如何使用MongoDB Aggregate进行排序？
使用aggregate()排序，在MongoDB中使用$sort。让我们创建一个包含文档的集合-
> db.demo164.insertOne({"StudentAge":24});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e36883d9e4f06af551997c8")
}
> db.demo164.insertOne({"StudentAge":25});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e3688409e4f06af551997c9")
}
> db.demo164.insertOne({"StudentAge":22});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e3688429e4f06af551997ca")
}
> db.demo164.insertOne({"StudentAge":21});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e3688439e4f06af551997cb")
}在find()方法的帮助下显示集合中的所有文档-
> db.demo164.find();
这将产生以下输出-
{ "_id" : ObjectId("5e36883d9e4f06af551997c8"), "StudentAge" : 24 }
{ "_id" : ObjectId("5e3688409e4f06af551997c9"), "StudentAge" : 25 }
{ "_id" : ObjectId("5e3688429e4f06af551997ca"), "StudentAge" : 22 }
{ "_id" : ObjectId("5e3688439e4f06af551997cb"), "StudentAge" : 21 }这是使用MongoDB的查询aggregate()-
> db.demo164.aggregate([{ $sort : { StudentAge : -1 } }]);这将产生以下输出-
{ "_id" : ObjectId("5e3688409e4f06af551997c9"), "StudentAge" : 25 }
{ "_id" : ObjectId("5e36883d9e4f06af551997c8"), "StudentAge" : 24 }
{ "_id" : ObjectId("5e3688429e4f06af551997ca"), "StudentAge" : 22 }
{ "_id" : ObjectId("5e3688439e4f06af551997cb"), "StudentAge" : 21 }