MongoDB：聚合和展平数组字段

Question

在使用关系数据库（SQL Server、MySQL、Oracle、Informix）工作了 28 年后，我转向了 MongoDB。过去两周进展缓慢。我想提交几个问题来确认我的想法。

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"}
    ]
}

我想展平时间点嵌套数组。以下工作，但是有没有办法在时间点中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，我必须修改代码，或者如果该子文档有很多属性，我需要列出每个属性而不是使用通配符，如果可能。

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}

我目前使用的是 MongoDB 3.2

Answer 1

从 MongoDb 3.4 开始，我们可以使用 $addFields 将顶级字段添加到嵌入文档中，并使用 $replaceRoot 来将嵌入文档提升到顶层。

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

样本输出

{ "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 }