[英]How to convert SQL query to optimized Rails Active Record query?
I have this SQL query that works properly when I run it but how do I represent it and get the same data via ActiveRecord? 我有这个SQL查询,在我运行它时可以正常运行,但是如何表示它并通过ActiveRecord获取相同的数据?
select concat(o.code, ' - ', u.name)
from users u join user_organizations uo on u.id = uo.user_id
join organizations o on uo.organization_id = o.id
where u.approved = true
and u.deleted_at is null
and o.deleted_at is null
and u.type = 'Banker'
order by o.code asc;
I tried this 我试过了
Banker.joins(:user_organizations => :organization)
.where(:approved => true)
.select("concat(organizations.code, users.name)")
.order("organizations.code asc")
but it didn't work and I had expected it to work. 但它没有用,我曾期望它能用。
There a couple of things to address and open questions which make it difficult to answer properly 有几件事情要解决和提出问题,使他们很难正确回答
Banker
and Organization
? 您如何在Banker
和Organization
之间建立关系? deleted_at is NULL
)? 您是否为该软删除功能使用了一些gem( deleted_at is NULL
)? (eg paranoia ) (例如妄想症 ) I assume that your setup looks sth like this 我认为您的设置看起来像这样
class User < ApplicationRecord
has_and_belongs_to_many :organizations
end
class Organization < ApplicationRecord
has_and_belongs_to_many :users
end
If so you should be able to do the following 如果是这样,您应该能够执行以下操作
User.select("concat(organizations.code, ' - ', users.name)")
.includes(:organizations)
.where('users.deleted_at is not null')
.where('organizations.deleted_at is not null')
.where('users.type = ?', 'Banker')
.order('organizations.code ASC')
In case you are using the paranoia gem mentioned above for users and organizations the *.deleted_at is not null
query parts will be added automatically which reduces the ruby query to sth like this. 如果您对用户和组织使用上面提到的偏执狂宝石,则*.deleted_at is not null
查询部分将被自动添加,这将ruby查询减少了。
User.select("concat(organizations.code, ' - ', users.name)")
.includes(:organizations)
.where('users.type = ?', 'Banker')
.order('organizations.code ASC')
For the rare case you don't know about the rails guides. 在极少数情况下,您对导轨不了解。 Here's the link to the article about associations . 这是有关关联的文章的链接。
盾可能对您的情况有用。
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