Question

我有三个表：users，members，projects。中间是一个连接表，表示其他两个表之间的has-many-through;它有一些感兴趣的属性，包括join_code和activated。

更广泛：

class User < ActiveRecord::Base
  has_many :members
  has_many :projects, :through => :members
end

class 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"
end

class Project < ActiveRecord::Base
  has_many :members
  has_many :users, :through => :members
end

目标：给定一个特定用户，我想要一个能够获得所有项目的查询，并且只需要那些将这些项目链接到用户的成员记录。

到目前为止，我在user.rb中使用此方法执行查询：

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操作：

<% 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，我通常会project.members，但在我的上下文中，我只对那些链接回用户的一个成员记录感兴趣。

以下是我认为原始SQL应该是什么样的

select projects.*, members.* 
from projects inner join members on projects.id = members.project_id
where members.user_id = X and members.join_code is null and members.activated = 't';

如何在Arel（或ActiveRecord）中执行此操作？

Answer 1

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.name
Member 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 %>" class="<%= 'active' if project == @project %>">
    <% member = project.members.detect { |member| member.user_id == current_user.id } %>
    (Do something with that member record here)
  </li>
<% end %>

在视图中获取成员记录的表达式非常难看，但select是Array方法，而不是查询，除了上面显示的两个以外，没有额外的数据库命中出现在Rails服务器的输出。因此我想我的n + 1问题已经解决了。

Answer 2

您似乎希望最多只有一个活动成员将每个用户链接到一个项目。如果是这种情况，则以下情况应该起作用：

在member.rb：

scope :live, where(:join_code => nil, :activated => true)

在user.rb：

def live_projects_with_members
  members.live.includes(:project).group_by(&:project)
end

在您看来：

<% current_user.live_projects_with_members.each do |project, members| %>
  <% member = members.first %>
  <li project_id="<%= project.id %>" class="<%= 'active' if project == @project %>">
      (Do something with that member record here)
  </li>
<% end %>

如果您想为您的使用统计数据添加额外的联接，您可以这样做：

def live_projects_with_members
  members.live.includes(:project, :stats).group_by(&:project)
end

Answer 3

在live_members类上添加名为Project的关联。

class Project < ActiveRecord::Base
  has_many :live_members, :class_name => "Member", 
                :conditions => {:join_code => nil, :activated => true}
  has_many :members
  has_many :users, :through => :members
end

在live_projects类上添加名为User的关联。

class User < ActiveRecord::Base
  has_many :members
  has_many :projects, :through => :members
  has_many :live_projects, :through => :members, :source => :project, 
             :include => :live_member, :order => "projects.name"
end

现在你可以：

user.live_projects

渴望加载“有很多通过” - 我需要Arel吗？

3 个答案: