简化 MongoDB 聚合

Question

我正在使用此类聚合按深层嵌套字段 ObjectId 对所有产品进行排序。

首先我填充了 catalogProduct 字段。

然后在 catalogProduct 中填充类别。

按类别 ID 对所有数据进行排序（return product if ids arr includes category._id）

倒序排序，return页，分页限制为8。

然后在没有分页和限制的情况下获取所有已排序产品的总数。

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

我得到这样的数据：

[
  { 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数，但我不知道如何。

Answer 1

您可以观察开始阶段，直到 $match by catalogProduct.category._id 在 2 $facet 中重复。因此，你可以简单地将它们分解出来，然后将后面的阶段分别放入$facet。

以下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"},
        ]
        }},
]);

简化 MongoDB 聚合

Simplify MongoDB aggregation

javascript

database

query-optimization

mongoose

mongodb