更广泛地:
class User < ActiveRecord::Base has_many :members has_many :projects,:through => :membersendclass Member < ActiveRecord::Base belongs_to :user belongs_to :project # has a column called join_code # has a column called activated # Note that this class can be thought of as "membership"endclass Project < ActiveRecord::Base has_many :members has_many :users,:through => :membersend
目标:给定一个特定的用户,我想要一个能够获得所有项目的查询,并且只需要加载将这些项目链接到用户的成员记录.
到目前为止,我在user.rb中有这个方法来执行查询:
def live_projects self.projects.order("projects.name").includes(:members).where(:members => {:join_code => nil,:activated => true})end
但这还不够.我希望能够在视图代码中执行此 *** 作:
<% current_user.live_projects.each do |project| %> <li project_ID="<%= project.ID %>"> <% member = project.member %> (Do something with that member record here) <%= project.name %> <% end %> </li><% end %>
在这里,通常,我有project.members,但在我的上下文中,我只对那个链接回用户的成员记录感兴趣.
这就是我认为原始sql应该是什么样子
select projects.*,members.* from projects inner join members on projects.ID = members.project_IDwhere members.user_ID = X and members.join_code is null and members.activated = 't';
如何在Arel(或ActiveRecord)中执行此 *** 作?
解决方法 我可能在这里有一些答案,即我写的ActiveRecord代码似乎很合理.再次,这是查询:def live_projects self.projects.order("projects.name").includes(:members).where(:members => {:join_code => nil,:activated => true})end
在使用示例数据运行UI时,它会从Rails服务器生成此输出:
Project Load (0.6ms) SELECT "projects".* FROM "projects" INNER JOIN "members" ON "projects".ID = "members".project_ID WHERE "members"."join_code" IS NulL AND "members"."activated" = 't' AND (("members".user_ID = 3)) ORDER BY projects.nameMember Load (2.0ms) SELECT "members".* FROM "members" WHERE ("members".project_ID IN (50,3,6,37,5,1))
然后在视图代码中我可以这样做:
<% current_user.live_projects.each do |project| %> <li project_ID="<%= project.ID %>" > <% member = project.members.detect { |member| member.user_ID == current_user.ID } %> (Do something with that member record here) </li><% end %>
获取成员记录的表达式在视图中非常难看,但select是一个Array方法,而不是查询,除了上面显示的两个以外,没有额外的DB命中出现在Rails服务器的输出中.因此,我想我的n 1问题已经解决了.
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