我有2个集合student_details和subject_details,其中每个学生可以将多个主题存储在student_details集合中作为参考数组。
现在,我需要获取学生详细信息以及已过滤的主题,其中subject_details.status = ACTIVE。
我该如何实现$elemMatch
对$ref
对象的使用。我正在使用类似下面的内容,但是它没有返回任何记录。
db.getCollection('student_details').find( { subjects: { $elemMatch: { $ref: "subject_details", status: 'ACTIVE' }}})
student_details
================
{
"_id" : "STD-1",
"name" : "XYZ",
"subjects" : [
{
"$ref" : "subject_details",
"$id" : "SUB-1"
},
{
"$ref" : "subject_details",
"$id" : "SUB-2"
},
{
"$ref" : "subject_details",
"$id" : "SUB-3"
}
]
}
subject_details
===============
{
"_id" : "SUB-1",
"name" : "MATHEMATICS",
"status" : "ACTIVE"
}
{
"_id" : "SUB-2",
"name" : "PHYSICS",
"status" : "ACTIVE"
}
{
"_id" : "SUB-3",
"name" : "CHEMISTRY",
"status" : "INACTIVE"
}
在查询中使用dbref时很麻烦。但是您可以使用以下聚合管道解决此问题:
db.student_details.aggregate([
{
$unwind: "$subjects"
},
{
$set: {
"fk": {
$arrayElemAt: [{
$objectToArray: "$subjects"
}, 1]
}
}
},
{
$lookup: {
"from": "subject_details",
"localField": "fk.v",
"foreignField": "_id",
"as": "subject"
}
},
{
$match: {
"subject.status": "ACTIVE"
}
},
{
$group: {
"_id": "$_id",
"name": {
$first: "$name"
},
"subjects": {
$push: {
$arrayElemAt: ["$subject", 0]
}
}
}
}
])
结果对象将如下所示:
{
"_id": "STD-1",
"name": "XYZ",
"subjects": [
{
"_id": "SUB-1",
"name": "MATHEMATICS",
"status": "ACTIVE"
},
{
"_id": "SUB-2",
"name": "PHYSICS",
"status": "ACTIVE"
}
]
}
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