Reputation: 109
I have 2 collections one is for group details and other is for group_members details, group_members have group's _id stored (as FK)
Like group : -> _id,name,type,date
group_members -> _id,group_id,members_id,role,status
so how can I get the groups that have most count of memebrs in that with order by group_membrs's count (top 10 groups with more members)
Upvotes: 0
Views: 1306
Reputation: 7568
So if the groupMembers
collection contains the group ID as you describe, e.g.:
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);
Then this is how you'd get the count in each group:
c=db.foo2.aggregate([
{$group: {_id: "$gid", n: {$sum:1} }}
,{$sort: {n: -1}}
,{$limit: 10}
]);
If we toss in the group master collection, e.g.
var r = [
{_id: "A", name: "GroupA"}
,{_id: "B", name: "GroupB"}
,{_id: "C", name: "GroupC"}
];
db.foo.insert(r);
Then we simply do a $lookup
to "join" the data:
db.foo2.aggregate([
{$group: {_id: "$gid", n: {$sum:1} }}
,{$sort: {n: -1}}
,{$limit: 10}
,{$lookup: { from: "foo", localField: "_id", foreignField: "_id", as: "X"}}
]);
Which will yield something like this:
{
"_id" : "C",
"n" : 4,
"X" : [
{
"_id" : "C",
"name" : "GroupC"
}
]
}
{
"_id" : "A",
"n" : 3,
"X" : [
{
"_id" : "A",
"name" : "GroupA"
}
]
}
Upvotes: 1