Cursor Pagination using ObjectID / Timestamp field in MongoDB - mongodb

I sort and paginate records by a last_message_id field which takes an ObjectID (not timestamp), though both of them are the same if we deep dive.
The problem is some of the records may be ignored in pagination if we paginate using timestamp, because timestamps are not unique and they can be duplicate.
In my project there are two collections, "Rooms" and "Messages". Whenever a user sends a message, The associated room of the message will be updated with the _id of the message.
I do this because i want to show recently active rooms to users.
Is there a solution to the problem?
Rooms:
_id
room_name
last_message_id
x
General
612a8e6ab075cf9f2b8c6f9d
y
Politics
612a8e6ab075cf9f2b8c6f9e
Messages:
_id
room_id
text
612a8e6ab075cf9f2b8c6f9e
y
...
612a8e6ab075cf9f2b8c6f9d
x
...

I solved the issue by adding a compound index:
['last_message_id', '_id']
The order of the fields must not be changed.
We can also use a timestamp (created_at) rather than an ObjectId (last_message_id)
Example:
// Pagination result: (limit 1)
[{_id: "613091804ae06dde2e507ff5", last_message_id: "612ff5e11d3130cc1fb1448e"}]
// to get more results:
// http://www.website.com/results?id=613091804ae06dde2e507ff5&last_message_id=612ff5e11d3130cc1fb1448e
// SudoQuery:
let id = req.query.id;
let last_message_id = req.query.last_message_id;
let query = {};
if (id && last_message_id) {
query = {
$or: [
{
last_message_id: {
$lt: last_message_id
}
},
{
last_message_id,
_id: {
$lt: id
}
}
]
};
}
db.items.find(query).sort({last_message_id: -1, _id: -1}).limit(1);

Related

How to get the particuler object fields using ReativeMongo without a case class [duplicate]

