Question

结构或多或少;

[   
    {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: []}
]

我希望在children数组不为空时将children id替换为名称。

因此查询的结果预期为;

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

我做了什么来实现这一目标;

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

用其父母填充儿童，这不是我想要的：）

{ "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" } ] }

我误解了什么？

Answer 1

在使用$lookup阶段之前，您应该为子数组使用$unwind，然后为子项使用$lookup。在$lookup阶段之后，您需要使用$group来获取带有名称的儿童数组，而不是 id

你可以尝试一下：

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

为什么使用$group？

例如：你的第一份文件

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

在$unwind文档看起来像

之后

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

在$lookup

之后

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

然后在$group

之后

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

Answer 2

使用当前的Mongo 3.4版本，您可以使用$graphLookup。

$maxDepth设置为0进行非递归查找。您可能希望在查找之前添加$match阶段以过滤没有子项的记录。

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

$ aggregation和$查找同一个集合 - mongodb

2 个答案: