如何在MongoDB中汇总集合和按字段计数分组？
在MongoDB中aggregate()，使用$group和聚合集合。让我们创建一个包含文档的集合-
> db.demo616.insertOne({"details":{"Name":"Chris","Age":21}});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e99bfac65492f6c60d00283")
}
> db.demo616.insertOne({"details":{"Name":"Chris","Age":22}});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e99bfb065492f6c60d00284")
}
> db.demo616.insertOne({"details":{"Name":"Bob","Age":23}});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e99bfb865492f6c60d00285")
}
> db.demo616.insertOne({"details":{"Name":"Sam","Age":21}});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e99bfbd65492f6c60d00286")
}
> db.demo616.insertOne({"details":{"Name":"Chris","Age":24}});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e99bfc165492f6c60d00287")
}在find()方法的帮助下显示集合中的所有文档-
> db.demo616.find();
这将产生以下输出-
{ "_id" : ObjectId("5e99bfac65492f6c60d00283"), "details" : { "Name" : "Chris", "Age" : 21 } }
{ "_id" : ObjectId("5e99bfb065492f6c60d00284"), "details" : { "Name" : "Chris", "Age" : 22 } }
{ "_id" : ObjectId("5e99bfb865492f6c60d00285"), "details" : { "Name" : "Bob", "Age" : 23 } }
{ "_id" : ObjectId("5e99bfbd65492f6c60d00286"), "details" : { "Name" : "Sam", "Age" : 21 } }
{ "_id" : ObjectId("5e99bfc165492f6c60d00287"), "details" : { "Name" : "Chris", "Age" : 24 } }以下是汇总集合和按字段计数分组的查询-
> db.demo616.aggregate(
... [
... {
... $group:{_id:"$details.Name",Total:{$sum:1}}}
... ]
... );这将产生以下输出-
{ "_id" : "Sam", "Total" : 1 }
{ "_id" : "Bob", "Total" : 1 }
{ "_id" : "Chris", "Total" : 3 }热门推荐
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