node.js - mongodb 中带有 _id 字段的不同集合

Question

我有一个这样的集合:

{
    "_id" : ObjectId("5d5aa6b5cced8d2cac81e7d7"),
    "stream" : "Programming",
    "coursename" : "java",
    "fee" : 50000,
    "date" : ISODate("2019-08-19T13:40:05.312Z"),
    "__v" : 0
}

{
    "_id" : ObjectId("5d5aa6d2cced8d2cac81e7d8"),
    "stream" : "Basic",
    "coursename" : "Msoffice",
    "fee" : 50000,
    "date" : ISODate("2019-08-19T13:40:34.560Z"),
    "__v" : 0
}

{
    "_id" : ObjectId("5d5aa6e0cced8d2cac81e7d9"),
    "stream" : "Basic",
    "coursename" : "Tally",
    "fee" : 80000,
    "date" : ISODate("2019-08-19T13:40:48.667Z"),
    "__v" : 0
}

{
    "_id" : ObjectId("5d5aa707cced8d2cac81e7da"),
    "stream" : "Programming",
    "coursename" : "C++",
    "fee" : 80000,
    "date" : ISODate("2019-08-19T13:41:27.556Z"),
    "__v" : 0
}

在这里您可以看到编程和基本流是重复的。我想区分流及其内部的 _id 和流字段。所以结果将是 2 个集合。例如:

{
    "_id" : ObjectId("5d5aa6b5cced8d2cac81e7d7"),
    "stream" : "Programming"
}
{
    "_id" : ObjectId("5d5aa6d2cced8d2cac81e7d8"),
    "stream" : "Basic"
}

distinct 仅返回数组内的stream。

Answer 1

使用 Mongodb 聚合 $group 和 $project 阶段将给出预期结果

db.exampleCollection.aggregate([{$group:{_id:"$stream",stream_id:{$first:"$_id"}}},{$project:{stream:"$_id",_id:'$stream_id'}}]);

结果示例 mongo_aggregation_result