pra*_*ech 22 mysql activerecord ruby-on-rails
我如何在mysql查询下写入rails activerecord
select
A.*,
B.*
from
raga_contest_applicants_songs AS A
join
raga_contest_applicants AS B
ON B.contest_applicant_id = A.contest_applicant_id
join
raga_contest_rounds AS C
ON C.contest_cat_id = B.contest_cat_id
WHERE
C.contest_cat_id = contest_cat_id
GROUP BY
C.contest_cat_id
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我知道如何在两个表上编写连接,对如何在3个表上使用连接不太自信.
jef*_*unt 67
要重写你在问题中得到的SQL查询,我认为它应该如下(尽管我很难完全可视化你的模型关系,所以这有点猜测):
RagaContextApplicantsSong.
joins(:raga_contest_applicants => [:raga_content_rounds], :contest_cat).
group('raga_contest_rounds.contest_cat_id')
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...这样该joins
方法既可以处理两个连接,也可以处理WHERE
子句,最后是group
调用.
作为参考更多:
如果您要将多个关联加入同一个模型,您只需列出它们:
Post.joins(:category, :comments)
Returns all posts that have a category and at least one comment
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如果要加入嵌套表,可以将它们列为哈希:
Post.joins(:comments => :guest)
Returns all comments made by a guest
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嵌套关联,多级:
Category.joins(:posts => [{:comments => :guest}, :tags])
Returns all posts with their comments where the post has at least one comment made by a guest
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您还可以链接ActiveRecord查询接口调用,以便:
Post.joins(:category, :comments)
...produces the same SQL as...
Post.joins(:category).joins(:comments)
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如果所有其他方法都失败了,您总是可以将SQL片段直接传递给joins
方法,作为从工作查询中获取更多以ARQI为中心的东西的垫脚石
Client.joins('LEFT OUTER JOIN addresses ON addresses.client_id = clients.id')
=> SELECT clients.* FROM clients LEFT OUTER JOIN addresses ON addresses.client_id = clients.id
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