628

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.

philipxy
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The Unknown
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    The last answer on this question is probably the most relevant, since MongoDB 3.2+ implemented a join solution called $lookup. Thought I would push it here because maybe not everyone will read to the bottom. http://stackoverflow.com/a/33511166/2593330 – thefourtheye Nov 22 '15 at 16:33
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    Correct, **$lookup** was introduced in MongoDB 3.2. Details can be found at https://docs.mongodb.org/master/reference/operator/aggregation/lookup/#pipe._S_lookup – NDB Nov 23 '15 at 13:20
  • Convert any query to mongo, check the answer: https://stackoverflow.com/questions/68155715/how-do-i-write-this-sql-query-in-mongodb-syntax/68549980#68549980 – KushalSeth Jul 27 '21 at 18:32
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    I believe the original intention of NoSQL was that you store data as you want to retrieve it.. So, store it that way and retrieve it that way – Robert Perry Feb 06 '22 at 13:02

19 Answers19

395

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.

Ahmad Baktash Hayeri
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Clayton Gulick
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165

We can merge/join all data inside only one collection with a easy function in few lines using the mongodb client console, and now we could be able of perform the desired query. Below a complete example,

.- Authors:

db.authors.insert([
    {
        _id: 'a1',
        name: { first: 'orlando', last: 'becerra' },
        age: 27
    },
    {
        _id: 'a2',
        name: { first: 'mayra', last: 'sanchez' },
        age: 21
    }
]);

.- Categories:

db.categories.insert([
    {
        _id: 'c1',
        name: 'sci-fi'
    },
    {
        _id: 'c2',
        name: 'romance'
    }
]);

.- Books

db.books.insert([
    {
        _id: 'b1',
        name: 'Groovy Book',
        category: 'c1',
        authors: ['a1']
    },
    {
        _id: 'b2',
        name: 'Java Book',
        category: 'c2',
        authors: ['a1','a2']
    },
]);

.- Book lending

db.lendings.insert([
    {
        _id: 'l1',
        book: 'b1',
        date: new Date('01/01/11'),
        lendingBy: 'jose'
    },
    {
        _id: 'l2',
        book: 'b1',
        date: new Date('02/02/12'),
        lendingBy: 'maria'
    }
]);

.- The magic:

db.books.find().forEach(
    function (newBook) {
        newBook.category = db.categories.findOne( { "_id": newBook.category } );
        newBook.lendings = db.lendings.find( { "book": newBook._id  } ).toArray();
        newBook.authors = db.authors.find( { "_id": { $in: newBook.authors }  } ).toArray();
        db.booksReloaded.insert(newBook);
    }
);

.- Get the new collection data:

db.booksReloaded.find().pretty()

.- Response :)

{
    "_id" : "b1",
    "name" : "Groovy Book",
    "category" : {
        "_id" : "c1",
        "name" : "sci-fi"
    },
    "authors" : [
        {
            "_id" : "a1",
            "name" : {
                "first" : "orlando",
                "last" : "becerra"
            },
            "age" : 27
        }
    ],
    "lendings" : [
        {
            "_id" : "l1",
            "book" : "b1",
            "date" : ISODate("2011-01-01T00:00:00Z"),
            "lendingBy" : "jose"
        },
        {
            "_id" : "l2",
            "book" : "b1",
            "date" : ISODate("2012-02-02T00:00:00Z"),
            "lendingBy" : "maria"
        }
    ]
}
{
    "_id" : "b2",
    "name" : "Java Book",
    "category" : {
        "_id" : "c2",
        "name" : "romance"
    },
    "authors" : [
        {
            "_id" : "a1",
            "name" : {
                "first" : "orlando",
                "last" : "becerra"
            },
            "age" : 27
        },
        {
            "_id" : "a2",
            "name" : {
                "first" : "mayra",
                "last" : "sanchez"
            },
            "age" : 21
        }
    ],
    "lendings" : [ ]
}

I hope this lines can help you.

