以下方法使用aggregation framework來提出更接近所需輸出的解決方案。這取決於第三個集合，可以看作是兩個集合survey和trackings之間的合併。
首先，假設你有一個測試文件以下集合根據你的問題的例子：
// survey collection
db.survey.insert({
_id: 1111,
name: "name",
questions: [
{_id: 1, text: "a,b, or c?", type: "multipleChoice", options: ["a", "b", "c",]},
{_id: 2, text: "what do you think", type: "freeform"}
],
participants: [{_id: 1, name: "user 1"}, {_id: 2, name: "user 2"}],
results: [{_id: 123, userId: 1, questionId: 1, answer: "a"},
{_id: 124, userId: 2, questionId: 1, answer: "b"},
{_id: 125, userId: 1, questionId: 2, answer: "this is some answer"},
{_id: 126, userId: 2, questionId: 2, answer: "this is another answer"}]
})
// trackings collection
db.trackings.insert([
{
_id:1,
surveyId: 1111,
userId: 1,
starttime: "2015-05-13 10:46:20.347Z",
endtime: "2015-05-13 10:59:20.347Z"
},
{
_id:2,
surveyId: 1111,
userId: 2,
starttime: "2015-05-13 10:13:06.176Z",
endtime: "2015-05-13 10:46:28.176Z"
}
])
要創建第三集合（讓叫它output_collection），你會需要遍歷在trackings集合上使用find()遊標的forEach()方法，將帶有日期字符串的字段轉換爲實際的ISODate對象，創建一個存儲survey結果的數組字段，然後將合併的對象保存到第三個集合中。下面演示了此操作：
db.trackings.find().forEach(function(doc){
var survey = db.survey.find({"_id": doc.surveyId}).toArray();
doc.survey = survey;
doc["starttime"] = ISODate(doc.starttime);
doc["endtime"] = ISODate(doc.endtime);
db.output_collection.save(doc);
});
兩個集合融入output_collection，與db.output_collection.findOne()查詢會產生後：
{
"_id" : 1,
"surveyId" : 1111,
"userId" : 1,
"starttime" : ISODate("2015-05-13T10:46:20.347Z"),
"endtime" : ISODate("2015-05-13T10:59:20.347Z"),
"survey" : [
{
"_id" : 1111,
"name" : "name",
"questions" : [
{
"_id" : 1,
"text" : "a,b, or c?",
"type" : "multipleChoice",
"options" : [
"a",
"b",
"c"
]
},
{
"_id" : 2,
"text" : "what do you think",
"type" : "freeform"
}
],
"participants" : [
{
"_id" : 1,
"name" : "user 1"
},
{
"_id" : 2,
"name" : "user 2"
}
],
"results" : [
{
"_id" : 123,
"userId" : 1,
"questionId" : 1,
"answer" : "a"
},
{
"_id" : 124,
"userId" : 2,
"questionId" : 1,
"answer" : "b"
},
{
"_id" : 125,
"userId" : 1,
"questionId" : 2,
"answer" : "this is some answer"
},
{
"_id" : 126,
"userId" : 2,
"questionId" : 2,
"answer" : "this is another answer"
}
]
}
]
}
還可以再敷在此集合聚集。聚合流水線應由四個運算符級組成，這些運算符級從輸入文檔解構數組併爲每個元素輸出文檔。每個輸出文檔用一個元素值替換數組。
下一$project操作器級重塑每個文檔的流中，例如通過增加一個新字段duration，其計算在所述開始時間和結束時間日期字段之間分鐘的時間差，並使用Arithmetic Operators做計算。
在此之後是操作者$group流水線級，其由"survey"密鑰組輸入文件，並且使用所述accumulator expression(s)到每個組。消耗所有輸入文檔併爲每個不同的組輸出一個文檔。
所以你聚集的管道應該是這樣的：
db.output_collection.aggregate([
{ "$unwind": "$survey" },
{ "$unwind": "$survey.questions" },
{ "$unwind": "$survey.participants" },
{ "$unwind": "$survey.results" },
{
"$project": {
"survey": 1,
"surveyId": 1,
"userId": 1,
"starttime": 1,
"endtime": 1,
"duration": {
"$divide": [
{ "$subtract": [ "$endtime", "$starttime" ] },
1000 * 60
]
}
}
},
{
"$group": {
"_id": "$surveyId",
"survey": { "$first": "$survey.name"},
"totalAverageTime": {
"$avg": "$duration"
},
"fastestTime": {
"$min": "$duration"
},
"slowestTime": {
"$max": "$duration"
},
"questions": {
"$addToSet": "$survey.questions"
},
"answers": {
"$addToSet": "$survey.results"
}
}
},
{
"$out": "survey_results"
}
])
db.survey_results.find()輸出
/* 0 */
{
"result" : [
{
"_id" : 1111,
"survey" : "name",
"totalAverageTime" : 23.18333333333334,
"fastestTime" : 13,
"slowestTime" : 33.36666666666667,
"questions" : [
{
"_id" : 2,
"text" : "what do you think",
"type" : "freeform"
},
{
"_id" : 1,
"text" : "a,b, or c?",
"type" : "multipleChoice",
"options" : [
"a",
"b",
"c"
]
}
],
"answers" : [
{
"_id" : 126,
"userId" : 2,
"questionId" : 2,
"answer" : "this is another answer"
},
{
"_id" : 124,
"userId" : 2,
"questionId" : 1,
"answer" : "b"
},
{
"_id" : 125,
"userId" : 1,
"questionId" : 2,
"answer" : "this is some answer"
},
{
"_id" : 123,
"userId" : 1,
"questionId" : 1,
"answer" : "a"
}
]
}
],
"ok" : 1
}
UPDATE
一旦獲得一個ggregation輸出到另一個集合，通過$out聚合管道說survey_results，然後你可以用find()光標forEach()方法運用一些本地JavaScript函數，以便得到最終的目標：
db.survey_results.find().forEach(function(doc){
var questions = [];
doc.questions.forEach(function(q){
var answers = [];
doc.answers.forEach(function(a){
if(a.questionId === q._id){
delete a.questionId;
answers.push(a);
}
});
q.answers = answers;
questions.push(q);
});
delete doc.answers;
doc.questions = questions;
db.survey_results.save(doc);
});
輸出：
/* 0 */
{
"_id" : 1111,
"survey" : "name",
"totalAverageTime" : 23.18333333333334,
"fastestTime" : 13,
"slowestTime" : 33.36666666666667,
"questions" : [
{
"_id" : 2,
"text" : "what do you think",
"type" : "freeform",
"answers" : [
{
"_id" : 126,
"userId" : 2,
"answer" : "this is another answer"
},
{
"_id" : 125,
"userId" : 1,
"answer" : "this is some answer"
}
]
},
{
"_id" : 1,
"text" : "a,b, or c?",
"type" : "multipleChoice",
"options" : [
"a",
"b",
"c"
],
"answers" : [
{
"_id" : 124,
"userId" : 2,
"answer" : "b"
},
{
"_id" : 123,
"userId" : 1,
"answer" : "a"
}
]
}
]
}
我有一個解決方案，涉及創建另一個輸出集合，將兩個模式連接起來，然後使用聚合框架計算所需的聚合。但是，如果您可以指定正在使用的MongoDB版本，這將會非常有幫助，因爲這會影響聚合操作，因爲它需要使用某些在更高版本中找到的運算符。 – chridam
我們目前在使用mongodb 3.8和mongoose 4 – jonnie
這是一個錯字，mongodb 3.8嗎？ – chridam