我需要能够使用ActiveRelation使用UNION
链接任意数量的子选择。
我对ARel的实现有点困惑,因为它似乎假设UNION
是二进制操作。
然而:
( select_statement_a ) UNION ( select_statement_b ) UNION ( select_statement_c )
是有效的SQL。如果没有讨厌的字符串替换,这可能吗?
答案 0 :(得分:11)
尽管他在正确的轨道上,你可以做得比Adam Lassek提出的要好一些。我刚刚解决了一个类似的问题,试图从社交网络模型中获取朋友列表。朋友可以通过各种方式自动获取,但我希望有一个可以处理进一步链接的ActiveRelation友好查询方法。所以我有
class User
has_many :events_as_owner, :class_name => "Event", :inverse_of => :owner, :foreign_key => :owner_id, :dependent => :destroy
has_many :events_as_guest, :through => :invitations, :source => :event
def friends
friends_as_guests = User.joins{events_as_guest}.where{events_as_guest.owner_id==my{id}}
friends_as_hosts = User.joins{events_as_owner}.joins{invitations}.where{invitations.user_id==my{id}}
User.where do
(id.in friends_as_guests.select{id}
) |
(id.in friends_as_hosts.select{id}
)
end
end
end
利用Squeels子查询支持。生成的SQL是
SELECT "users".*
FROM "users"
WHERE (( "users"."id" IN (SELECT "users"."id"
FROM "users"
INNER JOIN "invitations"
ON "invitations"."user_id" = "users"."id"
INNER JOIN "events"
ON "events"."id" = "invitations"."event_id"
WHERE "events"."owner_id" = 87)
OR "users"."id" IN (SELECT "users"."id"
FROM "users"
INNER JOIN "events"
ON "events"."owner_id" = "users"."id"
INNER JOIN "invitations"
ON "invitations"."user_id" =
"users"."id"
WHERE "invitations"."user_id" = 87) ))
通过对上述代码的轻微修改,演示了需要可变数量组件的替代模式
def friends
friends_as_guests = User.joins{events_as_guest}.where{events_as_guest.owner_id==my{id}}
friends_as_hosts = User.joins{events_as_owner}.joins{invitations}.where{invitations.user_id==my{id}}
components = [friends_as_guests, friends_as_hosts]
User.where do
components = components.map { |c| id.in c.select{id} }
components.inject do |s, i|
s | i
end
end
end
以下是关于OP确切问题的解决方案的粗略猜测
class Shift < ActiveRecord::Base
def self.limit_per_day(options = {})
options[:start] ||= Date.today
options[:stop] ||= Date.today.next_month
options[:per_day] ||= 5
queries = (options[:start]..options[:stop]).map do |day|
where{|s| s.scheduled_start >= day}.
where{|s| s.scheduled_start < day.tomorrow}.
limit(options[:per_day])
end
where do
queries.map { |c| id.in c.select{id} }.inject do |s, i|
s | i
end
end
end
end
答案 1 :(得分:4)
由于ARel访问者生成联合的方式,我在使用Arel::Nodes::Union
时一直遇到SQL错误。看起来像老式的字符串插值是实现这一点的唯一方法。
我有一个 Shift 模型,我希望得到一个给定日期范围的班次集合,每天限制为五班。这是Shift模型的类方法:
def limit_per_day(options = {})
options[:start] ||= Date.today
options[:stop] ||= Date.today.next_month
options[:per_day] ||= 5
queries = (options[:start]..options[:stop]).map do |day|
select{id}.
where{|s| s.scheduled_start >= day}.
where{|s| s.scheduled_start < day.tomorrow}.
limit(options[:per_day])
end.map{|q| "( #{ q.to_sql } )" }
where %{"shifts"."id" in ( #{queries.join(' UNION ')} )}
end
(除了ActiveRecord,我使用的是Squeel)
不得不求助于字符串插值很烦人,但至少正在对用户提供的参数进行正确的清理。我当然会欣赏使这个更清洁的建议。
答案 2 :(得分:3)
我喜欢Squeel。但是不要使用它。所以我来到这个解决方案(Arel 4.0.2)
def build_union(left, right)
if right.length > 1
Arel::Nodes::UnionAll.new(left, build_union(right[0], right[1..-1]))
else
Arel::Nodes::UnionAll.new(left, right[0])
end
end
managers = [select_manager_1, select_manager_2, select_manager_3]
build_union(managers[0], managers[1..-1]).to_sql
# => ( (SELECT table1.* from table1)
# UNION ALL
# ( (SELECT table2.* from table2)
# UNION ALL
# (SELECT table3.* from table3) ) )
答案 3 :(得分:1)
有一种方法可以使用arel:
来完成这项工作tc=TestColumn.arel_table
return TestColumn.where(tc[:id]
.in(TestColumn.select(:id)
.where(:attr1=>true)
.union(TestColumn.select(:id)
.select(:id)
.where(:attr2=>true))))