Orlando Becerra
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    i'm wondering if this same code can be ran using doctrine mongodb? – abbood May 30 '14 at 13:46
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    What happens when one of the references objects gets an update? Does that update automatically reflect in the book object? Or does that loop need to run again? – balupton Jun 04 '14 at 05:55
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    This is fine as long as your data is small. It is going to bring each book content to your client and then fetch each category, lending and authors one by one. The moment your books are in thousands, this would go really really slow. A better technique probably would be to use aggregation pipeline and output the merged data into a separate collection. Let me get back to it again. I will add that an answer. – Sandeep Giri Jun 19 '14 at 15:31
  • Can you adapt your algorithm to this other example? http://stackoverflow.com/q/32718079/287948 – Peter Krauss Sep 22 '15 at 14:13
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    @SandeepGiri how can i do the aggregate pipeline since i have really really intensive data in separated collection need join ?? – Yassine Abdul-Rahman Oct 07 '15 at 20:16
  • If you use this to create temporary collections, then you need to be wary of collection name collisions. This might run simultaneously from multiple application calls. One work around is to use a UUID for the collection name. – steampowered Nov 04 '15 at 19:58
  • I created a mongoose plugin which does this. It saves a fetched query using mongoose `populate` into a temporary collection, then allows you to query the temporary collection. I also provide a method to delete the temporary collection without exposing it's collection name https://github.com/jeremyml/mongoose-temp-collection – steampowered Nov 11 '15 at 15:12
  • Well this guys create new collection as the result of joining the document. – Faris Rayhan Dec 03 '17 at 06:35
  • Is this the efficient one ?? – Praveen Oct 04 '22 at 13:08
154

This page on the official mongodb site addresses exactly this question:

https://mongodb-documentation.readthedocs.io/en/latest/ecosystem/tutorial/model-data-for-ruby-on-rails.html

When we display our list of stories, we'll need to show the name of the user who posted the story. If we were using a relational database, we could perform a join on users and stores, and get all our objects in a single query. But MongoDB does not support joins and so, at times, requires bit of denormalization. Here, this means caching the 'username' attribute.

Relational purists may be feeling uneasy already, as if we were violating some universal law. But let’s bear in mind that MongoDB collections are not equivalent to relational tables; each serves a unique design objective. A normalized table provides an atomic, isolated chunk of data. A document, however, more closely represents an object as a whole. In the case of a social news site, it can be argued that a username is intrinsic to the story being posted.

Community
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William Stein
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    @dudelgrincen it's a paradigm shift from normalization and relational databases. The goal of a NoSQL is to read and write from the database very quickly. With BigData you're going to have scads of application and front end servers with lower numbers on DBs. You're expected to do millions of transactions a second. Offload the heavy lifting from the database and put it onto the application level. If you need deep analysis, you run an integration job that puts your data into an OLAP database. You shouldn't be getting many deep queries from your OLTP dbs anyway. – Snowburnt Nov 04 '13 at 01:53
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    @dudelgrincen I should also say that it's not for every project or design. If you have something that works in a SQL type database why change it? If you can't massage your schema to work with noSQL, then don't. – Snowburnt Nov 12 '13 at 00:30
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    Migrations and a constantly evolving schemas are also a lot easier to manage on a NoSQL system. – justin May 06 '14 at 20:09
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    What if the user has 3.540 posts in the website, and he does change his username in profile? Should every post be updated with the new username? – Ivo Pereira Mar 02 '16 at 17:39
  • @IvoPereira it's a design choice, you can also do a second query to take the user data. – Wédney Yuri May 11 '16 at 00:45
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    @IvoPereira Yes and that's exactly why one should avoid modeling data this way. There is an article that explains the same scenario and its consequences: [Why You Should Never Use MongoDB](http://www.sarahmei.com/blog/2013/11/11/why-you-should-never-use-mongodb/) – Omid Nov 30 '17 at 20:43
  • The original doc apparently was removed. It seems to be a copy in https://mongodb-documentation.readthedocs.io/en/latest/ecosystem/tutorial/model-data-for-ruby-on-rails.html – abentan Feb 21 '18 at 03:19
  • https://mongodb-documentation.readthedocs.io/en/latest/ecosystem/tutorial/model-data-for-ruby-on-rails.html#caching-to-avoid-n-1 – hojin Feb 12 '19 at 03:33
  • How do you handle updates to the username? What if a user renames? Do you need to update all of it's occurrences over other collections? – Korenz Mar 16 '21 at 12:10
  • @Korenz, yes, all documents have to be updated. But consider this: by storing the username in each post, you are optimizing for read performance. And reading posts happens A LOT more than someone changing its username. So it all depends on your own use-cases. In this case, prefer read performance over write performance, when that write "rarely" happens (changing a username) – Remco Ros Feb 01 '22 at 14:48
40

With right combination of $lookup, $project and $match, you can join mutiple tables on multiple parameters. This is because they can be chained multiple times.

