引用与现有集合不同的集合并从同一集合计数

Question

我必须收集 A 和 B，其中 A 的文档包含对象 ID，这些对象 ID 存在于字段中心和 gcentre 的 B 中。我目前正在输出包含父中心名称的结果，方法是从集合 A 的对象 id 引用到 B，然后引用 gcentre 的 id 来查找子中心和中心，并计算通过 javascript 后处理分配的文档。刚接触聚合管道，我不知道如何通过对象 id 和那种记录计数来引用。聚合管道可以吗？我已经尝试使用 $lookup，但它似乎没有按预期提供输出。

集合 A 中的文档：

{
  "_id": {
        "$oid": "5bbafa98d8c77251bea30f8c"
    },
    "parentCentre": {
        "$oid": "1cbafa99d8c77251bea30f11"
    },
    "childCentre": {
        "$oid": "5cbafa99d8c77251bea30f8d"
    },
},
{
  "_id": {
        "$oid": "5bbafa98d8c77251bea30f8c"
    },
    "parentCentre": {
        "$oid": "1cbafa99d8c77251bea30f11"
    },
    "childCentre": {
        "$oid": "5cbafa99d8c77251bea30f8d"
    },
},
{
  "_id": {
        "$oid": "5bbafa98d8c77251bea30f8c"
    },
    "parentCentre": {
        "$oid": "1cbafa99d8c77251bea30f11"
    },
    "childCentre": {
        "$oid": "5cbafa99d8c77251bea30f8d"
    },
},
{
  "_id": {
        "$oid": "5bbafa98d8c77251bea30f8c"
    },
    "parentCentre": {
        "$oid": "1cbafa99d8c77251bea30f21"
    },
    "childCentre": {
        "$oid": "5cbafa99d8c77251bea30f6d"
    },
}

集合 B 中的文档：

   {
        "_id": {
                "$oid": "1cbafa99d8c77251bea30f11"
            },
            "Type": "Parent",
            "Name": "Kris Labs"
        },
    {
        "_id": {
                "$oid": "1cbafa99d8c77251bea30f21"
            },
            "Type": "Parent",
            "Name": "DEX Labs"
        },
    {
        "_id": {
                "$oid": "5cbafa99d8c77251bea30f8d"
            },
            "Type": "Child",
            "Name": "Mili Labs"
        },
        {
        "_id": {
                "$oid": "5cbafa99d8c77251bea30f6d"
            },
            "Type": "Child",
            "Name": "Max Labs"
        }

结果：

{
  "parentCentreName":"Kris Labs",
  "Records":{
            {
            childCentreName: "Max Labs",
            recordCount: 3
              }
             }
},
{
  "parentCentreName":"DEX Labs",
  "Records":{
            {
            childCentreName: "Mili Labs",
            recordCount: 1
              }
             }
}

Answer 1

您可以使用以下聚合查询：

db.A.aggregate([
  {
    $group: {
      _id: {
        p: "$parentCentre",
        c: "$childCentre"
      },
      count: {
        $sum: 1
      }
    }
  },
  {
    $group: {
      _id: "$_id.p",
      Records: {
        $push: {
          childCentreName: "$_id.c",
          recordCount: "$count"
        }
      }
    }
  },
  {
    $unwind: "$Records"
  },
  {
    "$lookup": {
      "from": "B",
      "localField": "_id",
      "foreignField": "_id",
      "as": "p"
    }
  },
  {
    "$lookup": {
      "from": "B",
      "localField": "Records.childCentreName",
      "foreignField": "_id",
      "as": "c"
    }
  },
  {
    $unwind: "$c"
  },
  {
    $unwind: "$p"
  },
  {
    $project: {
      "parentCentreName": "$p.Name",
      "Records.childCentreName": "$c.Name",
      "Records.recordCount": 1,
      _id: 0
    }
  },
  {
    $group: {
      _id: "$parentCentreName",
      "Records": {
        $push: "$Records"
      }
    }
  },
  {
    $project: {
      "parentCentreName": "$_id",
      "Records": 1,
      _id: 0
    }
  }
])

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