page
                        { $skip: (page - 1) * 8 },
                        /// limit result by 8
                        { $limit: 8 },
                    ],
                    // total count for pagination, the same operations
                    "totalCount": [
                        { $lookup: {
                                from: "catalogs",
                                localField: "catalogProduct",
                                foreignField: "_id",
                                as: "catalogProduct"
                            } },
                        { $unwind: "$catalogProduct" },
                        { $lookup: {
                                from: "categories",
                                localField: "catalogProduct.category",
                                foreignField: "_id",
                                as: "catalogProduct.category"
                            } },
                        { $unwind: "$catalogProduct.category" },
                        { $match: {
                                "catalogProduct.category._id": { $in: ids }
                            } },
                        // get total count of sorted data, without limit and pagination
                        {$count : "totalCount"},
                    ]
                    }},
            ]);
            
            products = sortedProducts[0].sortedProducts
            totalProducts = sortedProducts[0].totalCount.totalCount

我正在获取这样的数据：

[
  { sortedProducts: [ [Object], [Object] ], totalCount: [ [Object] ] }
]

而且很好。但我认为，聚合可以简化，我不需要重复操作来获得总计数，但我不知道如何进行。

推荐答案

可以观察起始阶段，直到2中重复$matchbycatalogProduct.category._id。因此，您可以简单地将它们提取出来，然后将后续阶段分别放入$facet。

下面是我建议的代码版本：

StorageModel.aggregate([
    { $lookup: {
        from: "catalogs",
        localField: "catalogProduct",
        foreignField: "_id",
        as: "catalogProduct"
    } },
    // deconstruct this Arr, because we get only one Object
    { $unwind: "$catalogProduct" },
    // populate category ref by Id inside catalogProduct object
    { $lookup: {
            from: "categories",
            localField: "catalogProduct.category",
            foreignField: "_id",
            as: "catalogProduct.category"
        } },
    // deconstruct this Arr, because we get only one Object
    { $unwind: "$catalogProduct.category" },
    // returns product, if ids arr includes a catalogProduct.category._id
    { $match: {
        "catalogProduct.category._id": { $in: ids }
    } },
    // Unite products arr and totalCount of sorted products
    {$facet: {
        "sortedProducts": [
            // populate catalogProduct ref by Id
            
            // sort in reverse order
            { $sort: { _id: -1 } },
            // returns only *page
            { $skip: (page - 1) * 8 },
            /// limit result by 8
            { $limit: 8 },
        ],
        // total count for pagination, the same operations
        "totalCount": [
            // get total count of sorted data, without limit and pagination
            {$count : "totalCount"},
        ]
        }},
]);

这篇关于简化MongoDB聚合的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！

问题描述

推荐答案

基础教程推荐