聚合使用两个子 object arrays MongoDB

Question

在不同的 collections object 中有两个 arrays。 我想在一个 object 指定和常见的 uuid 中从两个 arrays 获取所有字段。

这是来自两个不同对象的两个 arrays：

{
    "_id" : ObjectId("5a1ea4a4a1a13eaecf571267"),
    "storage" : "events",
    "list" : [
        {
            "uuid" : "5a03c1e0e31a11e7b8c2e398bcd9f882",
            "type" : "TYPE",
            "desc" : "DESC",
            "when" : 1513934100000,
            "loc" : "LOC",
            "schedule" : [ ]
        },
        {
            "uuid" : "1b45a340e70911e7b8c2e398bcd9f882",
            "type" : "TYPE1",
            "desc" : "DESC1",
            "when" : 1514624400000,
            "loc" : "LOC1",
            "schedule" : []
        }
        }
}
{
    "_id" : ObjectId("5a297001840b2aba87a5b1eb"),
    "storage" : "control",
    "list" : [
        {
            "uuid" : "5a03c1e0e31a11e7b8c2e398bcd9f882",
            "missing" : [
                {
                    "user" : "user",
                    "instrument" : "inst"
                },
                {
                    "user" : "user",
                    "instrument" : "inst"
                }
            ],
            "when" : 1513934100000
        },
                {
            "uuid" : "1b45a340e70911e7b8c2e398bcd9f882",
            "missing" : [
                {
                    "user" : "user",
                    "instrument" : "inst"
                },
                {
                    "user" : "user",
                    "instrument" : "inst"
                }
            ],
            "when" : 1514624400000
        },
         }
}

期望的结果是：（因为给定的 uuid 是 5a03c1e0e31a11e7b8c2e398bcd9f882）

{
    "list" : [
        {
            "uuid" : "5a03c1e0e31a11e7b8c2e398bcd9f882",
            "type" : "TYPE",
            "desc" : "DESC",
            "when" : 1513934100000,
            "loc" : "LOC",
            "schedule" : [ ],
            "missing" : [
                {
                    "user" : "user",
                    "instrument" : "inst"
                },
                {
                    "user" : "user",
                    "instrument" : "inst"
                }
           ]
        }
  }

问题是如果没有两个匹配的 uuid，则结果必须是 null。 我有多个 $list，所以展开时必须有两个具有相同 uuid 的对象，如果不是，结果必须是 null。

Answer 1

以下查询可以得到我们预期的 output：

db.collection.aggregate([
    {
        $unwind:"$list"
    },
    {  
        $group:{
            "_id":"$list.uuid",
            "list":{
                $push:"$list"
            }
        }
    },
    {
       $project:{
            "merged":{
                $reduce:{
                    "input":"$list",
                    "initialValue":{},
                    "in":{
                        $mergeObjects:["$$this","$$value"]
                    }
                }
            }
        }
    },
    {
        $group:{
            "_id":null,
            "list":{
                $push: "$merged"
            }
        }
    },
    {
        $project:{
            "_id":0,
            "storage" : "result",
            "list":1
        }
    }
]).pretty()

数据集：

{
    "_id" : ObjectId("5a1ea4a4a1a13eaecf571267"),
    "storage" : "events",
    "list" : [
        {
            "uuid" : "5a03c1e0e31a11e7b8c2e398bcd9f882",
            "type" : "TYPE",
            "desc" : "DESC",
            "when" : 1513934100000,
            "loc" : "LOC",
            "schedule" : [ ]
        },
        {
            "uuid" : "1b45a340e70911e7b8c2e398bcd9f882",
            "type" : "TYPE1",
            "desc" : "DESC1",
            "when" : 1514624400000,
            "loc" : "LOC1",
            "schedule" : [ ]
        }
    ]
}
{
    "_id" : ObjectId("5a297001840b2aba87a5b1eb"),
    "storage" : "control",
    "list" : [
        {
            "uuid" : "5a03c1e0e31a11e7b8c2e398bcd9f882",
            "missing" : [
                {
                    "user" : "user",
                    "instrument" : "inst"
                },
                {
                    "user" : "user",
                    "instrument" : "inst"
                }
            ],
            "when" : 1513934100000
        },
        {
            "uuid" : "1b45a340e70911e7b8c2e398bcd9f882",
            "missing" : [
                {
                    "user" : "user",
                    "instrument" : "inst"
                },
                {
                    "user" : "user",
                    "instrument" : "inst"
                }
            ],
            "when" : 1514624400000
        }
    ]
}

Output：

{
    "list" : [
        {
            "uuid" : "1b45a340e70911e7b8c2e398bcd9f882",
            "missing" : [
                {
                    "user" : "user",
                    "instrument" : "inst"
                },
                {
                    "user" : "user",
                    "instrument" : "inst"
                }
            ],
            "when" : 1514624400000,
            "type" : "TYPE1",
            "desc" : "DESC1",
            "loc" : "LOC1",
            "schedule" : [ ]
        },
        {
            "uuid" : "5a03c1e0e31a11e7b8c2e398bcd9f882",
            "missing" : [
                {
                    "user" : "user",
                    "instrument" : "inst"
                },
                {
                    "user" : "user",
                    "instrument" : "inst"
                }
            ],
            "when" : 1513934100000,
            "type" : "TYPE",
            "desc" : "DESC",
            "loc" : "LOC",
            "schedule" : [ ]
        }
    ],
    "storage" : "result"
}

查询分析：我们首先展开list数组，然后根据list.uuid进行分组。 每个组包含具有相同 UUID 的所有列表元素的集合。 稍后，列表元素合并为一个。

Answer 2

`db.a9e34aa77f3474fbab5e9827942fbdd.aggregate([
    {
        $project: {
            'list': {
                $filter: {
                    input: '$list',
                    as: 'item',
                    cond: {
                        $eq: ['$$item.uuid', '15d055a0d5d711e8bd45bff08b1fb980']
                    }
                }
            }
        }
    },
    {
        $unwind:"$list"
    },
    {
        $group:{
            "_id":"$list.uuid",
            "list":{
                $push:"$list"
            }
        }
    },
    {
        $project:{
            "merged":{
                $reduce:{
                    "input":"$list",
                    "initialValue":{},
                    "in":{
                        $mergeObjects:["$$this","$$value"]
                    }
                }
            }
        }
    },
    {
        $group:{
            "_id":null,
            "list":{
                $push: "$merged"
            }
        }
    },
    {
        $project:{
            "_id":0,
            "list":1
        }
    }
]).pretty()`