In my MongoDB, I have a student collection with 10 records having fields name and roll. One record of this collection is:
{
"_id" : ObjectId("53d9feff55d6b4dd1171dd9e"),
"name" : "Swati",
"roll" : "80",
}
I want to retrieve the field roll only for all 10 records in the collection as we would do in traditional database by using:
SELECT roll FROM student
I went through many blogs but all are resulting in a query which must have WHERE clause in it, for example:
db.students.find({ "roll": { $gt: 70 })
The query is equivalent to:
SELECT * FROM student WHERE roll > 70
My requirement is to find a single key only without any condition. So, what is the query operation for that.
From the MongoDB docs:
A projection can explicitly include several fields. In the following operation, find() method returns all documents that match the query. In the result set, only the item and qty fields and, by default, the _id field return in the matching documents.
db.inventory.find( { type: 'food' }, { item: 1, qty: 1 } )
In this example from the folks at Mongo, the returned documents will contain only the fields of item, qty, and _id.
Thus, you should be able to issue a statement such as:
db.students.find({}, {roll:1, _id:0})
The above statement will select all documents in the students collection, and the returned document will return only the roll field (and exclude the _id).
If we don't mention _id:0 the fields returned will be roll and _id. The '_id' field is always displayed by default. So we need to explicitly mention _id:0 along with roll.
get all data from table
db.student.find({})
SELECT * FROM student
get all data from table without _id
db.student.find({}, {_id:0})
SELECT name, roll FROM student
get all data from one field with _id
db.student.find({}, {roll:1})
SELECT id, roll FROM student
get all data from one field without _id
db.student.find({}, {roll:1, _id:0})
SELECT roll FROM student
find specified data using where clause
db.student.find({roll: 80})
SELECT * FROM students WHERE roll = '80'
find a data using where clause and greater than condition
db.student.find({ "roll": { $gt: 70 }}) // $gt is greater than
SELECT * FROM student WHERE roll > '70'
find a data using where clause and greater than or equal to condition
db.student.find({ "roll": { $gte: 70 }}) // $gte is greater than or equal
SELECT * FROM student WHERE roll >= '70'
find a data using where clause and less than or equal to condition
db.student.find({ "roll": { $lte: 70 }}) // $lte is less than or equal
SELECT * FROM student WHERE roll <= '70'
find a data using where clause and less than to condition
db.student.find({ "roll": { $lt: 70 }}) // $lt is less than
SELECT * FROM student WHERE roll < '70'
I think mattingly890 has the correct answer , here is another example along with the pattern/commmand
db.collection.find( {}, {your_key:1, _id:0})
> db.mycollection.find().pretty();
{
"_id": ObjectId("54ffca63cea5644e7cda8e1a"),
"host": "google",
"ip": "1.1.192.1"
}
db.mycollection.find({},{ "_id": 0, "host": 1 }).pretty();
Here you go , 3 ways of doing , Shortest to boring :
db.student.find({}, 'roll _id'); // <--- Just multiple fields name space separated
// OR
db.student.find({}).select('roll _id'); // <--- Just multiple fields name space separated
// OR
db.student.find({}, {'roll' : 1 , '_id' : 1 ); // <---- Old lengthy boring way
To remove specific field use - operator :
db.student.find({}).select('roll -_id') // <--- Will remove id from result
While gowtham's answer is complete, it is worth noting that those commands may differ from on API to another (for those not using mongo's shell).
Please refer to documentation link for detailed info.
Nodejs, for instance, have a method called `projection that you would append to your find function in order to project.
Following the same example set, commands like the following can be used with Node:
db.student.find({}).project({roll:1})
SELECT _id, roll FROM student
Or
db.student.find({}).project({roll:1, _id: 0})
SELECT roll FROM student
and so on.
Again for nodejs users, do not forget (what you should already be familiar with if you used this API before) to use toArray in order to append your .then command.
Try the following query:
db.student.find({}, {roll: 1, _id: 0});
And if you are using console you can add pretty() for making it easy to read.
db.student.find({}, {roll: 1, _id: 0}).pretty();
Hope this helps!!
Just for educational purposes you could also do it with any of the following ways:
1.
var query = {"roll": {$gt: 70};
var cursor = db.student.find(query);
cursor.project({"roll":1, "_id":0});
2.
var query = {"roll": {$gt: 70};
var projection = {"roll":1, "_id":0};
var cursor = db.student.find(query,projection);
`
db.<collection>.find({}, {field1: <value>, field2: <value> ...})
In your example, you can do something like:
db.students.find({}, {"roll":true, "_id":false})
Projection
The projection parameter determines which fields are returned in the
matching documents. The projection parameter takes a document of the
following form:
{ field1: <value>, field2: <value> ... }
The <value> can be any of the following:
1 or true to include the field in the return documents.
0 or false to exclude the field.
NOTE
For the _id field, you do not have to explicitly specify _id: 1 to
return the _id field. The find() method always returns the _id field
unless you specify _id: 0 to suppress the field.
READ MORE
For better understanding I have written similar MySQL query.
Selecting specific fields
MongoDB : db.collection_name.find({},{name:true,email:true,phone:true});
MySQL : SELECT name,email,phone FROM table_name;
Selecting specific fields with where clause
MongoDB : db.collection_name.find({email:'you#email.com'},{name:true,email:true,phone:true});
MySQL : SELECT name,email,phone FROM table_name WHERE email = 'you#email.com';
This works for me,
db.student.find({},{"roll":1})
