我有如下嵌套的数据,MongoDB的总
{
"_id" : ObjectId("5a30ee450889c5f0ebc21116"),
"academicyear" : "2017-18",
"fid" : "be02",
"fname" : "ABC",
"fdept" : "Comp",
"degree" : "BE",
"class" : "1",
"sem" : "8",
"dept" : "Comp",
"section" : "Theory",
"subname" : "BDA",
"fbValueList" : [
{
"_id" : ObjectId("5a30eecd3e3457056c93f7af"),
"score" : 20,
"rating" : "Fair"
},
{
"_id" : ObjectId("5a30eefd3e3457056c93f7b0"),
"score" : 10,
"rating" : "Fair"
},
{
"_id" : ObjectId("5a337e53341bf419040865c4"),
"score" : 88,
"rating" : "Excellent"
},
{
"_id" : ObjectId("5a337ee2341bf419040865c7"),
"score" : 75,
"rating" : "Very Good"
},
{
"_id" : ObjectId("5a3380b583dde50ddcea350e"),
"score" : 72,
"rating" : "Very Good"
}
]
},
{
"_id" : ObjectId("5a3764f1bc19b77dd9fd9a57"),
"academicyear" : "2017-18",
"fid" : "be02",
"fname" : "ABC",
"fdept" : "Comp",
"degree" : "BE",
"class" : "1",
"sem" : "5",
"dept" : "Comp",
"section" : "Theory",
"subname" : "BDA",
"fbValueList" : [
{
"_id" : ObjectId("5a3764f1bc19b77dd9fd9a59"),
"score" : 88,
"rating" : "Excellent"
},
{
"_id" : ObjectId("5a37667aee64bce1b14747d2"),
"score" : 74,
"rating" : "Good"
},
{
"_id" : ObjectId("5a3766b3ee64bce1b14747dc"),
"score" : 74,
"rating" : "Good"
}
]
}
我们正在尝试使用它进行聚合,
db.fbresults.aggregate([{$match:{academicyear:"2017-18",fdept:'Comp'}},{$group:{_id: {fname: "$fname", rating:"$fbValueList.rating"},count: {"$sum":1}}}])
,我们得到结果一样,
{ "_id" : { "fname" : "ABC", "rating" : [ "Fair","Fair","Excellent","Very Good", "Very Good", "Excellent", "Good", "Good" ] }, "count" : 2 }
但我们期待的结果是,
{ "_id" : { "fname" : "ABC", "rating_group" : [
{
rating: "Excellent"
count: 2
},
{
rating: "Very Good"
count: 2
},
{
rating: "Good"
count: 2
},
{
rating: "Fair"
count: 2
},
] }, "count" : 2 }
我们希望通过他们的名字获得个别教职人员小组,并根据他们的评级回应和评级数量在该小组内进行评估。
我们已经试过这一个,但我们没有结果。 Mongodb Aggregate Nested Group
您正在使用哪个mongo版本? –