在MongoDB中的所有子文档中获取一对不同的对象？
要获得一对不同的对象，请使用$组。让我们创建一个包含文档的集合-
> db.demo522.insertOne({"Name":"John","Score":45});{
"acknowledged" : true,
"insertedId" : ObjectId("5e89b646b3fbf26334ef611b")
}
> db.demo522.insertOne({"Name":"Bob","Score":67});{
"acknowledged" : true,
"insertedId" : ObjectId("5e89b64eb3fbf26334ef611c")
}
> db.demo522.insertOne({"Name":"John","Score":55});{
"acknowledged" : true,
"insertedId" : ObjectId("5e89b655b3fbf26334ef611d")
}
> db.demo522.insertOne({"Name":"Bob","Score":33});{
"acknowledged" : true,
"insertedId" : ObjectId("5e89b65cb3fbf26334ef611e")
}在find()方法的帮助下显示集合中的所有文档-
> db.demo522.find();
这将产生以下输出-
{ "_id" : ObjectId("5e89b646b3fbf26334ef611b"), "Name" : "John", "Score" : 45 }
{ "_id" : ObjectId("5e89b64eb3fbf26334ef611c"), "Name" : "Bob", "Score" : 67 }
{ "_id" : ObjectId("5e89b655b3fbf26334ef611d"), "Name" : "John", "Score" : 55 }
{ "_id" : ObjectId("5e89b65cb3fbf26334ef611e"), "Name" : "Bob", "Score" : 33 }以下是查询以获取MongoDB中所有子文档的一对不同的对象-
> var query = [
... {
... "$group": {
... "_id": "$Name",
... "Score": { "$sum": "$Score" }
... }
... },
... {
... "$project": {
... "Name": "$_id", "_id": 0, "Score": 1
... }
... }
... ];
>
> db.demo522.aggregate(query);这将产生以下输出-
{ "Score" : 100, "Name" : "Bob" }
{ "Score" : 100, "Name" : "John" }