Suppose we want to do following (reference)

SELECT S.* FROM LeftTable S
LEFT JOIN RightTable R ON S.ID = R.ID AND S.MID = R.MID  
WHERE R.TIM > 0 AND S.MOB IS NOT NULL

Step 1: Link all tables

you can $lookup as many tables as you want.

$lookup - one for each table in query

$unwind - correctly denormalises data , else it'd be wrapped in arrays

Python code..

db.LeftTable.aggregate([
                        # connect all tables

                        {"$lookup": {
                          "from": "RightTable",
                          "localField": "ID",
                          "foreignField": "ID",
                          "as": "R"
                        }},
                        {"$unwind": "R"}
                   
                        ])

Step 2: Define all conditionals

$project : define all conditional statements here, plus all the variables you'd like to select.

Python Code..

db.LeftTable.aggregate([
                        # connect all tables

                        {"$lookup": {
                          "from": "RightTable",
                          "localField": "ID",
                          "foreignField": "ID",
                          "as": "R"
                        }},
                        {"$unwind": "R"},

                        # define conditionals + variables

                        {"$project": {
                          "midEq": {"$eq": ["$MID", "$R.MID"]},
                          "ID": 1, "MOB": 1, "MID": 1
                        }}
                        ])

Step 3: Join all the conditionals

$match - join all conditions using OR or AND etc. There can be multiples of these.

$project: undefine all conditionals

Complete Python Code..

db.LeftTable.aggregate([
                        # connect all tables

                        {"$lookup": {
                          "from": "RightTable",
                          "localField": "ID",
                          "foreignField": "ID",
                          "as": "R"
                        }},
                        {"$unwind": "$R"},

                        # define conditionals + variables

                        {"$project": {
                          "midEq": {"$eq": ["$MID", "$R.MID"]},
                          "ID": 1, "MOB": 1, "MID": 1
                        }},

                        # join all conditionals

                        {"$match": {
                          "$and": [
                            {"R.TIM": {"$gt": 0}}, 
                            {"MOB": {"$exists": True}},
                            {"midEq": {"$eq": True}}
                        ]}},

                        # undefine conditionals

                        {"$project": {
                          "midEq": 0
                        }}

                        ])

Pretty much any combination of tables, conditionals and joins can be done in this manner.

sbharti
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38

You have to do it the way you described. MongoDB is a non-relational database and doesn't support joins.

Otto Allmendinger
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    Seems wrong performance wise coming from a sql server background, but its maybe not that bad with a document db? – terjetyl Jul 15 '10 at 18:20
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    from a sql server background as well, I would appreciate MongoDB taking a 'result set' (with selected returned fields) as input for a new query in one go, much like nested queries in SQL – Stijn Sanders Nov 26 '10 at 23:17
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    @terjetyl You have to really plan for it. What fields are you going be presenting on the front end, if it's a limited amount in an individual view then you take those as embedded documents. The key is to not need to do joins. If you want to do deep analysis, you do it after the fact in another database. Run a job that transforms the data into an OLAP cube for optimal performance. – Snowburnt Nov 04 '13 at 01:56
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    From mongo 3.2 version left joins are supported. – Somnath Muluk Nov 26 '15 at 11:12
  • MongoDB is non-tabular, not not-relational. Many relationships can we well-represented with nested documents. – Matthew Read Jul 29 '22 at 19:51
23

You can join two collection in Mongo by using lookup which is offered in 3.2 version. In your case the query would be

db.comments.aggregate({
    $lookup:{
        from:"users",
        localField:"uid",
        foreignField:"uid",
        as:"users_comments"
    }
})

or you can also join with respect to users then there will be a little change as given below.

db.users.aggregate({
    $lookup:{
        from:"comments",
        localField:"uid",
        foreignField:"uid",
        as:"users_comments"
    }
})

It will work just same as left and right join in SQL.

Alex M
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jarry jafery
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19

As others have pointed out you are trying to create a relational database from none relational database which you really don't want to do but anyways, if you have a case that you have to do this here is a solution you can use. We first do a foreach find on collection A( or in your case users) and then we get each item as an object then we use object property (in your case uid) to lookup in our second collection (in your case comments) if we can find it then we have a match and we can print or do something with it. Hope this helps you and good luck :)

db.users.find().forEach(
function (object) {
    var commonInBoth=db.comments.findOne({ "uid": object.uid} );
    if (commonInBoth != null) {
        printjson(commonInBoth) ;
        printjson(object) ;
    }else {
        // did not match so we don't care in this case
    }
});
VishAl
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grepit
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17

Here's an example of a "join" * Actors and Movies collections:

https://github.com/mongodb/cookbook/blob/master/content/patterns/pivot.txt

It makes use of .mapReduce() method

* join - an alternative to join in document-oriented databases

JosephSlote
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antitoxic
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    -1, This is NOT joining data from two collections. It is using data from a single collection (actors) pivoting data around. So that things that were keys are now values and values are now keys... **very** different than a JOIN. – Evan Teran May 22 '12 at 17:44
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    This exactly what you have to do, MongoDB is not relational but document oriented. MapReduce allows to play with data with big performance (you can use cluster etc....) but even for simple cases, its very useful ! – Thomas Decaux Jun 17 '12 at 19:16
16

$lookup (aggregation)

Performs a left outer join to an unsharded collection in the same database to filter in documents from the “joined” collection for processing. To each input document, the $lookup stage adds a new array field whose elements are the matching documents from the “joined” collection. The $lookup stage passes these reshaped documents to the next stage. The $lookup stage has the following syntaxes:

Equality Match

To perform an equality match between a field from the input documents with a field from the documents of the “joined” collection, the $lookup stage has the following syntax:

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

The operation would correspond to the following pseudo-SQL statement:

SELECT *, <output array field>
FROM collection
WHERE <output array field> IN (SELECT <documents as determined from the pipeline>
                               FROM <collection to join>
                               WHERE <pipeline> );

Mongo URL

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GoutamS
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    sub-query is totally different than join, if your left side table is huge, sub-query means each row has to do a query itself. it will become very slow. join is very fast in sql. – yww325 Apr 30 '20 at 02:14
12

It depends on what you're trying to do.

You currently have it set up as a normalized database, which is fine, and the way you are doing it is appropriate.

However, there are other ways of doing it.

You could have a posts collection that has imbedded comments for each post with references to the users that you can iteratively query to get. You could store the user's name with the comments, you could store them all in one document.

The thing with NoSQL is it's designed for flexible schemas and very fast reading and writing. In a typical Big Data farm the database is the biggest bottleneck, you have fewer database engines than you do application and front end servers...they're more expensive but more powerful, also hard drive space is very cheap comparatively. Normalization comes from the concept of trying to save space, but it comes with a cost at making your databases perform complicated Joins and verifying the integrity of relationships, performing cascading operations. All of which saves the developers some headaches if they designed the database properly.

With NoSQL, if you accept that redundancy and storage space aren't issues because of their cost (both in processor time required to do updates and hard drive costs to store extra data), denormalizing isn't an issue (for embedded arrays that become hundreds of thousands of items it can be a performance issue, but most of the time that's not a problem). Additionally you'll have several application and front end servers for every database cluster. Have them do the heavy lifting of the joins and let the database servers stick to reading and writing.

TL;DR: What you're doing is fine, and there are other ways of doing it. Check out the mongodb documentation's data model patterns for some great examples. http://docs.mongodb.org/manual/data-modeling/

Snowburnt
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    "Normalization comes from the concept of trying to save space" I question this. IMHO normalization comes from the concept of avoiding redundancy. Say you store the name of a user along with a blogpost. What if she marries? In a not normalized model you will have to wade through all posts and change the name. In a normalized model you usually change ONE record. – DanielKhan Nov 27 '13 at 13:28
  • @DanielKhan preventing redundancy and saving space are similar concepts, but on re-analysis I do agree, redundancy is the root cause for this design. I'll reword. Thanks for the note. – Snowburnt Nov 27 '13 at 19:15
11

There is a specification that a lot of drivers support that's called DBRef.

DBRef is a more formal specification for creating references between documents. DBRefs (generally) include a collection name as well as an object id. Most developers only use DBRefs if the collection can change from one document to the next. If your referenced collection will always be the same, the manual references outlined above are more efficient.

Taken from MongoDB Documentation: Data Models > Data Model Reference > Database References

NDB
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Pickels
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10

Before 3.2.6, Mongodb does not support join query as like mysql. below solution which works for you.

 db.getCollection('comments').aggregate([
        {$match : {pid : 444}},
        {$lookup: {from: "users",localField: "uid",foreignField: "uid",as: "userData"}},
   ])
Anish Agarwal
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4

You can run SQL queries including join on MongoDB with mongo_fdw from Postgres.

tagurit
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metdos
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3

MongoDB does not allow joins, but you can use plugins to handle that. Check the mongo-join plugin. It's the best and I have already used it. You can install it using npm directly like this npm install mongo-join. You can check out the full documentation with examples.

(++) really helpful tool when we need to join (N) collections

(--) we can apply conditions just on the top level of the query

Example

var Join = require('mongo-join').Join, mongodb = require('mongodb'), Db = mongodb.Db, Server = mongodb.Server;
db.open(function (err, Database) {
    Database.collection('Appoint', function (err, Appoints) {

        /* we can put conditions just on the top level */
        Appoints.find({_id_Doctor: id_doctor ,full_date :{ $gte: start_date },
            full_date :{ $lte: end_date }}, function (err, cursor) {
            var join = new Join(Database).on({
                field: '_id_Doctor', // <- field in Appoints document
                to: '_id',         // <- field in User doc. treated as ObjectID automatically.
                from: 'User'  // <- collection name for User doc
            }).on({
                field: '_id_Patient', // <- field in Appoints doc
                to: '_id',         // <- field in User doc. treated as ObjectID automatically.
                from: 'User'  // <- collection name for User doc
            })
            join.toArray(cursor, function (err, joinedDocs) {

                /* do what ever you want here */
                /* you can fetch the table and apply your own conditions */
                .....
                .....
                .....


                resp.status(200);
                resp.json({
                    "status": 200,
                    "message": "success",
                    "Appoints_Range": joinedDocs,


                });
                return resp;


            });

    });
Michael Mior
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Aouidane Med Amine
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3

You can do it using the aggregation pipeline, but it's a pain to write it yourself.

You can use mongo-join-query to create the aggregation pipeline automatically from your query.

This is how your query would look like:

const mongoose = require("mongoose");
const joinQuery = require("mongo-join-query");

joinQuery(
    mongoose.models.Comment,
    {
        find: { pid:444 },
        populate: ["uid"]
    },
    (err, res) => (err ? console.log("Error:", err) : console.log("Success:", res.results))
);

Your result would have the user object in the uid field and you can link as many levels deep as you want. You can populate the reference to the user, which makes reference to a Team, which makes reference to something else, etc..

Disclaimer: I wrote mongo-join-query to tackle this exact problem.

Marcelo Lazaroni
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0

playORM can do it for you using S-SQL(Scalable SQL) which just adds partitioning such that you can do joins within partitions.

Dean Hiller
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-2

Nope, it doesn't seem like you're doing it wrong. MongoDB joins are "client side". Pretty much like you said:

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.

1) Select from the collection you're interested in.
2) From that collection pull out ID's you need
3) Select from other collections
4) Decorate your original results.

It's not a "real" join, but it's actually alot more useful than a SQL join because you don't have to deal with duplicate rows for "many" sided joins, instead your decorating the originally selected set.

There is alot of nonsense and FUD on this page. Turns out 5 years later MongoDB is still a thing.

Michael Cole
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  • *'you don't have to deal with duplicate rows for "many" sided joins'* - no idea what you mean by this. Can you clarify? – Mark Amery Sep 20 '15 at 20:23
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    @MarkAmery, sure. In SQL a n-n relationship will return duplicate rows. E.g. Friends. If Bob is friends with Mary and Jane, you'll get 2 rows for Bob: Bob,Mary and Bob,Jane. 2 Bobs is a lie, there is only one Bob. With client-side joins you can start with Bob and decorate how you like: Bob, "Mary and Jane". SQL let's you do this with subqueries, but that's doing work on the db server that could be done on the client. – Michael Cole Sep 21 '15 at 14:51
-3

I think, if You need normalized data tables - You need to try some other database solutions.

