Question

我有一个我想要聚合的集合，使特定字段不同，但也将其他字段数据添加到该聚合结果中。以下是我的收藏示例

   {
        "_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（＆＃34; mappedTracks＆＃34;））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，但我还想将其所属文档的_id，userID，artistname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，这假设您现在想要所有这四个字段的明确值组合。如果您只想在mappedTracks上区分，则必须决定如何处理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$group之后重命名某些字段，请参阅聚合运算符$project。

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

在mongodb中将一些数据添加到不同的查询中

1 个答案: