我可以从 MongoDb 中获得前 10 名的组列表吗，它是按成员数排序的

Question

我有 2 个集合，一个是组详细信息，另一个是 group_members 详细信息， group_members 已存储组的 _id（作为 FK）

赞组 : -> _id,name,type,date

group_members -> _id,group_id,members_id,角色,状态

那么我怎样才能得到其中拥有最多 memebrs 的组，并按 group_membrs 的计数排序（成员最多的前 10 个组）

Answer 1

因此，如果 groupMembers 集合包含您描述的组 ID，例如：

var r = [
 {_id: "M1", gid: "A", member: "Bob"}
 ,{_id: "M2", gid: "A", member: "Sally"}
 ,{_id: "M3", gid: "A", member: "Dan"}
 ,{_id: "M4", gid: "B", member: "Tess"}
 ,{_id: "M5", gid: "B", member: "George"}
 ,{_id: "M6", gid: "C", member: "P_1"}
 ,{_id: "M7", gid: "C", member: "P_2"}
 ,{_id: "M8", gid: "C", member: "P_3"}
 ,{_id: "M9", gid: "C", member: "P_4"}
 ];
 db.foo2.insert(r);

那么这就是您在每组中获得计数的方式：

c=db.foo2.aggregate([
  {$group: {_id: "$gid", n: {$sum:1} }}
  ,{$sort: {n: -1}}
  ,{$limit: 10}
]);

如果我们在群主合集里折腾，比如

var r = [
     {_id: "A", name: "GroupA"}
     ,{_id: "B", name: "GroupB"}
     ,{_id: "C", name: "GroupC"}
];
db.foo.insert(r);

然后我们简单地对"join"数据做一个$lookup：

db.foo2.aggregate([
  {$group: {_id: "$gid", n: {$sum:1} }}
  ,{$sort: {n: -1}}
  ,{$limit: 10}
  ,{$lookup: { from: "foo", localField: "_id", foreignField: "_id", as: "X"}}
]);

这将产生类似这样的结果：

{
"_id" : "C",
"n" : 4,
"X" : [
    {
        "_id" : "C",
        "name" : "GroupC"
    }
]
}
{
"_id" : "A",
"n" : 3,
"X" : [
    {
        "_id" : "A",
        "name" : "GroupA"
    }
]
}

我可以从 MongoDb 中获得前 10 名的组列表吗，它是按成员数排序的

Can I get top 10 group list from MongoDb which is order by members count

mongoose

mongodb

mongodb-query

mongoose-populate

mongoose-schema