ruby-on-rails - Activerecord rails 合并多个选择结果并在 View 中呈现行

Question

我是 Rails 的新手，正在做面试的测试。

几乎达到了预期的结果，您可以在下面找到代码。

数据示例:

Bookings
id,room_id,start_date,end_date,number_of_guests
1,1,2015-06-01,2015-06-07,1
2,3,2015-06-01,2015-06-07,2
3,5,2015-06-01,2015-06-07,2
2001,6,2015-06-01,2015-06-07,2
...

Hosts
id,name,address
1,Mr Host 1,1 Camden 
2,Mr Host 2,2 Camden Street
3,Mr Host 3,3 Camden Street
4,Mr Host 4,4 Camden Street
...

Rooms
id,host_id,capacity
1,1,2
2,1,2
3,2,2
4,2,2
...

事件记录:

class Room < ActiveRecord::Base
  belongs_to :host
  has_many :bookings
end

class Host < ActiveRecord::Base
  has_many :rooms
end

class Booking < ActiveRecord::Base
  belongs_to :room
end

Controller :

  start_date = '2015-06-01' # hardcoded value for testing purposes
  end_date = '2015-06-07'   # hardcoded value for testing purposes   

  @available_rooms = Room.select(:host_id, :name, :address, :id, :number_of_guests ,:capacity)
               .joins(:bookings, :host)
               .where("bookings.number_of_guests <> rooms.capacity 
                        AND ? < rooms.capacity 
                        AND ? <= bookings.start_date
                        AND ? <= bookings.end_date", params[:guests], start_date, end_date)

  @booked_rooms = Room.select(:host_id, :name, :address, :id, :number_of_guests, :capacity)
               .joins(:bookings, :host)

  @total_rooms = Room.select(:host_id, :name, :address, :id, "0 as number_of_guests", :capacity)
               .joins(:host)

  @rooms = ((@total_rooms - @booked_rooms) + @available_rooms).sort_by(&:id)

看法:

 <% if @rooms %>
<section class="rooms">
 <ul class="list-unstyled list-rooms">
  <% for room in @rooms %>
  <li class="clearfix">
    <article>
      <div class="description">
        <header>
          <h2>host#<%= "#{room.host_id}: #{room.name}" %></h2> 
          <h3><%= "#{room.address} "%></h3>
        </header>
        <p>room#<%= "#{room.id} is available (#{room.number_of_guests} booked, #{room.capacity - room.number_of_guests} free out of total #{room.capacity})" %></p>
      </div>
    </article>
  </li>
  <% end %>
</ul>
</section>
<% end %>

当前结果:

  host#1: Mr Host 1
  1 Camden Street
  room#1 is available (1 booked, 1 free out of 2 total)

  host#1: Mr Host 1
  1 Camden Street      
  room#2 is available (0 booked, 2 free out of 2 total)

  host#2: Mr Host 2
  2 Camden Street
  room#4 is available (0 booked, 2 free out of 2 total)

  host#4: Mr Host 4
  4 Camden Street
  room#8 is available (0 booked, 2 free out of 2 total)

  host#5: Mr Host 5
  5 Camden Street
  room#9 is available (1 booked, 1 free out of 2 total)

  host#5: Mr Host 5
  5 Camden Street
  room#10 is available (0 booked, 2 free out of 2 total)

预期结果:

  host#1: Mr Host 1
  1 Camden Street
  room#1 is available (1 booked, 1 free out of 2 total)
  room#2 is available (0 booked, 2 free out of 2 total)

  host#2: Mr Host 2
  2 Camden Street
  room#4 is available (0 booked, 2 free out of 2 total)

  host#4: Mr Host 4
  4 Camden Street
  room#8 is available (0 booked, 2 free out of 2 total)

  host#5: Mr Host 5
  5 Camden Street
  room#9 is available (1 booked, 1 free out of 2 total)
  room#10 is available (0 booked, 2 free out of 2 total)

