I want to calculate the number of users like this in SQL.
select count(*) as nbr_users from users
How can I do the same thing in mongodb?
db.nbr_users.count()
Check out the SQL-Comparision on MongoDB website, It has all the common select statements converted to MongoDB queries -
https://docs.mongodb.com/manual/reference/sql-comparison/#select
Related
I am working one use case poc, where I want to query multiple documents present in the same collection in MongoDB based on some condition? an example would be enough.
Looking for a query similar like SQL select query on two tables(following query) is it possible in MongoDB?
select t1.X, t1.Y_DT,t1.Z,t1.adj,t1.bjc,t1.jbc,t1.mnk,t2.adj1,t2.bjc1,t2.jbc1,t2.mnk1 from test250 t1, test350 t2 where t1.X = t2.X AND t1.Y_DT=t2.Y_DT AND t1.Z = t2.Z;
if possible, any example query greatly helps me out.
I've recently started using MongoDB, and while translating queries, I've stumbled upon a problem when trying to translate MySQL NOT IN query.
What I'm trying to do is find ids that are not present in another collection. I need a way to return how many of these ids there are and then a list of these ids.
In MySQL, these ids are in a table called songs and are named spotify_id. I am checking against a table named artists where the ids are called track_id.
MySQL Query:
SELECT spotify_id FROM songs WHERE spotify_id NOT IN (SELECT artists.track_id FROM artists)
My current code using MongoDB:
track_ids = artists.find({}, {'track_id': 1})
track_ids_list = [d.get('track_id') for d in track_ids]
# Getting amount of ids not present in artists
tracks_num = songs.count_documents({'spotify_id': {'$nin': track_ids_list}})
# Getting tracks not present in artists
tracks = songs.find({'spotify_id': {'$nin': track_ids_list}})
The MongoDB query seems to return a positive number of documents, even when all of the ids should be present in both collections. The MySQL query works perfectly on the same database.
Is there a better way to perform a NOT IN query in MongoDB?
Thank you!
I'm investigating how MongoDB would work for us. One of the most used queries is used to get latest (or from a given time) measurements for each station. There is thousands of stations and each station has tens of thousands of measurements.
So we plan to have one collection for stations and another for measurements.
In SQL we would do the query with
SELECT * FROM measurements
INNER JOIN (
SELECT max(meas_time) station_id
FROM measurements
WHERE meas_time <= 'time_to_query'
GROUP BY station_id
) t2 ON t2.station_id = measurements.station_id
AND t2.meas_time = measurements.meas_time
This returns one measurement for each station, and the measurement is the newest one before the 'time_to_query'.
What query should be used in MongoDB to produce the same result? We are really using Rails and MongoId, but it should not matter.
update:
This question is not about how to perform a JOIN in MongoDB. The fact that in SQL getting the right data out of the table requires a join doesn't necessary mean that in MongoDB we would also need a join. There is only one table used in the query.
We came up with this query
db.measurements.aggregate([{$group:{ _id:{'station_id':"$station_id"}, time:{$max:'$meas_time'}}}]);
with indexes
db.measurements.createIndex({ station_id: 1, meas_time: -1 });
Even though it seems to give the right data it is really slow. Takes roughly a minute to get a bit over 3000 documents from a collection of 65 million.
Just found that MongoDB is not using the index in this query even though we are using the 3.2 version.
I guess worst case solution would be something like this (out of my head):
meassures = []
StationId.all.each do |station|
meassurement = Meassurment.where(station_id: station.id, meas_time <= 'time_to_query').order_by(meas_time: -1).limit(1)
meassures << [station.name, meassurement.measure, ....]
end
It depends on how much time query can take. Data should anyway be indexed by station_id and meas_time.
How much time does the SQL query take?
Mongodb distinct command returns the list of distinct keys in a given document.
Is there any way to retrieve the whole document rather than only keys using mongodb java driver or morphia?
In the documentation it says:
When possible, the distinct command will use an index to find the documents in the query as well as to return the data.
How does this work, when I try using java driver it always returns the list of keys to my queries. And it doesn't take limit and order queries into consideration. Is it possible to give a order by and limit queries into distinct query?
Thanks for the feedback.
For example, we have two collections
users {userId, firstName, lastName}
votes {userId, voteDate}
I need a report of the name of all users which have more than 20 votes a day.
How can I write query to get data from MongoDB?
The easiest way to do this is to cache the number of votes for each user in the user documents. Then you can get the answer with a single query.
If you don't want to do that, the map-reduce the results into a results collection, and query that collection. You can then run incremental map-reduces that only calculate new votes to keep your results up to date: http://www.mongodb.org/display/DOCS/MapReduce#MapReduce-IncrementalMapreduce
You shouldn't really be trying to do joins with Mongo. If you are you've designed your schema in a relational manner.
In this instance I would store the vote as an embedded document on the user.
In some scenarios using embedded documents isn't feasible, and in that situation I would do two database queries and join the results at the client rather than using MapReduce.
I can't provide a fuller answer now, but you should be able to achieve this using MapReduce. The Map step would return the userIds of the users who have more than 20 votes, the reduce step would return the firstName and lastName, I think...have a look here.