我的员工有分数,如果给定的用户在顶级X用户中,我希望有一个有效的方法来查询。
# person.rb
class Person
scope :top_score, -> {order('score DESC')}
scope :page_limit, -> { limit(10) }
def self.in_top_score(id)
top_score.page_limit.something_something_soemthign?
end
end
以前正在这样做:
user.id.in?(top_score.page_limit.pluck(:id))
,但我宁愿将此支票移至数据库,以防止对象序列化数百/数千条记录。
Person.order('score DESC').select([:score, :id]).limit(1)
Person Load (0.5ms) SELECT score, id FROM `people` ORDER BY score DESC LIMIT 1
=> [#<Person id: "dxvrDy...", score: 35>]
现在检查该列表中是否存在另一个用户^^
Person.order('score DESC').select([:score, :id]).limit(1).exists?({id: "c_Tvr6..."})
Person Exists (0.3ms) SELECT 1 AS one FROM `people` WHERE `people`.`id` = 'c_Tvr6...' LIMIT 1
=> true
返回true,但应返回false
更新答案
对不起,我的原始答案不正确。(exists?
查询显然使用LIMIT 1
并从page_limit
范围覆盖LIMIT 10
,显然也抛出了ORDER BY
子句。完全错误!:p)
这呢?它的优雅程度不那么优雅,但我这次实际上测试了答案:p,似乎可以根据需要工作。
def self.in_top_score?(id)
where(id: id).where(id: Person.top_score.page_limit).exists?
end
这是我的测试(使用Rails 4.2.6)及其生成的SQL(使用子查询)的示例用法:
pry(main)> Person.in_top_score?(56)
Person Exists (0.4ms) SELECT 1 AS one FROM "people" WHERE "people"."id" = $1 AND "people"."id" IN (SELECT "people"."id" FROM "people" ORDER BY "people"."score" DESC LIMIT 10) LIMIT 1 [["id", 56]]
=> false
在我的测试中,与您的原始版本相比,这确实确实具有至少具有性能的提升。
原始答案
top_score.page_limit.exists?(user.id)
http://apidock.com/rails/activerecord/findermethods/exists?