Question

使用MongoDB聚合框架时是否可以合并数组字段？以下是我要解决的摘要问题：

汇总的示例输入文档：

{
  "Category" : 1,
  "Messages" : ["Msg1", "Msg2"],
  "Value" : 1
},
{
  "Category" : 1,
  "Messages" : [],
  "Value" : 10
},
{
  "Category" : 1,
  "Messages" : ["Msg1", "Msg3"],
  "Value" : 100
},
{
  "Category" : 2,
  "Messages" : ["Msg4"],
  "Value" : 1000
},
{
  "Category" : 2,
  "Messages" : ["Msg5"],
  "Value" : 10000
},
{
  "Category" : 3,
  "Messages" : [],
  "Value" : 100000
}

我们希望在总结“价值”和合并“消息”的同时按“类别”进行分组。我试过这个聚合管道：

{group : {
        _id : "$Category",
        Value : { $sum : "$Value"},
        Messages : {$push : "$Messages"}
    }
}, 
{$unwind : "$Messages"}, 
{$unwind : "$Messages"}, 
{$group : {
        _id : "$_id",
        Value : {$first : "$Value"},
        Messages : {$addToSet : "$Messages"}
    }
}

结果是：

"result" : [{
        "_id" : 1,
        "Value" : 111,
        "Messages" : ["Msg3", "Msg2", "Msg1"]
    }, 
    {
        "_id" : 2,
        "Value" : 11000,
        "Messages" : ["Msg5", "Msg4"]
    }
]

然而，这完全错过了类别3，因为“类别”为3的文档没有任何“消息”，并且它们在第二次展开时被删除。我们希望结果包括以下内容：

{
    "_id" : 3,
    "Value" : 100000,
    "Messages" : []
}

聚合框架是否有一种巧妙的方法来实现这一目标？

Answer 1

如果保证Messages是一个数组，那么你可以使用这个技巧：

> db.messages.find()
    { "Category" : 1, "Messages" : [  "Msg1",  "Msg2" ], "Value" : 1 }
    { "Category" : 1, "Messages" : [ ], "Value" : 10 }
    { "Category" : 1, "Messages" : [  "Msg1",  "Msg3" ], "Value" : 100 }
    { "Category" : 2, "Messages" : [  "Msg4" ], "Value" : 1000 }
    { "Category" : 2, "Messages" : [  "Msg5" ], "Value" : 10000 }
    { "Category" : 3, "Messages" : [ ], "Value" : 100000 }

> var group1 = {
    "$group":   {
        "_id":      "$Category",
        "Value":    {
            "$sum":     "$Value"
        },
        "Messages": {
            "$push":    "$Messages"
        }
    }
};

> var project1 = {
    "$project": {
        "Value":    1,
        "Messages": {
            "$cond":    [
                {
                    "$eq":  [
                        "$Messages",
                        [ [ ] ]
                    ]
                },
                [ [ null ] ],
                "$Messages"
            ]
        }
    }
};

> db.messages.aggregate( group1, project1 )
    { "_id" : 3, "Value" : 100000, "Messages" : [  [  null ] ] }
    { "_id" : 2, "Value" : 11000, "Messages" : [  [  "Msg4" ],  [  "Msg5" ] ] }
    { "_id" : 1, "Value" : 111, "Messages" : [  [  "Msg1",  "Msg2" ],  [ ],  [  "Msg1",  "Msg3" ] ] }

现在展开两次并重新分组以获得单个Messages数组。

> var unwind = {"$unwind":"$Messages"};

> var group2 = {
    $group: {
        "_id":      "$_id", 
        "Value":    {
            "$first":       "$Value"
        }, 
        "Messages": {
            "$addToSet":    "$Messages"
        }
    }
};

> var project2 = {
    "$project": {
        "Category": "$_id",
        "_id":      0,
        "Value":    1,
        "Messages": {
            "$cond":    [
                {
                    "$eq":  [
                        "$Messages",
                        [ null ]
                    ]
                },
                [ ],
                "$Messages"
            ]
        }
    }
};

> db.messages.aggregate(group1, project1, unwind, unwind, group2 ,project2 )
    { "Value" : 111, "Messages" : [  "Msg3",  "Msg2",  "Msg1" ], "Category" : 1 }
    { "Value" : 11000, "Messages" : [  "Msg5",  "Msg4" ], "Category" : 2 }
    { "Value" : 100000, "Messages" : [ ], "Category" : 3 }

Answer 2

正如其中一条评论中已经提到的，原始问题的最简单答案是将preserveNullAndEmptyArrays添加到$ unwind阶段。

合并MongoDB聚合中的数组字段

2 个答案: