Mongodb – How to perform the SQL Join equivalent in MongoDB

join;mongodb

How do I perform the SQL Join equivalent in MongoDB?

For example say you have two collections (users and comments) and I want to pull all the comments with pid=444 along with the user info for each.

comments
  { uid:12345, pid:444, comment="blah" }
  { uid:12345, pid:888, comment="asdf" }
  { uid:99999, pid:444, comment="qwer" }

users
  { uid:12345, name:"john" }
  { uid:99999, name:"mia"  }

Is there a way to pull all the comments with a certain field (eg. …find({pid:444}) ) and the user information associated with each comment in one go?

At the moment, I am first getting the comments which match my criteria, then figuring out all the uid's in that result set, getting the user objects, and merging them with the comment's results. Seems like I am doing it wrong.

Best Answer

As of Mongo 3.2 the answers to this question are mostly no longer correct. The new $lookup operator added to the aggregation pipeline is essentially identical to a left outer join:

https://docs.mongodb.org/master/reference/operator/aggregation/lookup/#pipe._S_lookup

From the docs:

{
   $lookup:
     {
       from: <collection to join>,
       localField: <field from the input documents>,
       foreignField: <field from the documents of the "from" collection>,
       as: <output array field>
     }
}

Of course Mongo is not a relational database, and the devs are being careful to recommend specific use cases for $lookup, but at least as of 3.2 doing join is now possible with MongoDB.

Related Solutions

Sql – the difference between “INNER JOIN” and “OUTER JOIN”

Assuming you're joining on columns with no duplicates, which is a very common case:

An inner join of A and B gives the result of A intersect B, i.e. the inner part of a Venn diagram intersection.
An outer join of A and B gives the results of A union B, i.e. the outer parts of a Venn diagram union.

Examples

Suppose you have two tables, with a single column each, and data as follows:

Note that (1,2) are unique to A, (3,4) are common, and (5,6) are unique to B.

Inner join

An inner join using either of the equivalent queries gives the intersection of the two tables, i.e. the two rows they have in common.

select * from a INNER JOIN b on a.a = b.b;
select a.*, b.*  from a,b where a.a = b.b;

a | b
--+--
3 | 3
4 | 4

Left outer join

A left outer join will give all rows in A, plus any common rows in B.

select * from a LEFT OUTER JOIN b on a.a = b.b;
select a.*, b.*  from a,b where a.a = b.b(+);

a |  b
--+-----
1 | null
2 | null
3 |    3
4 |    4

Right outer join

A right outer join will give all rows in B, plus any common rows in A.

select * from a RIGHT OUTER JOIN b on a.a = b.b;
select a.*, b.*  from a,b where a.a(+) = b.b;

a    |  b
-----+----
3    |  3
4    |  4
null |  5
null |  6

Full outer join

A full outer join will give you the union of A and B, i.e. all the rows in A and all the rows in B. If something in A doesn't have a corresponding datum in B, then the B portion is null, and vice versa.

select * from a FULL OUTER JOIN b on a.a = b.b;

 a   |  b
-----+-----
   1 | null
   2 | null
   3 |    3
   4 |    4
null |    6
null |    5

SQL JOIN – WHERE clause vs. ON clause

They are not the same thing.

Consider these queries:

SELECT *
FROM Orders
LEFT JOIN OrderLines ON OrderLines.OrderID=Orders.ID
WHERE Orders.ID = 12345

and

SELECT *
FROM Orders
LEFT JOIN OrderLines ON OrderLines.OrderID=Orders.ID 
    AND Orders.ID = 12345

The first will return an order and its lines, if any, for order number 12345. The second will return all orders, but only order 12345 will have any lines associated with it.

With an INNER JOIN, the clauses are effectively equivalent. However, just because they are functionally the same, in that they produce the same results, does not mean the two kinds of clauses have the same semantic meaning.

Best Answer

Related Solutions

Sql – the difference between “INNER JOIN” and “OUTER JOIN”

SQL JOIN – WHERE clause vs. ON clause

Related Topic