How to join query in mongodb? - join

I have user document collection like this:
User {
id:"001"
name:"John",
age:30,
friends:["userId1","userId2","userId3"....]
}
A user has many friends, I have the following query in SQL:
select * from user where in (select friends from user where id=?) order by age
I would like to have something similar in MongoDB.

To have everything with just one query using the $lookup feature of the aggregation framework, try this :
db.User.aggregate(
[
// First step is to extract the "friends" field to work with the values
{
$unwind: "$friends"
},
// Lookup all the linked friends from the User collection
{
$lookup:
{
from: "User",
localField: "friends",
foreignField: "_id",
as: "friendsData"
}
},
// Sort the results by age
{
$sort: { 'friendsData.age': 1 }
},
// Get the results into a single array
{
$unwind: "$friendsData"
},
// Group the friends by user id
{
$group:
{
_id: "$_id",
friends: { $push: "$friends" },
friendsData: { $push: "$friendsData" }
}
}
]
)
Let's say the content of your User collection is the following:
{
"_id" : ObjectId("573b09e6322304d5e7c6256e"),
"name" : "John",
"age" : 30,
"friends" : [
"userId1",
"userId2",
"userId3"
]
}
{ "_id" : "userId1", "name" : "Derek", "age" : 34 }
{ "_id" : "userId2", "name" : "Homer", "age" : 44 }
{ "_id" : "userId3", "name" : "Bobby", "age" : 12 }
The result of the query will be:
{
"_id" : ObjectId("573b09e6322304d5e7c6256e"),
"friends" : [
"userId3",
"userId1",
"userId2"
],
"friendsData" : [
{
"_id" : "userId3",
"name" : "Bobby",
"age" : 12
},
{
"_id" : "userId1",
"name" : "Derek",
"age" : 34
},
{
"_id" : "userId2",
"name" : "Homer",
"age" : 44
}
]
}

Edit: this answer only applies to versions of MongoDb prior to v3.2.
You can't do what you want in just one query. You would have to first retrieve the list of friend user ids, then pass those ids to the second query to retrieve the documents and sort them by age.
var user = db.user.findOne({"id" : "001"}, {"friends": 1})
db.user.find( {"id" : {$in : user.friends }}).sort("age" : 1);

https://docs.mongodb.org/manual/reference/operator/aggregation/lookup/
This is the doc for join query in mongodb , this is new feature from version 3.2.
So this will be helpful.

You can use in Moongoose JS .populate() and { populate : { path : 'field' } }.
Example:
Models:
mongoose.model('users', new Schema({
name:String,
status: true,
friends: [{type: Schema.Types.ObjectId, ref:'users'}],
posts: [{type: Schema.Types.ObjectId, ref:'posts'}],
}));
mongoose.model('posts', new Schema({
description: String,
comments: [{type: Schema.Types.ObjectId, ref:'comments'}],
}));
mongoose.model('comments', new Schema({
comment:String,
status: true
}));
If you want to see your friends' posts, you can use this.
Users.find(). //Collection 1
populate({path:'friends', //Collection 2
populate:{path:'posts' //Collection 3
}})
.exec();
If you want to see your friends' posts and also bring all the comments, you can use this and too, you can indentify the collection if this not find and the query is wrong.
Users.find(). //Collection 1
populate({path:'friends', //Collection 2
populate:{path:'posts', //Collection 3
populate:{path:'commets, model:Collection'//Collection 4 and more
}}})
.exec();
And to finish, if you want get only some fields of some Collection, you can use the propiertie select Example:
Users.find().
populate({path:'friends', select:'name status friends'
populate:{path:'comments'
}})
.exec();

MongoDB doesn't have joins, but in your case you can do:
db.coll.find({friends: userId}).sort({age: -1})

one kind of join a query in mongoDB, is ask at one collection for id that match , put ids in a list (idlist) , and do find using on other (or same) collection with $in : idlist
u = db.friends.find({"friends": ? }).toArray()
idlist= []
u.forEach(function(myDoc) { idlist.push(myDoc.id ); } )
db.friends.find({"id": {$in : idlist} } )

