我有以下模式：
我想要选择为foreign_keys(author_id和editor_id)以及单独的一个(例如author_proposals和editor_proposals)调用投标,我需要选择懒惰或加载它们(例如User.includes(：提案)或没有连接).
更新：
#I have the scopes which is like this: class User < ActiveRecord::Base has_many :author_proposals,class_name: 'Proposal',foreign_key: :author_id has_many :editor_proposals,foreign_key: :editor_id end class Proposal < ActiveRecord::Base belongs_to :author,class_name: 'User',foreign_key: :author_id belongs_to :editor,foreign_key: :editor_id end
但我需要一个通用的,它将给我所有的建议(author_proposals和editor_proposals),它也将加载它们.我应该在has_many上使用条件吗？
解决方法
你可以这样做：
class User < ActiveRecord::Base
has_many :authored_proposals,foreign_key: :author_id
has_many :editored_proposals,foreign_key: :editor_id
def proposals
authored_proposals | editored_proposals
end
end
class Proposal < ActiveRecord::Base
belongs_to :author,foreign_key: :editor_id
def users
author | editor
end
end
您可以通过执行以下操作来加载提案：User.includes(：authored_proposals,：editored_proposals).这不是纯粹的铁路方式,但对我来说似乎更清洁.
你也可以做：
class User < ActiveRecord::Base has_many :authored_proposals,foreign_key: :editor_id has_many : proposals,finder_sql: proc { "SELECT * FROM proposals WHERE (proposals.author_id = #{id} or proposals. editor_id = #{id})" } end