在 mongodb 中将一些数据添加到不同的查询中

Question

我有一个要聚合的集合，使特定字段与众不同，但还将其他字段数据添加到该聚合结果中。下面是我收藏的一个例子

   {
        "_id": ObjectId("57d6bc99e4b014fc13cf9579"),
        "_class": "hbservlet.FinalSubmissions",
        "active": true,
        "mappedTracks": 
        [
            { "position": 0, "title": "01. Ain't No Time.mp3", "url": "/published/music/tr157d6bc99e4b014fc13cf9579_.aac" }, 
            { "position": 1, "title": "02. In Her Mouth.mp3", "url": "/published/music/tr257d6bc99e4b014fc13cf9579_.aac" }, 
            { "position": 2, "title": "03. Maybach.mp3", "url": "/published/music/tr357d6bc99e4b014fc13cf9579_.aac" }
        ],
        "createdBy": ObjectId("57d6bb99b17ee01a5427af08"),
        "userId": ObjectId("57d6bb99b17ee01a5427af08"),
        "artistname": "test",
    }, {
        "_id": ObjectId("14d6bc99ebc942fc13cf9579"),
        "_class": "hbservlet.FinalSubmissions",
        "active": false,
        "mappedTracks": [
            { "position": 0, "title": "partysong.mp3", "url": "/published/music/tr114d6bc99ebc942fc13cf9579_.aac" }, 
            { "position": 1, "title": "outside", "url": "/published/music/tr214d6bc99ebc942fc13cf9579_.aac" },],
        "createdBy": ObjectId("57d6bb99b17ee01a5427af08"),
        "userId": ObjectId("57d6bb99b17ee01a5427af08"),
        "artistname": "mynameismyname",
    }

我使用不同的查询 (db.published.distinct("mappedTracks")) 来收集所有的 mappedTracks，所以我得到了这个

{ "position": 0, "title": "01. Ain't No Time.mp3", "url": "/published/music/tr157d6bc99e4b014fc13cf9579_.aac" }, 
{ "position": 1, "title": "02. In Her Mouth.mp3", "url": "/published/music/tr257d6bc99e4b014fc13cf9579_.aac" }, 
{ "position": 2, "title": "03. Maybach.mp3", "url": "/published/music/tr357d6bc99e4b014fc13cf9579_.aac" },
{ "position": 0, "title": "partysong.mp3", "url": "/published/music/tr114d6bc99ebc942fc13cf9579_.aac" }, 
{ "position": 1, "title": "outside", "url": "/published/music/tr214d6bc99ebc942fc13cf9579_.aac" }

这是我想要的结果，但我还想将它所属的文档的_id、用户ID、艺术家名添加到创建的新对象中。

Answer 1

您可以尝试MongoDB aggregation 运算符$unwind 和$group。例如：

db.collection.aggregate([
          {$unwind:"$mappedTracks"}, 
          {$group:{_id:
                       {mappedTracks:"$mappedTracks", 
                        id:"$_id", 
                        userId:"$userId", 
                        artistname:"$artistname"
                        }
                   }
           }
])

请注意，这是假设您现在想要所有这四个字段的不同值组合。如果您只想区分mappedTracks，您必须决定如何处理userId 和artistname 的重复值。如果是这种情况，请参阅$first 运算符以在重复的情况下使用第一个值。

如果您想在$group 之后重命名某些字段，另请参阅聚合运算符$project。

如果此查询是您用例的常用查询，我建议您重新考虑您的Data Modelling or Document Structure。