Question

嗨，我有一个userAnswer schema，看起来像这样

var userAnswer = mongoose.Schema({
'questionId' : { 'type' : mongoose.Schema.Types.ObjectId, 'ref': 'Question'},
'userId' : { 'type': mongoose.Schema.Types.ObjectId, 'ref' : 'User'},
'score' : 1 
})

然后是一个userAttempt schema，其中包含与此问题有关的尝试

var userAttempt = mongoose.Schema({
'userAnswerId' : { 'type' : mongoose.Schema.Types.ObjectId, 'ref' : 'UserAnswer'},
'attemptData' : [{ 
    'userInputMCQ' : String,
    'time' : Date 
    }],
})

最后一个usersnapshots schema保存有关attemptData的{{1}}字段中每个元素的快照

看起来像这样

userAttempt Schema

现在，我想以var userAttemptSnapshot= mongoose.Schema({ userAttemptDataId : {type: mongoose.Schema.Types.ObjectId }, count : {type: Number, default: 1}, userData :[ { value: String, timeOfCreation : { type : Date, default : Date.now } }] }, { timestamps: true } );字段data中的所有question的方式提取与documents有关的所有attempData

UserAttempt Schema

使用'attemptData' : [{ 'userInputMCQ' : String, 'time' : Date }],作为单独的attempt和其各自的snapshots来获取

我的aggregation是

aggregation query

编辑： 我的db.useranswers.aggregate([ { $match : { 'questionId' : ObjectId("5a2dfefd6dc3de029e488961"),'userId' :ObjectId("5a2cea6d6dc3de029e488814") }}, { '$lookup' : {from: "userattempts", localField: "_id", foreignField: "userAnswerId", as: "attempts"}}, { '$unwind' : '$attempts'}, {'$unwind' : '$attempts.attemptData'} //after this stage i will use lookup to get snapshot data but i dont get those attempts as individual documents so that i could use lookup here })收藏是

userAnswer

{ "_id":ObjectId("5b4f359244781a45bcc98fcb") , "userId":ObjectId("5a2cea6d6dc3de029e488814"), "questionId": ObjectId("5a631937e2509a3288171c37"), }集合是

UserAttempt

我想要这样的输出

{
    "_id":ObjectId("5b4f359204a7251ebcabef24"),
    "attemptData":[
    {
        "_id" : ObjectId("5b4f359204a7251ebcabef23"),
        "userInputMCQ" : "0"
        "timeOfCreation" : ISODate("2018-07-18T18:11:54.855+05:30"),
    },
    {
        "_id" : ObjectId("5b4f35bb04a7251ebcabef27"),
        "userInputMCQ" : "0",
        "timeOfCreation" : ISODate("2018-07-18T18:12:35.765+05:30"),
    } ]
}

UserAttemptSnapshot collection is
{
    "_id":ObjectId("5b4f359204a7251ebcabef25"),
    "userAttemptId":ObjectId("5b4f359204a7251ebcabef23"),
    "count" : 2,
    "userData" : [ { 
                     'value' : "Some String" ,
                     "_id":ObjectId("5b4f359204a7251ebcabef26")  
                   },
                   { 
                     'value' : "Some String" ,
                     "_id": ObjectId("5b4f359204a7251ebcabef27")  
                   } ]
},
{
    "_id":ObjectId("5b4f359204a7251ebcabef25"),
    "userAttemptId":ObjectId("5b4f359204a7251ebcabef27"),
    "count" : 1,
    "userData" : [ { 
                     'value' : "Some String" ,
                     "_id":ObjectId("5b4f359204a7251ebcabeaee")  
                   },
                 ]
}

Answer 1

您可以在3.6中使用以下汇总。

db.useranswers.aggregate([ 
{"$lookup":{
    "from": "userattempts",
    "let": {"id":"$_id"},
    "pipeline":[
      {"$match":{"$expr":{"$eq":["$$id","$userAnswerId"]}}},
      {"$unwind":"$attemptData"},
      {"$lookup":{
        "from": "userattemptsnapshots",
        "localField": "attemptData._id",
        "foreignField":"userAttemptId",
        "as": "attemptData.attemptsnapshot"
      }},
      {"$unwind":"$attemptData.attemptsnapshot"},
      {"$group":{
         "_id":"$userAnswerId",
         "attemptData":{
            "$push":"$attemptData"
         }
      }}
    ],
    "as": "attempts"
 }},
{"$unwind":"$attempts"},
])

更新3.4版本

db.useranswers.aggregate([ 
{"$lookup":{
    "from": "userattempts",
    "localField": "_id",
    "foreignField":"userAnswerId",
    "as": "attempts"
}},
{"$unwind":"$attempts"},
{"$unwind":"$attempts.attemptData"},
{"$lookup":{
    "from": "userattemptsnapshots",
    "localField": "attempts.attemptData._id",
    "foreignField":"userAttemptId",
    "as": "attempts.attemptData.attemptsnapshot"
}},
{"$unwind":"$attempts.attemptData.attemptsnapshot"}
])

如何展开进一步包含数组的数组字段

1 个答案: