This only does the calculations for users created this year - that is harder to do with pre-aggregation. I prefer this over pre-aggregation because of the filtering on the user table. The right indexes are item_sold(sold_user_id) and item_bought(bought_user_id). (select count(*) from item_bought b where u.user_id = b.bought_user_id) (select count(*) from item_sold s where u.user_id = s.sold_user_id), In this case, I prefer the correlated subquery solution (assuming the right indexes are available): SELECT u.user_id, The solution is to pre-aggregate the results or use a correlated subquery in the select. The following Venn diagram displays the PostgreSQL inner join where we. In PostgreSQL, it as a default Join, therefore it is not compulsory to use the Inner Join keyword with the query. If someone has 100 items bought and 200 items sold, then the join produces 20,000 intermediate rows. The PostgreSQL Inner Join is used to return only those records from the tables, which are equivalent to the defined condition and hides other rows and columns. However, the query is doing unnecessary work. LEFT JOIN item_bought ON bought_user_id=user_id LEFT JOIN item_sold ON sold_user_id=user_id In part III of the series, I'll continue with some examples, this time explaining how we can combine multiple joins in the same query.The easy fix to your query is to use distinct: SELECT user_id, COUNT(distinct item_sold_id), COUNT(distinct item_bought_id) This will produce exactly the same SQL and output as before: (:bookings) If you're on rails 6 you could write a more semantic version for this solution using the ActiveRecord's missing method. The SQL: SELECT COUNT(*) FROM "accommodations" LEFT OUTER JOIN "bookings" ON "bookings"."accommodation_id" = "accommodations"."id" WHERE "bookings"."id" IS NULL Accommodation.left_joins(:bookings).where(bookings: ) So knowing that, we can use where to get only the accommodations with a null bookings.id. Because of my work with PostGIS (and FOSS4G) I became friends with Paul Ramsey.We are now co-workers at Crunchy Data and he is helping me up my SQL-fu. I learned just enough SQL to get the queries to return the right answers. So how can we filter that list to return only the no-bookings accommodations? Looking at the left outer join table from the earlier exercise, we can say that what differentiates the two groups is that the accommodations with bookings will have an attribute bookings.id with an integer value but the no-bookings accommodations will have a null value for that same attribute. My introduction to databases and PostgreSQL was for web application development and statistical analysis. We now know that left_joins will include the accommodations that have no bookings but it will also include all the others that do. As already mentioned, avoiding the NULL rows, change LEFT JOIN to JOIN. WHERE b.id IS NULL This extends to multiple LEFT JOINs and multiple tests ANDd together in the WHERE. Let's now say that we only want to get those accommodations that do not have any bookings. If you are trying to avoid the NOT NULL rows, this is the pattern: SELECT. Get all the accommodations with no bookings Will have to be written as: Accommodation.joins('LEFT OUTER JOIN bookings ON bookings.accommodation_id = acommodations.id") This means that, Accommodation.left_joins(:bookings) If you're working with earlier versions of rails you'll have to customize the rails joins method by passing an SQL string with the desired join method. Left_joins and left_outer_joins methods are only available from rails 5.2.3 onwards. Here's our domain model: class Accommodation 2, "Charming House"=>1, "Lisbon Flat"=>0, "Le Petit Chalet"=>1, "Farm House"=>0} Let's again follow along with a couple of exercises: Contrarily to an inner join, the left outer join will always return all the entries of the left table even if the join condition does not find any match on the right table. What is a left outer join and when should it be used?Ī left outer join (or just left join) is used to query a table based on matching and non-matching entries from a related table.
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