CODEWITHSUNDEEP
mysqlsqlruby-on-railsactiverecordsubquery
Dmitri
Dmitri

Reputation: 2457

Rails select subquery (without finder_sql, if possible)

I have an model called Object (doesn't really matter what it is)

It has a default price (column is called "price").

And then there is a Schedule object that allows to override the price for specific dates.

I want to be able to determine the MINIMUM price (which is by definition the MINIMUM between the default and "current" price) during the SQL-query just in order to be able to ORDER BY the calculated minimum price

I want to make my search query as efficient as possible and I was wondering if I can do something like that: 

Object.select("id AS p_id, id, (SELECT MIN(`schedules`.`price`) FROM `schedules` WHERE `schedules`.`object_id` = p_id`) AS objects.min_price").limit(5)

But, it generates an odd SQL that looks like this: 

SELECT `objects`.`id` AS t0_r0, `objects`.`title` AS t0_r1, `objects`.`created_at` AS t0_r2, `objects`.`updated_at` AS t0_r3, `objects`.`preferences` AS t0_r4 ........ (a lot of columns here) ... ` WHERE `objects`.`id` IN (1, 2, 3, 4 ....)

So, as you can see it doesn't work. First of all - it loads all the columns from the objects table, and second of all - it looks horrible.

The reason why I don't want to use finder_sql is that I have a lot of optional parameters and stuff, so using the AR::Relation object is highly preferred prior to fetching the results themselves.

In addition to abovementioned, I have a lot of records in the DB, and I think that loading them all into the memory is not a good idea and that is the main reason why I want to perform this subquery - just to filter-out as many records as possible.

Can someone help me how to do it more efficiently ?

Upvotes: 3

Views: 8560

Answers (1)

PinnyM
PinnyM

Reputation: 35531

You can make this easier if you generate the subquery separately and use a join instead of a correlated subquery:

subquery = Schedule.select('MIN(price) as min_price, object_id')
                   .group(:object_id)
                   .to_sql
Object.joins("JOIN (#{subquery}) schedules ON objects.p_id = schedules.object_id")
      .select('objects.*, schedules.min_price')
      .limit(5)

Upvotes: 18

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