Only populate array friends.
User.findOne({ _id: "userId"})
.populate('friends')
.exec((err, user) => {
//do something
});
Result is same like this:
{
"_id" : "userId",
"name" : "John",
"age" : 30,
"friends" : [
{ "_id" : "userId1", "name" : "Derek", "age" : 34 }
{ "_id" : "userId2", "name" : "Homer", "age" : 44 }
{ "_id" : "userId3", "name" : "Bobby", "age" : 12 }
]
}
Same this: Mongoose - using Populate on an array of ObjectId

You can use playOrm to do what you want in one Query(with S-SQL Scalable SQL).

var p = db.sample1.find().limit(2) ,
h = [];
for (var i = 0; i < p.length(); i++)
{
h.push(p[i]['name']);
}
db.sample2.find( { 'doc_name': { $in : h } } );
it works for me.

You can do it in one go using mongo-join-query. Here is how it would look like:
const joinQuery = require("mongo-join-query");
joinQuery(
mongoose.models.User,
{
find: {},
populate: ["friends"],
sort: { age: 1 },
},
(err, res) => (err ? console.log("Error:", err) : console.log("Success:", res.results))
);
The result will have your users ordered by age and all of the friends objects embedded.
How does it work?
Behind the scenes mongo-join-query will use your Mongoose schema to determine which models to join and will create an aggregation pipeline that will perform the join and the query.

Related

firebase database filtering is not working. Need some assistance

I have a simple test database that I cant get to filter. I indexed the category in the rules:
"questions":{
".indexOn": ["category"]
},
My filter for the quiz app:
/questions.json?orderBy="category"&equalTo="Basics"&print=pretty
and my database:
"-MKoucSP33zm4jC43AnY" : {
"title" : {
"answers" : [ {
"score" : 30,
"text" : "Pineapple"
}, {
"score" : 5,
"text" : "Ham"
}, {
"score" : 20,
"text" : "Yogurt"
}, {
"score" : 10,
"text" : "Crab"
} ],
"category" : "Basics",
"questionId" : "101",
"questionImage" : "",
"questionLink" : "",
"questionText" : "What topping do you like the best on pizza?"
}
}
The category property is nested under the title node, so the property you need to order/filter on is title/category:
/questions.json?orderBy="title/category"&equalTo="Basics"&print=pretty
You'll also need to update your index definition for that path, so:
"questions": { ".indexOn": "title/category" }
Working example: https://stackoverflow.firebaseio.com/64596200/questions.json?orderBy="title/category"&equalTo="Basics"

Firebase, get property by email

I have this database on Firebase:
{
"issues" : {
"-L04771_EjrLlv5u1-GU" : {
"issue" : "Test insert 1",
"last_edit" : "d8QICgTG5xR20RBzAXfzfu8gLgw2",
"owner" : "d8QICgTG5xR20RBzAXfzfu8gLgw2",
"owner_email" : "example1#gmail.com",
"status" : 1,
"url" : "http://www.example.com/example.html"
},
"-L047pIoqxkj4saaTYyQ" : {
"issue" : "Test insert 2",
"last_edit" : "d8QICgTG5xR20RBzAXfzfu8gLgw2",
"owner" : "d8QICgTG5xR20RBzAXfzfu8gLgw2",
"owner_email" : "example2#gmail.com",
"status" : 1,
"url" : "http://www.example.com/example.html"
}
}
}
I have to extract only those who have owner_email "example1#gmail.com".
Is possible?
You're probably looking for Firebase Queries and especially the equalTo-Query.
Your code would be something like this:
// Find all issues with owner_email = example1#gmail.com
var ref = firebase.database().ref("issues");
ref.orderByChild("owner_email").equalTo("example1#gmail.com").on("value", function(snapshot) {
// Loops through the matching issues
snapshot.forEach(function(child) {
console.log(child.key);
});
});

