2014-03-19 63 views
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SELECT rs.id as rs_id, 
sum(rs.amount_per_month) as rs_am1, 
sum(rs.amount_per_month_per_sqft) as rs_am2, 
rs.from_date,rs.to_date,cc.charge_type as cc 
FROM lease_rent_rolls as lrr 
INNER JOIN leases as l ON lrr.lease_id = l.id 
INNER JOIN rents as r ON l.id = r.lease_id 
INNER JOIN rent_schedules as rs ON r.id = rs.rent_id 
INNER JOIN charge_codes as cc ON rs.charge_code_id = cc.id 
WHERE lrr.id = 449443 
AND DATE(NOW()) BETWEEN rs.from_date AND rs.to_date 
GROUP BY cc.charge_type 

我在模型中使用此SQL查询中使用find_by_sql,但我应该得到这个同样的结果在Ruby中加入WHERE条件在模型中没有的帮助SQL。如何写的Ruby这个sql查询加入查询(Ruby on Rails的)

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您能否向我们展示您的模型,包括它们之间的关联? – BroiSatse

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请分享您的模特协会,然后它会更有帮助。 – uma

回答

0

事情是这样的:

@goods =Good.joins(:user).where("name like ?", "%#{@searched_good}%").where("category_id = ?", params[:category]).where("postal_code == ?", @searched_good_postal_code) 

有: - CATEGORY_ID和名称,是属于好阵列 - POSTAL_CODE属于用户阵列

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感谢帖子,但是m需要加入多个表格。 – Janu

1

如果你提供的关联那么这将是更有利于提供解。现在

,我正在考虑:

LeaseRentRoll有一个或多个租赁。

租赁有很多租金。

租有很多rent_schedules。

和RentSchedule有很多charge_codes。

如果LeaseRentRoll有一个租赁:

LeaseRentRoll.where(:id => 449443).first.lease.joins(:rents => [:rent_schedules => charge_codes]).select("rent_schedules.id as rs_id, sum(rent_schedules.amount_per_month) as rs_am1, sum(rent_schedules.amount_per_month_per_sqft) as rs_am2, 

rent_schedules.from_date,rent_schedules.to_date,charge_codes.charge_type为CC“)

如果LeaseRentRoll有许多租赁:

LeaseRentRoll.where(:id => 449443).first.leases.joins(:rents => [:rent_schedules => charge_codes]).select("rent_schedules.id as rs_id, sum(rent_schedules.amount_per_month) as rs_am1, sum(rent_schedules.amount_per_month_per_sqft) as rs_am2, rent_schedules.from_date,rent_schedules.to_date,charge_codes.charge_type as cc") 

请看看附图,这将有助于更多地了解上述查询。

enter image description here

1
RentSchedule.joins([{:rent => {:lease => :lease_rent_rolls}}, :charge_code]).where("lease_rent_rolls.id = #{lease_rr.id} 
and rent_schedules.from_date < '#{Date.today.strftime("%Y-%m-%d %H:%M:%s")}' 
and rent_schedules.to_date > '#{Date.today.strftime("%Y-%m-%d %H:%M%s")}'").group("charge_codes.charge_type") 

得到了结果这个Query.Thanks大家!

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随时接受你自己的答案。 – Pavan