Reputation: 2059
I have two collections in MongoDB. The first contains information about some football coaches and the second contains data about teams. For example, this is a document of coach collection:
{
"_id" : ObjectId("556caaac9262ab4f14165fca"),
"name" : "Luis",
"surname" : "Enrique Martinez Garcia",
"age" : 45,
"date_Of_birth" : {
"day" : 8,
"month" : 5,
"year" : 1970
},
"place_Of_birth" : "Gijòn",
"nationality" : "Spanish",
"preferred_formation" : "4-3-3 off",
"coached_Team" : [
{
"team_id" : "Bar.43",
"in_charge" : {
"from" : "01/july/2014"
},
"matches" : 59
},
{
"team_id" : "Cel.00",
"in_charge" : {
"from" : "9/june/2013",
"to" : "30/june/2014"
},
"matches" : 40
},
{
"team_id" : "Rom.01",
"in_charge" : {
"from" : "7/june/2011",
"to" : "10/may/2012"
},
"matches" : 41
}
Here is a document of team collection:
{
"_id" : "Bar.43",
"official_name" : "Futbol Club Barcelona",
"country" : "Spain",
"started_by" : {
"day" : 28,
"month" : 11,
"year" : 1899
},
"stadium" : {
"name" : "Camp Nou",
"capacity" : 99354
},
"palmarès" : {
"La Liga" : 23,
"Copa del Rey" : 27,
"Supercopa de Espana" : 11,
"UEFA Champions League" : 4,
"UEFA Cup Winners Cup" : 4,
"UEFA Super Cup" : 4,
"FIFA Club World cup" : 2
},
"uniform" : "blue and dark red"
}
Well, I know mongo does not support join between collections. Now suppose I saved the return of a query on team collection in an array called x. For example:
var x = db.team.find({_id:"Bar.43"}).toArray()
Now I want to use this array x to query coach collection and find coaches that coached the team with that id. I tried in some ways, but they don't work:
[1]
db.coach.aggregate([{$unwind:"$coached_Team"},{$match:{"coached_Team.team_id:"x[0]._id"}}])
[2]
db.team.find({"x[0]._id":{$in:coached_Team}})
P.S. I looked for similar questions in the forum, and the answers don't reply to mine.
This, for example, does not work.
Upvotes: 3
Views: 1253
Reputation: 7840
First you find all distinct
team id as
var teamId = db.team.distinct("_id")
teamId
contains array of team id . An use this aggregation for coach collection
db.coach.aggregate({"$unwind":"$coached_Team"},{"$match":{"coached_Team.team_id":{"$in":teamId}}}).pretty()
without aggregation use this
db.coach.find({"coached_Team":{"$elemMatch":{"team_id":{"$in":teamId}}}},{"coached_Team.$.team_id":1})
Or
db.coach.find({"coached_Team.team_id":{"$in":teamId}},{"coached_Team.$.team_id":1})
or if you want only specific team id change above distinct as :
var teamId = db.team.distinct("_id",{"_id":"Bar.43"})
Upvotes: 0
Reputation: 20150
you need to remove the quotes "
around your variable x[0]._id
. Otherwise this is encoded as a String and the content of the variable will not be looked up and filled in.
var x = db.team.find({_id:"Bar.43"}).toArray();
db.coach.find({"coached_Team.team_id":x[0]._id});
Upvotes: 1
Reputation: 46341
That is a bit simpler actually:
var x = db.team.find({_id:"Bar.43"}).toArray();
var coaches = db.coach.find( { "coached_Team.team_id" : x[0]._id } );
A slightly cleaner approach (that is required when you want multiple criteria) is using $elemMatch
:
var coaches = db.coach.find({ 'coached_Team' : {
'$elemMatch' : { 'team_id': x[0]._id /*, optionally more criteria */ } } })
Upvotes: 2