How to use join in finding data in MongoDB database? - mongodb

Question

how to join schema in mongodb?
Example :
a collection
{
ObjectId : ObjectId("ABCDE")
userName : "Jason",
level : 30,
money : 200
}
b collection
{
Id : ObjectId("AAACC"),
userId : ObjectId("ABCDE"),
item : "sword"
}
b.aggregate....
i want result is
id : ObjectId("AAACC"), userName : "Jason", item : "sword"

Answer 1

You should use the aggregation pipeline to join two collections and select the required data. I assume here that you have proper identity fields named _id instead of ObjectId and Id as in your sample:
db.items.aggregate([
{
$lookup:
{
from: "users",
localField: "userId",
foreignField: "_id", // ObjectId in your sample
as: "user"
}
},
{ $unwind: "$user" },
{
$project:
{
"item": 1,
"userName": "$user.userName"
// Id: 1 if you will use your names, because only _id is selected by default
}
}
])
The first step is lookup which joins items and users collections on userId field equals _id field in users collection.
Then you should unwind results, because lookup puts all matched users into user field as an array of user documents.
And last step - project result documents to the desired format.
Now sample. If you have following documents in items collection:
{
"_id" : ObjectId("5c18df3e5d85eb27052a599c"),
"item" : "sword",
"userId" : ObjectId("5c18ded45d85eb27052a5988")
},
{
"_id" : ObjectId("5c18df4f5d85eb27052a599e"),
"item" : "helmet",
"userId" : ObjectId("5c18ded45d85eb27052a5988")
},
{
"_id" : ObjectId("5c18e2da5d85eb27052a59ee"),
"item" : "helmet"
}
And you have two users:
{
"_id" : ObjectId("5c18ded45d85eb27052a5988"),
"userName" : "Jason",
"level" : 30,
"money" : 200
},
{
"_id" : ObjectId("5c18dee35d85eb27052a598a"),
"userName" : "Bob",
"level" : 70,
"money" : 500
}
Then the query above will produce
{
"_id" : ObjectId("5c18df3e5d85eb27052a599c"),
"item" : "sword",
"userName" : "Jason"
},
{
"_id" : ObjectId("5c18df4f5d85eb27052a599e"),
"item" : "helmet",
"userName" : "Jason"
},
{
"_id" : ObjectId("5c18e2da5d85eb27052a59ee"),
"item" : "helmet"
}
NOTE: Usually user names should be unique. Consider to use them as identity for users collection. That will also give you desired result in items collection without any joins.

Answer 2

You can use the lookup aggregation operator to join both collections, and then project only the field from collection a that you are interested in:
db.b.aggregate([
{
$lookup: {
from: "a",
localField: "userId",
foreignField: "ObjectId",
as: "user"
}
},
{
$unwind: "$user"
},
{
$project: {
Id: 1
userName: "$user.userName",
item: 1
}
}
]);
I assume a.ObjectId should in fact be called a._id and b.Id should be b._id? Either way the same principle applies.
EDIT: had forgotten the unwind stage. You need this since your lookup will return the new joined field as an array (albeit with one element), so you need this to get rid of the square brackets.