Query object in child array

I am trying to use a query to retrieve data of student who are only 4 years old from my database but i cant figure out how to use Firebase to query the age (The array in each child).
{
"student" : {
"-Kv2RVDDsI-v6V7g_LBn" : [ {
"name" : "sam",
"age" : "6"
}, {
"name" : "tom",
"age" : "4"
}
],
"-hguyu-v6V7g_LBn" : [ {
"name" : "Tim",
"age" : "12"
}, {
"name" : "tom",
"age" : "4"
}
]
}
}
This is my code but it does no return anything.
ref.child("student").queryOrdered(byChild: "age").queryEqual(toValue: 4).observe(.childAdded, with: { (snapshot) in
let value = snapshot.value
print(value)
}) { (error) in
print(error.localizedDescription)
}
However, if i remove the queryOrdered(byChild: "age") it works.
Thanks.
You are referencing "Student" when actually your JSON database format child path name is "Students".
Also make sure that you are querying by key. Refer to the Firebase documentation to see how to query by key. I don't understand your structure though why you have 2 children per key.
Why not have a new key for each child?

Rails & Mongoid unique results

Consider the following example of mongo collection:
{"_id" : ObjectId("4f304818884672067f000001"), "hash" : {"call_id" : "1234"}, "something" : "AAA"}
{"_id" : ObjectId("4f304818884672067f000002"), "hash" : {"call_id" : "1234"}, "something" : "BBB"}
{"_id" : ObjectId("4f304818884672067f000003"), "hash" : {"call_id" : "1234"}, "something" : "CCC"}
{"_id" : ObjectId("4f304818884672067f000004"), "hash" : {"call_id" : "5555"}, "something" : "DDD"}
{"_id" : ObjectId("4f304818884672067f000005"), "hash" : {"call_id" : "5555"}, "something" : "CCC"}
I would like to query this collection and get only the first entry for each "call_id", in other words i'm trying to get unique results based on "call_id".
I tried to use .distinct method:
#result = Myobject.all.distinct('hash.call_id')
but the resulting array will contain only the unique call_id fields:
["1234", "5555"]
and I need all the other fields too.
Is it possible to make a query like this one?:
#result = Myobject.where('hash.call_id' => Myobject.all.distinct('hash.call_id'))
Thanks
You cannot simply return the document(or subset) by using the distinct. As per the documentation it only returns the distinct array of values based on the given key. But you can achieve this by using map-reduce
var _map = function () {
emit(this.hash.call_id, {doc:this});
}
var _reduce = function (key, values) {
var ret = {doc:[]};
var doc = {};
values.forEach(function (value) {
if (!doc[value.doc.hash.call_id]) {
ret.doc.push(value.doc);
doc[value.doc.hash.call_id] = true; //make the doc seen, so it will be picked only once
}
});
return ret;
}
The above code is self explanatory, on map function i am grouping it by key hash.call_id and returning the whole doc so it can be processed by reduce funcition.
On reduce function, just loop through the grouped result set and pick only one item from the grouped set (among the multiple duplicate key values - distinct simulation).
Finally create some test data
> db.disTest.insert({hash:{call_id:"1234"},something:"AAA"})
> db.disTest.insert({hash:{call_id:"1234"},something:"BBB"})
> db.disTest.insert({hash:{call_id:"1234"},something:"CCC"})
> db.disTest.insert({hash:{call_id:"5555"},something:"DDD"})
> db.disTest.insert({hash:{call_id:"5555"},something:"EEE"})
> db.disTest.find()
{ "_id" : ObjectId("4f30a27c4d203c27d8f4c584"), "hash" : { "call_id" : "1234" }, "something" : "AAA" }
{ "_id" : ObjectId("4f30a2844d203c27d8f4c585"), "hash" : { "call_id" : "1234" }, "something" : "BBB" }
{ "_id" : ObjectId("4f30a2894d203c27d8f4c586"), "hash" : { "call_id" : "1234" }, "something" : "CCC" }
{ "_id" : ObjectId("4f30a2944d203c27d8f4c587"), "hash" : { "call_id" : "5555" }, "something" : "DDD" }
{ "_id" : ObjectId("4f30a2994d203c27d8f4c588"), "hash" : { "call_id" : "5555" }, "something" : "EEE" }
and running this map reduce
> db.disTest.mapReduce(_map,_reduce, {out: { inline : 1}})
{
"results" : [
{
"_id" : "1234",
"value" : {
"doc" : [
{
"_id" : ObjectId("4f30a27c4d203c27d8f4c584"),
"hash" : {
"call_id" : "1234"
},
"something" : "AAA"
}
]
}
},
{
"_id" : "5555",
"value" : {
"doc" : [
{
"_id" : ObjectId("4f30a2944d203c27d8f4c587"),
"hash" : {
"call_id" : "5555"
},
"something" : "DDD"
}
]
}
}
],
"timeMillis" : 2,
"counts" : {
"input" : 5,
"emit" : 5,
"reduce" : 2,
"output" : 2
},
"ok" : 1,
}
You get the first document of the distinct set. You can do the same in mongoid by first stringify the map/reduce functions and call mapreduce like this
MyObject.collection.mapreduce(_map,_reduce,{:out => {:inline => 1},:raw=>true })
Hope it helps

MongoDB/Mongoid: Can one query for ObjectID in embedded documents?

For the record, I'm a bit of a newbie when it comes to Rails and MongoDB.
I'm using Rails+Mongoid+MongoDB to build an app and I've noticed that Mongoid adds ObjectID to embedded documents for some reason.
Is there any way to query all documents in a collection by ObjectID both the main documents and nested ones?
If I run this command
db.programs.findOne( { _id: ObjectId( "4d1a035cfa87b171e9000002" ) } )
I get these results which is normal since I'm querying for the ObjectID at root level.
{
"_id" : ObjectId("4d1a035cfa87b171e9000002"),
"created_at" : "Tue Dec 28 2010 00:00:00 GMT+0000 (GMT)",
"name" : "program",
"routines" : [
{
"name" : "Day 1",
"_id" : ObjectId("4d1a7689fa87b17f50000020")
},
{
"name" : "Day 2",
"_id" : ObjectId("4d1a7695fa87b17f50000022")
},
{
"name" : "Day 3",
"_id" : ObjectId("4d1a76acfa87b17f50000024")
},
{
"name" : "Day 4",
"_id" : ObjectId("4d1a76ecfa87b17f50000026")
},
{
"name" : "Day 5",
"_id" : ObjectId("4d1a7708fa87b17f50000028")
},
{
"name" : "Day 6",
"_id" : ObjectId("4d1a7713fa87b17f5000002a")
},
{
"name" : "Day 7",
"_id" : ObjectId("4d1a7721fa87b17f5000002c")
}
],
"user_id" : ObjectId("4d190cdbfa87b15c2900000a")
}
Now if I try to query with an ObjectID with one of the embedded document (routines) I get null like so.
db.programs.findOne( { _id: ObjectId( "4d1a7689fa87b17f50000020" ) } )
null
I know one can query embedded objects like so
db.postings.find( { "author.name" : "joe" } );
but that seems a bit redundant if you get passed an ObjectID of some sort and want to find in what document that ObjectID resides.
So I guess my question is this...
Is it possible, with some method I'm not familiar with, to query by ObjectID and search the ObjectID's in the embedded documents?
Thanks.
You can't query ObjectIDs globally like that. You would have to do
db.programs.find({"routines._id": ObjectId("4d1a7689fa87b17f50000020")})
no, you can search only by field like { "routines._id" : ObjectId("4d1a7689fa87b17f50000020")}
If you want to get the matched sub-document only you can use $elemMatch with '$' operator like below:
db.programs.find({"_id" : ObjectId("4d1a035cfa87b171e9000002"),
routines:{$elemMatch:{"_id" : ObjectId("4d1a7689fa87b17f50000020")}}},{"routines.$":1})
It will return you only that matched sub-document instead of the complete sub-document.

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