MongoDB聚合框架与组查询示例？
为此，请在MongoDB聚合中使用$group。让我们创建一个包含文档的集合-
> db.demo639.insertOne(
... {
... "_id" : 1,
... "CountryName" : "US",
... "Info1" : {
... "Name" : "Chris",
... "SubjectName" : "MySQL",
... "Marks" : 78
... },
... "Info2" : {
... "Name" : "Chris",
... "SubjectName" : "MySQL",
... "Marks" : 78
... }
... }
... );
{ "acknowledged" : true, "insertedId" : 1 }
> db.demo639.insertOne(
... {
... "_id" : 2,
... "CountryName" : "UK",
... "Info1" : {
... "Name" : "Chris",
... "SubjectName" : "MySQL",
... "Marks" : 79
... },
... "Info2" : {
... "Name" : "Chris",
... "SubjectName" : "MySQL",
... "Marks" : 88
... }
... }
... );
{ "acknowledged" : true, "insertedId" : 2 }在find()方法的帮助下显示集合中的所有文档-
> db.demo639.find();
这将产生以下输出-
{ "_id" : 1, "CountryName" : "US", "Info1" : { "Name" : "Chris", "SubjectName" : "MySQL", "Marks" : 78 }, "Info2" : { "Name" : "Chris", "SubjectName" : "MySQL", "Marks" : 78 } }
{ "_id" : 2, "CountryName" : "UK", "Info1" : { "Name" : "Chris", "SubjectName" : "MySQL", "Marks" : 79 }, "Info2" : { "Name" : "Chris", "SubjectName" : "MySQL", "Marks" : 88 } }以下是使用$group查询的聚合框架查询-
> db.demo639.aggregate({$group : {_id : {Name1:"$Info1.Name",Name2:"$Info2.Name"},
...
... Marks:{$first:"$Info2.Marks"},
... Marks:{$sum:"$Info2.Marks"}
... }}).pretty();这将产生以下输出-
{ "_id" : { "Name1" : "Chris", "Name2" : "Chris" }, "Marks" : 166 }