我正在嘗試撰寫一個查詢來獲取存盤在 mongo 中的一些調查資料的所有結果。棘手的部分是有些問題是單選題,有些問題是多選型別的問題,有些是需要平均的值,所以我想根據問題的型別執行不同的聚合。
結果存盤在這樣的模式中,陣列中的每個專案都是調查回應。
[
{
metaData: {
survey: new ObjectId("62206ea0b31be3535abac547")
},
answers: {
'question1': 'a',
'question2': 'a',
'question3': ['a','c'],
'question4': 3
},
createdAt: 2022-03-03T07:30:40.517Z,
},
{
metaData: {
survey: new ObjectId("62206ea0b31be3535abac547"),
},
answers: {
'question1': 'a',
'question2': 'b',
'question3': ['a','c'],
'question4': 2
},
createdAt: 2022-03-03T07:30:40.518Z,
},
{
metaData: {
survey: new ObjectId("62206ea0b31be3535abac547"),
},
answers: {
'question1': 'b',
'question2': 'c',
'question3': ['b']
'question4': 1
},
createdAt: 2022-03-03T07:30:40.518Z,
}
]
question1 和 question2 是單選題,所以只能有 1 個答案,而問題 3 是多選題,所以用戶可以有多個答案。問題 4 是一個需要平均的值。
我認為有一些方法可以在單個聚合管道中通過一些方面、分組、過濾器、投影等的組合來實作這一點,但我被卡住了。
我想得到一個看起來像這樣的最終結果
{
'question1' : {
'a' : 2,
'b' : 1
},
'question2' : {
'a' : 1,
'b' : 1,
'c' : 1,
},
'question3' : {
'a' : 2,
'b' : 1,
'c' : 2,
},
'question4' : 2 //avg (3 2 1)/3
}
或者更好:
{
'radio': {
'question1' : {
'a' : 2,
'b' : 1
},
'question2' : {
'a' : 1,
'b' : 1,
'c' : 1,
},
},
'multi': {
'question3' : {
'a' : 2,
'b' : 1,
'c' : 2,
}
},
'avg' : {
'question4' : 2
}
}
我的管道看起來像這樣:
Response.aggregate([
{ $match: { 'metaData.survey': surveyId} }, // filter only for the specific survey
{ $project: { // I assume I have to turn the answers into an array
"answers": { $objectToArray: "$answers" },
"createdAt": "$createdAt"
}
},
// maybe facet here?
// conceptually, In the next stage I'd want to bucket the questions
// by type with something like below, then perform the right type of
// aggregation depending on the question type
// if $in [$$answers.k ['question1, 'question2']] group by k, v and count
// if $in [$$answers.k ['question3']] unwind and count each unique value?
// { $facet : { radio: [], multi:[]}}
])
基本上,我知道哪個問題 ID 是單選或多選,我只是想弄清楚如何格式化管道以根據問題 ID 在已知陣列中實作所需的輸出。
如果我能弄清楚如何根據 createdAt 時間對按天/月進行分組,則可以加分
uj5u.com熱心網友回復:
db.collection.aggregate([
{
$match: {}
},
{
$project: { answers: { $objectToArray: "$answers" } }
},
{
$unwind: "$answers"
},
{
$unwind: "$answers.v"
},
{
$group: {
_id: "$answers",
c: { "$sum": 1 }
}
},
{
$group: {
_id: "$_id.k",
v: { "$push": { k: "$_id.v", v: "$c" } }
}
},
{
$group: {
_id: null,
v: { "$push": { k: "$_id", v: { "$arrayToObject": "$v" } } }
}
},
{
$set: { v: { $arrayToObject: "$v" } }
},
{
$replaceWith: "$v"
}
])
mongoplayground
db.collection.aggregate([
{
$match: {}
},
{
$project: { answers: { $objectToArray: "$answers" } }
},
{
$unwind: "$answers"
},
{
$set: {
"answers.type": {
$switch: {
branches: [
{
case: { $isArray: "$answers.v" },
then: "multi"
},
{
case: { $eq: [ { $type: "$answers.v" }, "string" ] },
then: "radio"
},
{
case: { $isNumber: "$answers.v" },
then: "avg"
}
],
default: "other"
}
}
}
},
{
$unwind: "$answers.v"
},
{
$group: {
_id: "$answers",
c: { $sum: 1 }
}
},
{
$group: {
_id: "$_id.k",
type: { $first: "$_id.type" },
v: {
$push: {
k: { $toString: "$_id.v" },
v: "$c"
}
}
}
},
{
$group: {
_id: "$type",
v: {
$push: {
k: "$_id",
v: { $arrayToObject: "$v" }
}
}
}
},
{
$group: {
_id: null,
v: {
$push: {
k: "$_id",
v: { $arrayToObject: "$v" }
}
}
}
},
{
$set: { v: { $arrayToObject: "$v" } }
},
{
$replaceWith: "$v"
},
{
$set: {
avg: {
$arrayToObject: {
$map: {
input: { $objectToArray: "$avg" },
as: "s",
in: {
k: "$$s.k",
v: {
$avg: {
$map: {
input: { $objectToArray: "$$s.v" },
as: "x",
in: { $multiply: [ { $toInt: "$$x.k" }, "$$x.v" ] }
}
}
}
}
}
}
}
}
}
])
mongoplayground
轉載請註明出處,本文鏈接:https://www.uj5u.com/caozuo/439670.html
