深 mongodb 聚合以在对象数组中创建具有公共字段的组？答案

【问题标题】：deep mongodb aggregate to make Groups with the common fields in array of objects?深 mongodb 聚合以在对象数组中创建具有公共字段的组？
【发布时间】：2020-07-31 13:11:59
【问题描述】：

我想使用 mongodb 聚合来创建组。我想在我的项目中实现这一点，但陷入了困境。没有找到更好的方法来做到这一点。


{
    "_id" : ObjectId("5e9a21868ed974259c0da402"),
    "shopId" : "5e975cc7be7c1b546b7abb17",
    "shopType" : "Medium Store",
        "products": [{
            "isPackedProduct" : true,
            "_id" : ObjectId("5e92ff706af877294d63098e"),
            "brand" : "ABC",
            "category" : "CAT1",
            "productName" : "P1",
            "subCategory" : "SUB1",
        },
        {
            "isPackedProduct" : true,
            "_id" : ObjectId("5e92ff706af877294d63098f"),
            "brand" : "EFG",
            "category" : "CAT1",
            "productName" : "P2",
            "subCategory" : "SUB2",
        },
        {
            "isPackedProduct" : true,
            "_id" : ObjectId("5e92ff706af84d630977298f"),
            "brand" : "EFG",
            "category" : "CAT2",
            "productName" : "P3",
            "subCategory" : "SUB1",
        }
        ....
     ]    
}

从这组 json 中我想将数据显示为：


   {
     "_id" : ObjectId("5e9a21868ed974259c0da402"),
     "shopId" : "5e975cc7be7c1b546b7abb17",
     "CAT1":{
       "SUB1":{
          "products": [{
            ...ALL the Products which have CAT1 and SUB1
          }]
        }
     },
     "CAT2":{
       "SUB1":{
          "products": [{
            ...ALL the Products which have CAT2 and SUB1
          }]
        }
     }
     ...
   }

到目前为止，我尝试了但没有接近解决方案：

db.shopproducts.aggregate([{$unwind: {path: '$products'}}, {$group: {_id: 'products.category'}}, {$project: {'products.category': 1, 'products.productName': 1}}])

此外，如果有更好的方法可以在不使用聚合的情况下执行此操作，那么欢迎提出建议。

提前致谢。

【问题讨论】：

标签： node.js mongodb mongoose aggregate

【解决方案1】：

我们需要应用几个$group 阶段。要将products.category 和products.subCategory 转换为对象字段，我们需要使用$arrayToObject 运算符。

[                                        {
  { "k" : "CAT1", "v" : "SUB1" }, ----\     "CAT1" : "SUB1",
  { "k" : "CAT2", "v" : "SUB1" }  ----/     "CAT2" : "SUB1"
]                                        }

试试这个：

db.shopproducts.aggregate([
  {
    $unwind: "$products"
  },
  {
    $group: {
      _id: {
        _id: "$_id",
        shopId: "$shopId",
        category: "$products.category",
        subCategory: "$products.subCategory"
      },
      products: {$push: "$products"}
    }
  },
  {
    $group: {
      _id: {
        _id: "$_id._id",
        shopId: "$_id.shopId",
        category: "$_id.category"
      },
      products: {
        $push: {
          k: "$_id.subCategory",
          v: {products: "$products"}
        }
      }
    }
  },
  {
    $group: {
      _id: {
        _id: "$_id._id",
        shopId: "$_id.shopId"
      },
      products: {
        $push: {
          k: "$_id.category",
          v: {$arrayToObject: "$products"}
        }
      }
    }
  },
  {
    $replaceRoot: {
      newRoot: {
        $mergeObjects: [
          {
            _id: "$_id._id",
            shopId: "$_id.shopId"
          },
          {
            $arrayToObject: "$products"
          }
        ]
      }
    }
  }
])

MongoPlayground

【讨论】：

【解决方案2】：

db.shopproducts.aggregate([
  {
    $unwind: {
      path: "$products"
    }
  },
  {
    $group: {
      _id: {
        cat: "$products.category",
        sub_cat: "$products.subCategory"
      },
      products: {
        $addToSet: "$products"
      }
    }
  }
])

我猜这就是你想要的，不是吗？

Mongoplayground

【讨论】：