Question

我简化了这些模型：

class Game::Champ < ActiveRecord::Base
  has_one :contract, :class_name => "Game::ChampTeamContract", :dependent => :destroy
  has_one :team, :through => :contract
  # Attributes: :avg => integer
end
#
class Game::Team < ActiveRecord::Base
  has_many :contracts, :class_name => "Game::ChampTeamContract", :dependent => :destroy
  has_many :champs, :through => :contracts
end
#
class Game::ChampTeamContract < ActiveRecord::Base
  belongs_to :champ
  belongs_to :team
  # Attributes: :expired => bool, :under_negotiation => bool
end
#

所以我想在这里做的是找到没有 Game :: ChampTeamContract所有或者有的所有Game :: Champs，但是（不：under_negociation OR是：已过期），按Champ.avg ASC排序

我有点坚持使用两个查询，查询结果并对其进行排序。我希望有一个更好的方式来更多地“铁路”

更新：刚刚添加了一个约束：expired

Answer 1

我刚用一个超级简单的查询进行测试：

@bars1 = Bar.where(:something => 1)
@bars2 = Bar.where(:something => 2)

@bars = @bars1 + @bars2

不确定它是否正确，但它有效......

Answer 2

尝试类似：

Game::Champs.
  joins("left outer join game_champ_team_contracts on game_champ_team_contracts.champ_id = game_champs.id").
  where("game_champ_team_contracts.id is null or (game_champ_team_contracts.state != ? or game_champ_team_contracts.state = ?)", :under_negotiation, :expired).
  order("game_champs.avg ASC")

如果保持原样，这是一个非常讨厌的行，所以如果你使用它，它需要整理。使用范围或方法尽可能地将其拆分！

（Rails 3）将两个查询合并为一个

2 个答案: