Question

使用关系数据库（SQL Server，MySQL，Oracle，Informix）28年后，我已转移到MongoDB。过去两周一直很缓慢。我想提出几个问题来证实我的想法。

我的文档如下所示（忽略此问题的分组）：

{
    "_id": "xyz-800",
    "site": "xyz",
    "user": 800,
    "timepoints": [
        {"timepoint": 0, "a": 1500, "b": 700},
        {"timepoint": 2, "a": 1000, "b": 200},
        {"timepoint": 4, "a": 3500, "b": 1500}
    ],
    "groupings": [
        {"type": "MNO", "group": "<10%", "raw": "1"},
        {"type": "IJK", "group": "Moderate", "raw": "23"}
    ]
}

我想展平嵌套数组的时间点。以下是有效的，但是有没有办法在时间点中对属性进行通配符而不是列出每个属性？原因可能是如果将新属性（例如，'c'）添加到子文档中，则必须修改代码，或者如果此子文档具有许多属性，我需要列出每个属性而不是使用通配符，如果可能的。

db.records.aggregate( {$unwind : "$timepoints"}, 
                      {$project: {_id: 1, site: 1, user: 1, 
                                  'timepoint': '$timepoints.timepoint', 
                                  'a': '$timepoints.a', 
                                  'b': '$timepoints.b'}})

结果：

{"id":"xyz-800", "site":"xyz", "user":800, "timepoint": 0, "a":1500, "b":700}
{"id":"xyz-800", "site":"xyz", "user":800, "timepoint": 2, "a":1000, "b":200}
{"id":"xyz-800", "site":"xyz", "user":800, "timepoint": 4, "a":3500, "b":1500}

我目前正在使用MongoDB 3.2

Answer 1

启动MongoDb 3.4，我们可以使用 $addFields 将顶级字段添加到嵌入式文档中，并使用 $replaceRoot < / strong>将嵌入式文档推广到顶级。

db.records.aggregate({ $unwind: "$timepoints" }, { $addFields: { "timepoints._id": "$_id", "timepoints.site": "$site", "timepoints.user": "$user" } }, { $replaceRoot: { newRoot: "$timepoints" } })

示例输出

{ "timepoint" : 0, "a" : 1500, "b" : 700, "_id" : "xyz-800", "site" : "xyz", "user" : 800 } { "timepoint" : 2, "a" : 1000, "b" : 200, "_id" : "xyz-800", "site" : "xyz", "user" : 800 } { "timepoint" : 4, "a" : 3500, "b" : 1500, "_id" : "xyz-800", "site" : "xyz", "user" : 800 }

MongoDB：聚合并展平数组字段

1 个答案: