Question

我在MySql中进行了查询，但是在Rails和mysql2 gem中工作。

以下是信息：

http://sqlfiddle.com/#!2/9adb8/6

查询正常运行没有问题并显示此结果：

 UNIT  V1  A1  N1   V2 A2  N2   V3  A3  N3   V4  A4  N4   V5  A5  N5
 LIFE  2   0   0    1   2  0    0   0   0     0   0   0   0    0   0
 ROB   0   1   0    0   1  2    0   0   0     0   0   0   0    0   0

- 为rails 2.3.8安装了mysql2 gem

gem install mysql2 -v0.2.6

- 处理控制器：

class PolicyController < ApplicationController

   def index
      @result =  ActiveRecord::Base.connection.execute("select distinct @sql := concat('SELECT pb.name as unit,',group_concat(concat('SUM(CASE WHEN p.state =0 AND ce.id=',id,' THEN 1 ELSE 0 END ) AS v',id,',SUM(CASE WHEN p.state =1 AND ce.id=',id,' THEN 1 ELSE 0 END ) AS a',id,',SUM(CASE WHEN p.state =2 AND ce.id=',id,' THEN 1 ELSE 0 END ) AS n',id)),' FROM cia_ensures ce LEFT JOIN policies p on ce.id = p.cia_ensure_id INNER JOIN policy_business_units pb ON pb.id = p.policy_business_unit_id  INNER JOIN comercial_areas ca ON ca.id = pb.comercial_area_id AND ca.id=1  Group by p.policy_business_unit_id') from cia_ensures where id in(1,2,3,4,5);")
      @result2 = ActiveRecord::Base.connection.execute("prepare stmt from @sql;")
      @result3 = ActiveRecord::Base.connection.execute("execute stmt;")
   end

end

这是日志：

SQL (0.9ms)   select distinct @sql := concat('SELECT pb.name as unit,',group_concat(concat('SUM(CASE WHEN p.state =0 AND ce.id=',id,' THEN 1 ELSE 0 END ) AS v',id,',SUM(CASE WHEN p.state =1 AND ce.id=',id,' THEN 1 ELSE 0 END ) AS a',id,',SUM(CASE WHEN p.state =2 AND ce.id=',id,' THEN 1 ELSE 0 END ) AS n',id)),' FROM cia_ensures ce LEFT JOIN policies p on ce.id = p.cia_ensure_id INNER JOIN policy_business_units pb ON pb.id = p.policy_business_unit_id INNER JOIN comercial_areas ca ON ca.id = pb.comercial_area_id AND ca.id=1 Group by p.policy_business_unit_id') from cia_ensures where id in(1,2,3,4,5);
SQL (0.9ms)   prepare stmt from @sql;
SQL (0.2ms)   execute stmt;

这是视图（工作正常但没有问题，但似乎太长写了几次相同的代码）

<table>
  <% @result3.each do |policy| %>
  <tr>
      <td><%= policy[0] %></td>
      <td><%= policy[1] %></td>
      <td><%= policy[2] %></td>
      <td><%= policy[3] %></td>
      <td><%= policy[4] %></td>
      <td><%= policy[5] %></td>
      ...
  </tr>
  <%end%> 
</table>

我尝试使用inspect但是它显示了一个td中的所有信息而不是每个td：

<% @result3.each do |policy| %>
<tr>      
 <td align="center"><%= policy.inspect %></td>
</tr>
<%end%>

如何在不写大量行的情况下显示所有这些内容？

有可能在一行中制作这个吗？无需编写＆lt;％= policy [#NUMBER]％＆gt;

请有人帮我这个吗？

我真的很感激。

Answer 1

嗯，可能误解了你的问题，因为它看起来很简单，你想缩短它吗？

变化：

<td><%= policy[0] %></td>
<td><%= policy[1] %></td>
<td><%= policy[2] %></td>
<td><%= policy[3] %></td>
<td><%= policy[4] %></td>
<td><%= policy[5] %></td>

要：

<% policy.each do |p| %>
  <td><%= p %></td>
<% end %>

Answer 2

这样做

  <% @result3.each do |policy| %>
  <tr>
      <% policy.each { |p| raw "<td>#{p}</td>" } %>
  </tr>
  <%end%>

Answer 3

只需遍历每个policy对象中的每个集合。

<table>
  <% @result3.each do |policy| %>
      <tr>
          <%policy.each do |item| %>
              <td><%= item %></td>
          <%end%>
      </tr>
  <%end%> 
</table>

如何在rails视图中迭代数组？

3 个答案: