$ aggregation和$查找同一個集合 - mongodb

Question

結構或多或少像;

[   
    {id: 1, name: "alex" , children: [2, 4, 5]},
    {id: 2, name: "felix", children: []},
    {id: 3, name: "kelly", children: []},
    {id: 4, name: "hannah", children: []},
    {id: 5, name: "sonny", children: [6]},
    {id: 6, name: "vincenzo", children: []}
]

當children數組不為空時，我想用名字替換children ID。

所以查詢的結果是預期的;

[   {id: 1, name: "alex" , children: ["felix", "hannah" , "sonny"]}
    {id: 5, name: "sonny", children: ["vincenzo"]}
]

我做了什么來實現這一目標;

db.list.aggregate([
  {$lookup: { from: "list", localField: "id", foreignField: "children", as: "children" }},
  {$project: {"_id" : 0, "name" : 1, "children.name" : 1}},
])

用它的父母填充孩子，這不是我想要的:)

{ "name" : "alex", "parent" : [ ] }
{ "name" : "felix", "parent" : [ { "name" : "alex" } ] }
{ "name" : "kelly", "parent" : [ ] }
{ "name" : "hannah", "parent" : [ { "name" : "alex" } ] }
{ "name" : "sonny", "parent" : [ { "name" : "alex" } ] }
{ "name" : "vincenzo", "parent" : [ { "name" : "sonny" } ] }

我誤解了什么？

Answer 1

在使用$lookup階段之前，你應該使用$unwind for children數組然后$lookup for children。 在$lookup階段之后，你需要使用$group來獲取帶有name而不是id的 children數組

你可以試試：

db.list.aggregate([
    {$unwind:"$children"},
    {$lookup: { 
        from: "list",
        localField: "children",
        foreignField: "id",
        as: "childrenInfo" 
      }
    },
    {$group:{
       _id:"$_id",
       children:{$addToSet:{$arrayElemAt:["$childrenInfo.name",0]}},
       name:{$first:"$name"}
      }
    }
]);

// can use $push instead of $addToSet if name can be duplicate

為何使用$group ？

例如：您的第一份文件

{id: 1, name: "alex" , children: [2, 4, 5]}

在$unwind你的文件會是這樣的

{id: 1, name: "alex" , children: 2},
{id: 1, name: "alex" , children: 4},
{id: 1, name: "alex" , children: 5}

在$lookup

{id: 1, name: "alex" , children: 2,
  "childrenInfo" : [ 
        {
            "id" : 2,
            "name" : "felix",
            "children" : []
        }
    ]},
//....

然后在$group

 {id: 1, name: "alex" , children: ["felix", "hannah" , "sonny"]}

Answer 2

使用當前的Mongo 3.4版本，您可以使用$graphLookup 。

對於非遞歸查找， $maxDepth設置為0 。 您可能希望在查找之前添加$match階段以過濾沒有子項的記錄。

db.list.aggregate([{
    $graphLookup: {
        from: "list",
        startWith: "$children",
        connectFromField: "children",
        connectToField: "id",
        as: "childrens",
        maxDepth: 0,
    }
}, {
    $project: {
        "_id": 0,
        "name": 1,
        "childrenNames": "$childrens.name"
    }
}]);