javascript - mongodb - 汇总和展开外国引用文件

Question

因此对于我的示例数据库设置:

db.lists.insertMany([
    { _id: "1", name: "list1", included_lists: ["2"], items: ["i1"] },
    { _id: "2", name: "list2", included_lists: [], items: ["i2", "i3"] }
])


db.items.insertMany([
    { _id: "i1", name: "item1", details: [{}, {}, {}] },
    { _id: "i2", name: "item2", details: [{}, {}, {}] },
    { _id: "i3", name: "item3", details: [{}, {}, {}] }
])

我目前通过以下方式获取我的项目数据:

db.lists.aggregate([
    { "$match": { "_id": { "$in": ["1", "2"] } } },
    {
        "$lookup": {
            "from": "items",
            "localField": "items",
            "foreignField": "_id",
            "as": "item"
        }
    },
    { "$unwind": "$item" },
    {
        "$facet": {
            "results": [
                { "$skip": 0 },
                { "$limit": 10 },
                {
                    "$project": {
                        name: 1,
                        item: 1
                    }
                }
            ],
            "total": [
                { "$count": "total" },
            ]
        }
    }
]).pretty()

返回:

{
    "results" : [
        {
            "_id" : "1",
            "name" : "list1",
            "item" : {
                "_id" : "i1",
                "name" : "item1",
                "details" : [
                    {

                    },
                    {

                    },
                    {

                    }
                ]
            }
        },
        {
            "_id" : "2",
            "name" : "list2",
            "item" : {
                "_id" : "i2",
                "name" : "item2",
                "details" : [
                    {

                    },
                    {

                    },
                    {

                    }
                ]
            }
        },
        {
            "_id" : "2",
            "name" : "list2",
            "item" : {
                "_id" : "i3",
                "name" : "item3",
                "details" : [
                    {

                    },
                    {

                    },
                    {

                    }
                ]
            }
        }
    ],
    "total" : [
        {
            "total" : 3
        }
    ]
}

我想做的是删除 { "$match": { "_id": { "$in": ["1", "2"] } } }, 因为我想删除获取 id 数组所需的查询，而只是从列表 _id 及其 included_lists id 中获取所有 id。然后像我的结果一样返回所有 items。

这个问题类似于:mongodb - unwinding nested subdocuments但由于歧义和缺少数据库文件，我已经重新询问了。

Answer 1

你可以通过图形查找然后分组来做到这一点

db.lists.aggregate([
    { "$match": { "_id": { "$in": ["1"] } } },
    {
      $graphLookup: {
      from: "lists",
      startWith: "$_id" ,
      connectFromField: "included_lists",
      connectToField: "_id",
      as: "connected",
   }
        },
  
    {$unwind:"$connected"},
    { $group:{_id:"$connected._id",items:{$first:'$connected.items'},name:{$first:'$connected.name'}}},
     {
        "$lookup": {
            "from": "items",
            "localField": "items",
            "foreignField": "_id",
            "as": "item"
        }
    },
    { "$unwind": "$item" },
    {
        "$facet": {
            "results": [
                { "$skip": 0 },
                { "$limit": 10 },
                {
                    "$project": {
                        name: 1,
                        item: 1
                    }
                }
            ],
            "total": [
                { "$count": "total" },
            ]
        }
    }
    
    

   
]).pretty()