no condition in where clause i.e., inside first curly braces.
inside next curly braces: list of projection field names to be needed in the result and 1 indicates particular field is the part of the query result
getting name of the student
student-details = db.students.find({{ "roll": {$gt: 70} },{"name": 1, "_id": False})
getting name & roll of the student
student-details = db.students.find({{ "roll": {$gt: 70}},{"name": 1,"roll":1,"_id": False})
I just want to add to the answers that if you want to display a field that is nested in another object, you can use the following syntax
db.collection.find( {}, {{'object.key': true}})
Here key is present inside the object named object
{ "_id" : ObjectId("5d2ef0702385"), "object" : { "key" : "value" } }
var collection = db.collection('appuser');
collection.aggregate(
{ $project : { firstName : 1, lastName : 1 } },function(err, res){
res.toArray(function(err, realRes){
console.log("response roo==>",realRes);
});
});
it's working
Use the Query like this in the shell:
1. Use database_name
e.g: use database_name
2. Which returns only assets particular field information when matched , _id:0 specifies not to display ID in the result
db.collection_name.find( { "Search_Field": "value" },
{ "Field_to_display": 1,_id:0 } )
If u want to retrieve the field "roll" only for all 10 records in the collections.
Then try this.
In MongoDb :
db.students.find( { } , { " roll " : { " $roll " })
In Sql :
select roll from students
The query for MongoDB here fees is collection and description is a field.
db.getCollection('fees').find({},{description:1,_id:0})
Apart from what people have already mentioned I am just introducing indexes to the mix.
So imagine a large collection, with let's say over 1 million documents and you have to run a query like this.
The WiredTiger Internal cache will have to keep all that data in the cache if you have to run this query on it, if not that data will be fed into the WT Internal Cache either from FS Cache or Disk before the retrieval from DB is done (in batches if being called for from a driver connected to database & given that 1 million documents are not returned in 1 go, cursor comes into play)
Covered query can be an alternative. Copying the text from docs directly.
When an index covers a query, MongoDB can both match the query conditions and return the results using only the index keys; i.e. MongoDB does not need to examine documents from the collection to return the results.
When an index covers a query, the explain result has an IXSCAN stage that is not a descendant of a FETCH stage, and in the executionStats, the totalDocsExamined is 0.
Query : db.getCollection('qaa').find({roll_no : {$gte : 0}},{_id : 0, roll_no : 1})
Index : db.getCollection('qaa').createIndex({roll_no : 1})
If the index here is in WT Internal Cache then it would be a straight forward process to get the values. An index has impact on the write performance of the system thus this would make more sense if the reads are a plenty compared to the writes.
If you are using the MongoDB driver in NodeJs then the above-mentioned answers might not work for you. You will have to do something like this to get only selected properties as a response.
import { MongoClient } from "mongodb";
// Replace the uri string with your MongoDB deployment's connection string.
const uri = "<connection string uri>";
const client = new MongoClient(uri);
async function run() {
try {
await client.connect();
const database = client.db("sample_mflix");
const movies = database.collection("movies");
// Query for a movie that has the title 'The Room'
const query = { title: "The Room" };
const options = {
// sort matched documents in descending order by rating
sort: { "imdb.rating": -1 },
// Include only the `title` and `imdb` fields in the returned document
projection: { _id: 0, title: 1, imdb: 1 },
};
const movie = await movies.findOne(query, options);
/** since this method returns the matched document, not a cursor,
* print it directly
*/
console.log(movie);
} finally {
await client.close();
}
}
run().catch(console.dir);
This code is copied from the actual MongoDB doc you can check here.
https://docs.mongodb.com/drivers/node/current/usage-examples/findOne/
db.student.find({}, {"roll":1, "_id":0})
This is equivalent to -
Select roll from student
db.student.find({}, {"roll":1, "name":1, "_id":0})
This is equivalent to -
Select roll, name from student
In mongodb 3.4 we can use below logic, i am not sure about previous versions
select roll from student ==> db.student.find(!{}, {roll:1})
the above logic helps to define some columns (if they are less)
Using Studio 3T for MongoDB, if I use .find({}, { _id: 0, roll: true }) it still return an array of objects with an empty _id property.
Using JavaScript map helped me to only retrieve the desired roll property as an array of string:
var rolls = db.student
.find({ roll: { $gt: 70 } }) // query where role > 70
.map(x => x.roll); // return an array of role
Not sure this answers the question but I believe it's worth mentioning here.
There is one more way for selecting single field (and not multiple) using db.collection_name.distinct();
e.g.,db.student.distinct('roll',{});
Or, 2nd way: Using db.collection_name.find().forEach(); (multiple fields can be selected here by concatenation)
e.g., db.collection_name.find().forEach(function(c1){print(c1.roll);});
_id = "123321"; _user = await likes.find({liker_id: _id},{liked_id:"$liked_id"}); ;
let suppose you have liker_id and liked_id field in the document so by putting "$liked_id" it will return _id and liked_id only.
For Single Update :
db.collection_name.update({ field_name_1: ("value")}, { $set: { field_name_2 : "new_value" }});
For MultiUpdate :
db.collection_name.updateMany({ field_name_1: ("value")}, { $set: {field_name_2 : "new_value" }});
Make sure indexes are proper.

How to make a query without nested document in MongoDB? [duplicate]

In my MongoDB, I have a student collection with 10 records having fields name and roll. One record of this collection is:
{
"_id" : ObjectId("53d9feff55d6b4dd1171dd9e"),
"name" : "Swati",
"roll" : "80",
}
I want to retrieve the field roll only for all 10 records in the collection as we would do in traditional database by using:
SELECT roll FROM student
I went through many blogs but all are resulting in a query which must have WHERE clause in it, for example:
db.students.find({ "roll": { $gt: 70 })
The query is equivalent to:
SELECT * FROM student WHERE roll > 70
My requirement is to find a single key only without any condition. So, what is the query operation for that.
From the MongoDB docs:
A projection can explicitly include several fields. In the following operation, find() method returns all documents that match the query. In the result set, only the item and qty fields and, by default, the _id field return in the matching documents.
db.inventory.find( { type: 'food' }, { item: 1, qty: 1 } )
In this example from the folks at Mongo, the returned documents will contain only the fields of item, qty, and _id.
Thus, you should be able to issue a statement such as:
db.students.find({}, {roll:1, _id:0})
The above statement will select all documents in the students collection, and the returned document will return only the roll field (and exclude the _id).
If we don't mention _id:0 the fields returned will be roll and _id. The '_id' field is always displayed by default. So we need to explicitly mention _id:0 along with roll.
get all data from table
db.student.find({})
SELECT * FROM student
get all data from table without _id
db.student.find({}, {_id:0})
SELECT name, roll FROM student
get all data from one field with _id
db.student.find({}, {roll:1})
SELECT id, roll FROM student
get all data from one field without _id
db.student.find({}, {roll:1, _id:0})
SELECT roll FROM student
find specified data using where clause
db.student.find({roll: 80})
SELECT * FROM students WHERE roll = '80'
find a data using where clause and greater than condition
db.student.find({ "roll": { $gt: 70 }}) // $gt is greater than
SELECT * FROM student WHERE roll > '70'
find a data using where clause and greater than or equal to condition
db.student.find({ "roll": { $gte: 70 }}) // $gte is greater than or equal
SELECT * FROM student WHERE roll >= '70'
find a data using where clause and less than or equal to condition
db.student.find({ "roll": { $lte: 70 }}) // $lte is less than or equal
SELECT * FROM student WHERE roll <= '70'
find a data using where clause and less than to condition
db.student.find({ "roll": { $lt: 70 }}) // $lt is less than
SELECT * FROM student WHERE roll < '70'
I think mattingly890 has the correct answer , here is another example along with the pattern/commmand
db.collection.find( {}, {your_key:1, _id:0})
> db.mycollection.find().pretty();
{
"_id": ObjectId("54ffca63cea5644e7cda8e1a"),
"host": "google",
"ip": "1.1.192.1"
}
db.mycollection.find({},{ "_id": 0, "host": 1 }).pretty();
Here you go , 3 ways of doing , Shortest to boring :
db.student.find({}, 'roll _id'); // <--- Just multiple fields name space separated
// OR
db.student.find({}).select('roll _id'); // <--- Just multiple fields name space separated
// OR
db.student.find({}, {'roll' : 1 , '_id' : 1 ); // <---- Old lengthy boring way
To remove specific field use - operator :
db.student.find({}).select('roll -_id') // <--- Will remove id from result
While gowtham's answer is complete, it is worth noting that those commands may differ from on API to another (for those not using mongo's shell).
Please refer to documentation link for detailed info.
Nodejs, for instance, have a method called `projection that you would append to your find function in order to project.
Following the same example set, commands like the following can be used with Node:
db.student.find({}).project({roll:1})
SELECT _id, roll FROM student
Or
db.student.find({}).project({roll:1, _id: 0})
SELECT roll FROM student
and so on.
Again for nodejs users, do not forget (what you should already be familiar with if you used this API before) to use toArray in order to append your .then command.
Try the following query:
db.student.find({}, {roll: 1, _id: 0});
And if you are using console you can add pretty() for making it easy to read.
db.student.find({}, {roll: 1, _id: 0}).pretty();
Hope this helps!!
Just for educational purposes you could also do it with any of the following ways:
1.
var query = {"roll": {$gt: 70};
var cursor = db.student.find(query);
cursor.project({"roll":1, "_id":0});
2.
var query = {"roll": {$gt: 70};
var projection = {"roll":1, "_id":0};
var cursor = db.student.find(query,projection);
`
db.<collection>.find({}, {field1: <value>, field2: <value> ...})
In your example, you can do something like:
db.students.find({}, {"roll":true, "_id":false})
Projection
The projection parameter determines which fields are returned in the
matching documents. The projection parameter takes a document of the
following form:
{ field1: <value>, field2: <value> ... }
The <value> can be any of the following:
1 or true to include the field in the return documents.
0 or false to exclude the field.
NOTE
For the _id field, you do not have to explicitly specify _id: 1 to
return the _id field. The find() method always returns the _id field
unless you specify _id: 0 to suppress the field.
READ MORE
For better understanding I have written similar MySQL query.
Selecting specific fields
MongoDB : db.collection_name.find({},{name:true,email:true,phone:true});
MySQL : SELECT name,email,phone FROM table_name;
Selecting specific fields with where clause
MongoDB : db.collection_name.find({email:'you#email.com'},{name:true,email:true,phone:true});
MySQL : SELECT name,email,phone FROM table_name WHERE email = 'you#email.com';
This works for me,
db.student.find({},{"roll":1})
no condition in where clause i.e., inside first curly braces.
inside next curly braces: list of projection field names to be needed in the result and 1 indicates particular field is the part of the query result
getting name of the student
student-details = db.students.find({{ "roll": {$gt: 70} },{"name": 1, "_id": False})
getting name & roll of the student
student-details = db.students.find({{ "roll": {$gt: 70}},{"name": 1,"roll":1,"_id": False})
I just want to add to the answers that if you want to display a field that is nested in another object, you can use the following syntax
db.collection.find( {}, {{'object.key': true}})
Here key is present inside the object named object
{ "_id" : ObjectId("5d2ef0702385"), "object" : { "key" : "value" } }
var collection = db.collection('appuser');
collection.aggregate(
{ $project : { firstName : 1, lastName : 1 } },function(err, res){
res.toArray(function(err, realRes){
console.log("response roo==>",realRes);
});
});
it's working
Use the Query like this in the shell:
1. Use database_name
e.g: use database_name
2. Which returns only assets particular field information when matched , _id:0 specifies not to display ID in the result
db.collection_name.find( { "Search_Field": "value" },
{ "Field_to_display": 1,_id:0 } )
If u want to retrieve the field "roll" only for all 10 records in the collections.
Then try this.
In MongoDb :
db.students.find( { } , { " roll " : { " $roll " })
In Sql :
select roll from students
The query for MongoDB here fees is collection and description is a field.
db.getCollection('fees').find({},{description:1,_id:0})
Apart from what people have already mentioned I am just introducing indexes to the mix.
So imagine a large collection, with let's say over 1 million documents and you have to run a query like this.
The WiredTiger Internal cache will have to keep all that data in the cache if you have to run this query on it, if not that data will be fed into the WT Internal Cache either from FS Cache or Disk before the retrieval from DB is done (in batches if being called for from a driver connected to database & given that 1 million documents are not returned in 1 go, cursor comes into play)
Covered query can be an alternative. Copying the text from docs directly.
When an index covers a query, MongoDB can both match the query conditions and return the results using only the index keys; i.e. MongoDB does not need to examine documents from the collection to return the results.
When an index covers a query, the explain result has an IXSCAN stage that is not a descendant of a FETCH stage, and in the executionStats, the totalDocsExamined is 0.
Query : db.getCollection('qaa').find({roll_no : {$gte : 0}},{_id : 0, roll_no : 1})
Index : db.getCollection('qaa').createIndex({roll_no : 1})
If the index here is in WT Internal Cache then it would be a straight forward process to get the values. An index has impact on the write performance of the system thus this would make more sense if the reads are a plenty compared to the writes.
If you are using the MongoDB driver in NodeJs then the above-mentioned answers might not work for you. You will have to do something like this to get only selected properties as a response.
import { MongoClient } from "mongodb";
// Replace the uri string with your MongoDB deployment's connection string.
const uri = "<connection string uri>";
const client = new MongoClient(uri);
async function run() {
try {
await client.connect();
const database = client.db("sample_mflix");
const movies = database.collection("movies");
// Query for a movie that has the title 'The Room'
const query = { title: "The Room" };
const options = {
// sort matched documents in descending order by rating
sort: { "imdb.rating": -1 },
// Include only the `title` and `imdb` fields in the returned document
projection: { _id: 0, title: 1, imdb: 1 },
};
const movie = await movies.findOne(query, options);
/** since this method returns the matched document, not a cursor,
* print it directly
*/
console.log(movie);
} finally {
await client.close();
}
}
run().catch(console.dir);
This code is copied from the actual MongoDB doc you can check here.
https://docs.mongodb.com/drivers/node/current/usage-examples/findOne/
db.student.find({}, {"roll":1, "_id":0})
This is equivalent to -
Select roll from student
db.student.find({}, {"roll":1, "name":1, "_id":0})
This is equivalent to -
Select roll, name from student
In mongodb 3.4 we can use below logic, i am not sure about previous versions
select roll from student ==> db.student.find(!{}, {roll:1})
the above logic helps to define some columns (if they are less)
Using Studio 3T for MongoDB, if I use .find({}, { _id: 0, roll: true }) it still return an array of objects with an empty _id property.
Using JavaScript map helped me to only retrieve the desired roll property as an array of string:
var rolls = db.student
.find({ roll: { $gt: 70 } }) // query where role > 70
.map(x => x.roll); // return an array of role
Not sure this answers the question but I believe it's worth mentioning here.
There is one more way for selecting single field (and not multiple) using db.collection_name.distinct();
e.g.,db.student.distinct('roll',{});
Or, 2nd way: Using db.collection_name.find().forEach(); (multiple fields can be selected here by concatenation)
e.g., db.collection_name.find().forEach(function(c1){print(c1.roll);});
_id = "123321"; _user = await likes.find({liker_id: _id},{liked_id:"$liked_id"}); ;
let suppose you have liker_id and liked_id field in the document so by putting "$liked_id" it will return _id and liked_id only.
For Single Update :
db.collection_name.update({ field_name_1: ("value")}, { $set: { field_name_2 : "new_value" }});
For MultiUpdate :
db.collection_name.updateMany({ field_name_1: ("value")}, { $set: {field_name_2 : "new_value" }});
Make sure indexes are proper.

