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.
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.
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.
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.
You have to do it the way you described. MongoDB is a non-relational database and doesn't support joins.
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.
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.
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
}
});
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
$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
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/
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
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"}},
])
You can run SQL queries including join on MongoDB with mongo_fdw from Postgres.
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;
});
});
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.
playORM can do it for you using S-SQL(Scalable SQL) which just adds partitioning such that you can do joins within partitions.
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.
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" ] } }
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.
How would I go about displaying the best reviews and the worst reviews at the top of the page.
I think the user's "useful" and "notUseful" votes should have an effect on the result.
I have reviews and if people click on the useful and notUseful buttons their Id gets added to the appropriate array (useful or notUseful).
you can tell what a positive or a negative score is by the "overall" score. that is 1 through 5. so 1 would be the worst and 5 would be the best.
I guess If someone gave a review with a 5 overall score but only got one useful but someone gave a score with a 4 overall and 100 people clicking on "useful" the one with 100 people should be shown as the best positive?
I only want to show 2 reviews at the top of the page the best and the worst worst review if there are ties with the overall scores the deciding factor should be the usefulness. so if there are 2 reviews with the same overall score and one of them has 5 usefuls and 10 notUsefuls that would be -5 usefuls and in the other review someone has 5 usefuls and and 4 notUsefuls that would be 1 usefuls so that would be shown to break the tie.
I'm hopping to do it with one mongoose query and not aggregation but I think the answer will be aggregation.
I guess there could be a cut off like scores greater than 3 is a positive review and lower is negative review.
I use mongoose.
Thanks in advance for your help.
some sample data.
{
"_id" : ObjectId("5929f89a54aa92274c4e4677"),
"compId" : ObjectId("58d94c441eb9e52454932db6"),
"anonId" : ObjectId("5929f88154aa92274c4e4675"),
"overall" : 3,
"titleReview" : "53",
"reviewText" : "53",
"companyName" : "store1",
"replies" : [],
"version" : 2,
"notUseful" : [ObjectId("58d94c441eb9e52454932db6")],
"useful" : [],
"dateCreated" : ISODate("2017-05-27T22:07:22.207Z"),
"images" : [],
"__v" : 0
}
{
"_id" : ObjectId("5929f8dfa1435135fc5e904b"),
"compId" : ObjectId("58d94c441eb9e52454932db6"),
"anonId" : ObjectId("5929f8bab0bc8834f41e9cf8"),
"overall" : 3,
"titleReview" : "54",
"reviewText" : "54",
"companyName" : "store1",
"replies" : [],
"version" : 1,
"notUseful" : [ObjectId("5929f83bf371672714bb8d44"), ObjectId("5929f853f371672714bb8d46")],
"useful" : [],
"dateCreated" : ISODate("2017-05-27T22:08:31.516Z"),
"images" : [],
"__v" : 0
}
{
"_id" : ObjectId("5929f956a692e82398aaa2f2"),
"compId" : ObjectId("58d94c441eb9e52454932db6"),
"anonId" : ObjectId("5929f93da692e82398aaa2f0"),
"overall" : 3,
"titleReview" : "56",
"reviewText" : "56",
"companyName" : "store1",
"replies" : [],
"version" : 1,
"notUseful" : [],
"useful" : [],
"dateCreated" : ISODate("2017-05-27T22:10:30.608Z"),
"images" : [],
"__v" : 0
}
If I am reading your question correctly then it appears you want a calculated difference of the "useful" and "nonUseful" votes to also be taken into account when sorting on the "overall" score of the documents.
The better option here is include that calculation in your stored documents, but for totality we will cover both options.
Aggregation
Without changes to your schema and other logic, then aggregation is indeed required to do that calculation. This is best presented as:
Model.aggregate([
{ "$addFields": {
"netUseful": {
"$subtract": [
{ "$size": "$useful" },
{ "$size": "$notUseful" }
]
}
}},
{ "$sort": { "overall": 1, "netUseful": -1 } }
],function(err, result) {
})
So you are basically getting the difference between the two arrays, where more "useful" responses have a positive impact boosting the ranking ans more "notUseful" will reduce that impact. Depending on the MongoDB version you have available you use either $addFields with only the additional field or $project with all the fields you need to return.