我正在考虑使用 jquery 来获得预期的结果，但我必须有一种方法可以通过事件记录或在 View 中调用某种帮助程序来实现这一点。提前致谢。

理想情况下，我想重写查询以检索如下行的结果:

  host_id, address, total_rooms, room_id[0], booked[0], capacity[0], ... , room_id[n], booked[n], capacity[n]

因此，对于前两行，我们将有:

  1, 1 Camden Street, 2, 1, 1, 2, 2, 0, 2
  2, 2 Camden Street, 1, 4, 0, 2

相当于前端的结果:

主持人#1:主持人1先生
1 卡姆登街
房间#1 可用(预订了 1 个，总共 2 个中有 1 个免费)
房间#2 可用(0 个预订，2 个免费，共 2 个)

主持人#2:主持人先生2
2 卡姆登街
房间#4 可用(0 个预订，2 个免费，共 2 个)

Answer 1

您不能在 SQL 级别执行此操作。 SQL 返回统一的行，您可以按照自己的喜好对它们进行排序，但在 SQL 中对行进行分组不能这样工作。您不能使用它来返回一对多关系的部分行；它将为每一行返回完整的连接记录。

如果您只想输出顶级记录(在本例中为“主机”)，然后多次输出关联记录(“房间”)，则需要在前端对结果进行分组:

#...
@rooms = ((@total_rooms - @booked_rooms) + @available_rooms).sort_by(&:id)

# Produce a mapping of hosts to their rooms
@rooms_by_host = @rooms.group_by do |room|
  { id: host_id, address: room.address, name: room.name }
end

group_by 将对块返回相同值的元素进行分组。在这种情况下，任何具有相同 host_id 的房间, address和 name将在返回的散列中组合成一个嵌套数组。

在您看来，您现在将有两个循环:一个循环遍历主机 => 房间映射的外部循环，以及循环遍历其房间的每个主机的内部循环:

<section class="rooms">
  <ul class="list-unstyled list-rooms">
    <% @rooms_by_host.each do |host, rooms| %>
      <li class="clearfix">
        <article>
          <div class="description">
            <header>
              <h2>host#<%= host[:id] %> - <%= host[:name] %></h2> 
              <h3><%= host[:address] %></h3>
            </header>
            <ul class="rooms">
              <% rooms.each do |room %>
                <li>room#<%= room.id %> is available (<%= room.number_of_guests %> booked, <%= room.capacity - room.number_of_guests %> free out of total <%= room.capacity %>)</li>
              <% end %>
            </ul>
          </div>
        </article>
      </li>
    <% end %>
  </ul>
</section>

如果您担心在面试中如何展示自己，您应该考虑以下几点:

不要使用 for x in y在 ruby 。惯用的循环方式是 y.each do |x| .使用 for x in y尖叫缺乏经验。

<% if @rooms %>是无用的支票，@rooms是一个数组，它永远不能是一个假值。您需要 if @rooms.present?或 if @rooms.any?

不要同时使用 <%= %>和 "#{...}" .内部#{}是完全多余的。你做过的每个地方...

<%= "#{room.address} "%>

你应该这样做

<%= room.address %>

让 ActiveRecord 构建尽可能多的 SQL 会更简洁。而不是这个...

.where("bookings.number_of_guests <> rooms.capacity 
                            AND ? < rooms.capacity 
                            AND ? <= bookings.start_date
                            AND ? <= bookings.end_date", params[:guests], start_date, end_date)

用这个

.where('bookings.number_of_guests <> rooms.capacity')
.where('? < rooms.capactiy', params[:guests])
.where('? <= bookings.start_date', start_date)
.where('? <= bookings.end_date', end_date)

它还使您的参数更接近它们的使用位置。

如果您想稍微清理一下您的 View ，请使用部分。最内部的循环可能是:

    <ul class="rooms">
      <%= render rooms %>
    </ul>

然后将循环体移动到 app/views/rooms/_room.html.erb

<li>room#<%= room.id %> is available (<%= room.number_of_guests %> booked, <%= room.capacity - room.number_of_guests %> free out of total <%= room.capacity %>)</li>