MongoDB REST API dynamic data pagination

I am implementing REST API with Express and Mongoose. I am stuck with a task to paginate results for a collection with large number of documents. Document is a social media post which has fields like: created_at, updated_at and score.
Ordinary paging implementation using limit and page number doesn't work because the nature of data. Posts can be removed or added while user is browsing the results.
I could implement range-based pagination as suggested in MongoDB documentation.
Last items in a page created_at value could be used to query next page.
Post.find({ created_at: { $gt: req.query.next } })
But what if I want to sort results by score which is not unique or all scores are the same?
There are multiple ways to do this, but in your case i would suggest to create the find / sort conditions dynamicly and use them in your find statement.
Outta my head just to give you an idea:
var MyFilter = {};
var MySort = {};
if (req.query.created_at != undefined) {
MyFilter = { created_at : { $gte : req.query.created_at) } };
}
if (req.query.sortOn != undefined){
MySort = { req.query.sort : 1 };
}
Post.find(MyFilter)
.sort(MySort)
.exec(function (err, posts) {
===do your stuff===
});
https://groups.google.com/forum/#!searchin/mongodb-user/paging%7Csort:relevance/mongodb-user/_Jrv9vI1lDE/Wg0eGV2AkZYJ
I just found the same topic in google groups with perfect answer.
You could try creating a compound index on age and _id, and sort by both fields to get the first three documents. To get the subsequent set of users, your query should return documents that satisfy either of the following requirements:
have the same age as the last document and have a greater _id
or
have a greater age
For example (lastAge and lastID are the age and _id of the latest user):
db.collection.find( { $or : [ { age : lastAge, _id : {$gt : lastID } } , { age : {$gt : lastID } } ] } )

documents with tags in mongodb: getting tag counts

I have a collection1 of documents with tags in MongoDB. The tags are an embedded array of strings:
{
name: 'someObj',
tags: ['tag1', 'tag2', ...]
}
I want to know the count of each tag in the collection. Therefore I have another collection2 with tag counts:
{
{
tag: 'tag1',
score: 2
}
{
tag: 'tag2',
score: 10
}
}
Now I have to keep both in sync. It is rather trivial when inserting to or removing from collection1. However when I update collection1 I do the following:
1.) get the old document
var oldObj = collection1.find({ _id: id });
2.) calculate the difference between old and new tag arrays
var removedTags = $(oldObj.tags).not(obj.tags).get();
var insertedTags = $(obj.tags).not(oldObj.tags).get();
3.) update the old document
collection1.update(
{ _id: id },
{ $set: obj }
);
4.) update the scores of inserted & removed tags
// increment score of each inserted tag
insertedTags.forEach(function(val, idx) {
// $inc will set score = 1 on insert
collection2.update(
{ tag: val },
{ $inc: { score: 1 } },
{ upsert: true }
)
});
// decrement score of each removed tag
removedTags.forEach(function(val, idx) {
// $inc will set score = -1 on insert
collection2.update(
{ tag: val },
{ $inc: { score: -1 } },
{ upsert: true }
)
});
My questions:
A) Is this approach of keeping book of scores separately efficient? Or is there a more efficient one-time query to get the scores from collection1?
B) Even if keeping book separately is the better choice: can that be done in less steps, e.g. letting mongoDB calculate what tags are new / removed?
The solution, as nickmilion correctly states, would be an aggregation. Though I would do it with a nack: we'll save its results in a collection. What will do is to trade real time results for an extreme speed boost.
How I would do it
More often than not, the need for real time results is overestimated. Hence, I'd go with precalculated stats for the tags and renew it every 5 minutes or so. That should be well enough, since most of such calls are requested async by the client and hence some delay in case the calculation has to be made on a specific request is negligible.
db.tags.aggregate(
{$unwind:"$tags"},
{$group: { _id:"$tags", score:{"$sum":1} } },
{$out:"tagStats"}
)
db.tagStats.update(
{'lastRun':{$exists:true}},
{'lastRun':new Date()},
{upsert:true}
)
db.tagStats.ensureIndex({lastRun:1}, {sparse:true})
Ok, here is the deal. First, we unwind the tags array, group it by the individual tags and increment the score for each occurrence of the respective tag. Next, we upsert lastRun in the tagStats collection, which we can do since MongoDB is schemaless. Next, we create a sparse index, which only holds values for documents in which the indexed field exists. In case the index already exists, ensureIndex is an extremely cheap query; however, since we are going to use that query in our code, we don't need to create the index manually. With this procedure, the following query
db.tagStats.find(
{lastRun:{ $lte: new Date( ISODate().getTime() - 300000 ) } },
{_id:0, lastRun:1}
)
becomes a covered query: A query which is answered from the index, which tends to reside in RAM, making this query lightning fast (slightly less than 0.5 msecs median in my tests). So what does this query do? It will return a record when the last run of the aggregation was run more than 5 minutes ( 5*60*1000 = 300000 msecs) ago. Of course, you can adjust this to your needs.
Now, we can wrap it up:
var hasToRun = db.tagStats.find(
{lastRun:{ $lte: new Date( ISODate().getTime() - 300000 ) } },
{_id:0, lastRun:1}
);
if(hasToRun){
db.tags.aggregate(
{$unwind:"$tags"},
{$group: {_id:"$tags", score:{"$sum":1} } },
{$out:"tagStats"}
)
db.tagStats.update(
{'lastRun':{$exists:true}},
{'lastRun':new Date()},
{upsert:true}
);
db.tagStats.ensureIndex({lastRun:1},{sparse:true});
}
// For all stats
var tagsStats = db.tagStats.find({score:{$exists:true}});
// score for a specific tag
var scoreForTag = db.tagStats.find({score:{$exists:true},_id:"tag1"});
Alternative approach
If real time results really matter and you need the stats for all the tags, simply use the aggregation without saving it to another collection:
db.tags.aggregate(
{$unwind:"$tags"},
{$group: { _id:"$tags", score:{"$sum":1} } },
)
If you only need the results for one specific tag at a time, a real time approach could be to use a special index, create a covered query and simply count the results:
db.tags.ensureIndex({tags:1})
var numberOfOccurences = db.tags.find({tags:"tag1"},{_id:0,tags:1}).count();
answering your questions:
B): you don't have to calculate the dif yourself use $addToSet
A): you can get the counts via aggregation framework with a combination of $unwind and $count

mongoose compund unique index does not work

I have following index:
PaymentSchema.index({ driver_id: 1, year: 1, month: 1 },{ unique: true });
So I want this collection to hold just one record for each different combination of fields driver_id, year and month. I want to update collection with upsert option:
var query = {
driver_id: req.params.driver_id,
year: req.params.year,
month: req.params.month,
amount: req.params.old_value
};
var update = {
$set: {
amount: req.params.new_value
}
};
var options = {
upsert: true
};
Payment.update(query,update,options,function(err,rows){
if(err) return next(err);
res.json({});
});
So what I want is to update document with given unique key (driver_id+year+month) and with additional condition amount = .... If query conditions are ok document should be updated - and it works. If document is not found according to this conditions and document with unique index does not exist it is created. But if document with unique index exists (only amount condition is incorrect) then a new document in created with same unique index (driver_id + year + month). It is strange because I declared unique index on those 3 fields (driver_id+year+month) and I can see in mongoshell that there exist two documents with those fields the same...
Solved: I had to restart mongod and delete database (probably reindex option would work too).