But I've foun that sollution for MOngo on Git By the way, in inserts code - it has movie's name, but noi movie's ID.

Problem

You have a collection of Actors with an array of the Movies they've done.

You want to generate a collection of Movies with an array of Actors in each.

Some sample data

 db.actors.insert( { actor: "Richard Gere", movies: ['Pretty Woman', 'Runaway Bride', 'Chicago'] });
 db.actors.insert( { actor: "Julia Roberts", movies: ['Pretty Woman', 'Runaway Bride', 'Erin Brockovich'] });

Solution

We need to loop through each movie in the Actor document and emit each Movie individually.

The catch here is in the reduce phase. We cannot emit an array from the reduce phase, so we must build an Actors array inside of the "value" document that is returned.

The code
map = function() {
  for(var i in this.movies){
    key = { movie: this.movies[i] };
    value = { actors: [ this.actor ] };
    emit(key, value);
  }
}

reduce = function(key, values) {
  actor_list = { actors: [] };
  for(var i in values) {
    actor_list.actors = values[i].actors.concat(actor_list.actors);
  }
  return actor_list;
}

Notice how actor_list is actually a javascript object that contains an array. Also notice that map emits the same structure.

Run the following to execute the map / reduce, output it to the "pivot" collection and print the result:

printjson(db.actors.mapReduce(map, reduce, "pivot")); db.pivot.find().forEach(printjson);

Here is the sample output, note that "Pretty Woman" and "Runaway Bride" have both "Richard Gere" and "Julia Roberts".

{ "_id" : { "movie" : "Chicago" }, "value" : { "actors" : [ "Richard Gere" ] } }
{ "_id" : { "movie" : "Erin Brockovich" }, "value" : { "actors" : [ "Julia Roberts" ] } }
{ "_id" : { "movie" : "Pretty Woman" }, "value" : { "actors" : [ "Richard Gere", "Julia Roberts" ] } }
{ "_id" : { "movie" : "Runaway Bride" }, "value" : { "actors" : [ "Richard Gere", "Julia Roberts" ] } }

Max Sherbakov
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  • Note that most of the content of this answer (i.e. the bit that's in comprehensible English) is copied from the MongoDB cookbook at the GitHub link the answerer provided. – Mark Amery Oct 05 '15 at 14:23
-6

We can merge two collection by using mongoDB sub query. Here is example, Commentss--

`db.commentss.insert([
  { uid:12345, pid:444, comment:"blah" },
  { uid:12345, pid:888, comment:"asdf" },
  { uid:99999, pid:444, comment:"qwer" }])`

Userss--

db.userss.insert([
  { uid:12345, name:"john" },
  { uid:99999, name:"mia"  }])

MongoDB sub query for JOIN--

`db.commentss.find().forEach(
    function (newComments) {
        newComments.userss = db.userss.find( { "uid": newComments.uid } ).toArray();
        db.newCommentUsers.insert(newComments);
    }
);`

Get result from newly generated Collection--

db.newCommentUsers.find().pretty()

Result--

`{
    "_id" : ObjectId("5511236e29709afa03f226ef"),
    "uid" : 12345,
    "pid" : 444,
    "comment" : "blah",
    "userss" : [
        {
            "_id" : ObjectId("5511238129709afa03f226f2"),
            "uid" : 12345,
            "name" : "john"
        }
    ]
}
{
    "_id" : ObjectId("5511236e29709afa03f226f0"),
    "uid" : 12345,
    "pid" : 888,
    "comment" : "asdf",
    "userss" : [
        {
            "_id" : ObjectId("5511238129709afa03f226f2"),
            "uid" : 12345,
            "name" : "john"
        }
    ]
}
{
    "_id" : ObjectId("5511236e29709afa03f226f1"),
    "uid" : 99999,
    "pid" : 444,
    "comment" : "qwer",
    "userss" : [
        {
            "_id" : ObjectId("5511238129709afa03f226f3"),
            "uid" : 99999,
            "name" : "mia"
        }
    ]
}`

Hope so this will help.

Krishna
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    Why did you basically copy this nearly identical, one-year-old answer? http://stackoverflow.com/a/22739813/4186945 – hackel May 05 '15 at 20:45