The $sort is then performed on the combination of the "overall" score in ascending order as per your rules, and the new field of "netUseful" in descending order ranking "positive" to "negative".
Re-Modelling
Foregoing aggregation altogether, you get a faster result from the plain query. But this of course means maintaining that "score" in the document as you add members to the array.
In basic options, you are using the $inc update operator along with $push to change the score.
So for a "useful" entry, you would do something like this:
Model.update(
{ "_id": docId, "useful": { "$ne": userId } },
{
"$push": { "useful": userId },
"$inc": { "netUseful": 1 }
},
function(err, status) {
}
)
And for a "notUseful" you do the opposite by "decrementing" with a negative value to $inc:
Model.update(
{ "_id": docId, "nonUseful": { "$ne": userId } },
{
"$push": { "nonUseful": userId },
"$inc": { "netUseful": -1 }
},
function(err, status) {
}
)
To cover all cases including where a vote is "changed" from "useFul" to "nonUseful" then you would expand on the logic and implement the appropriate reverse actions with $pull. But this should give the general idea.
N.B The reason we do not use the $addToSet operation here is because we want to make sure the user id is not present in the array when "incrementing" or "decrementing". Thus instead the $ne operator is used to test the value does not exist. If it does, then we do not attempt to modify the array or affect the "netUseful" value. The same applies to the reverse case of "removing" the user from those votes.
Since the calculation is always maintained with each update, you simply perform as query with a standard .sort()
Model.find().sort({ "overall": 1, "netUseful": -1 }).exec(function(err,results) {
})
So by moving the "cost" into the maintenance of the "votes", you remove the overhead of running the aggregation later. For my money, where this is a regular operation and the "sort" does not rely on other run-time parameters which force the calculation to be dynamic, then you use the stored result instead.
my schema looks like
{
qty:{
property1:{
//something
}
property2:[{
size:40,
color:"black",
enabled:"true"
}]
}
}
property 2 is array what i want to do is update those array object whose enabled is true in single query
I tried writing the following query
db.col.update({
"qty.property2.enabled" = "true"
}, {
"qty.property2.color" = "green"
}, callback)
but it is not working
error:
[main] Error: can't have . in field names [qty.pro.size]
db.col.update({"qty.property2.enabled":"true"},{$set: {'qty.property2.$.color': 'green'}}, {multi: true})
this is the way to update element inside array.
equal sign '=' cannot be used inside object
updating array is done using $
Alternative solution for multiple conditions:
db.foo.update({
_id:"i1",
replies: { $elemMatch:{
_id: "s2",
update_password: "abc"
}}
},
{
"$set" : {"replies.$.text" : "blah"}
}
);
Why
So I was looking for similar solution as this question, but in my case I needed array element to match multiple conditions and using currently provided answers resulted in changes to wrong fields.
If you need to match multiple fields, for example let say we have element like this:
{
"_id" : ObjectId("i1"),
"replies": [
{
"_id" : ObjectId("s1"),
"update_password": "abc",
"text": "some stuff"
},
{
"_id" : ObjectId("s2"),
"update_password": "abc",
"text": "some stuff"
}
]
}
Trying to do update by
db.foo.update({
_id:"i1",
"replies._id":"s2",
"replies.update_password": "abc"
},
{
"$set" : {"replies.$.text" : "blah"}
}
);
Would result in updating to field that only matches one condition, for example it would update s1 because it matches update_password condition, which is clearly wrong. I might have did something wrong, but $elemMatch solution solved any problems like that.
Suppose your documet looks like this.
{
"_id" : ObjectId("4f9808648859c65d"),
"array" : [
{"text" : "foo", "value" : 11},
{"text" : "foo", "value" : 22},
{"text" : "foobar", "value" : 33}
]
}
then your query will be
db.foo.update({"array.value" : 22}, {"$set" : {"array.$.text" : "blah"}})
where first curly brackets represents query criteria and second one sets the new value.