Question

我有一个select输出，输出如下：

pid-------Name------------------response
147------Factory Influenced-----Fish Paste
147------Factory Influenced-----Contract Packer
147------Factory Influenced-----Cereals
147------Factory Influenced-----Biscuits
154------Factory Influenced-----Werda
154------Factory Influenced-----Worcester
158------Factory Influenced-----Moirs
158------Factory Influenced-----Oats
158------Factory Influenced-----Noogy
157------Factory Influenced-----Safari Nuts
158------Factory Influenced-----Savoury Foods

我想做的是让“工厂影响”成为一个专栏，并将“响应”作为字段。但是，我想将这些值连接如下：

pid-----Factory Influenced          
147-----Fish Paste, Contract Packer, Cereals, Biscuits          
154-----Werda, Worcester            
158-----Moirs, Oats, Noogy, Safari Nuts, Savoury Foods

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Answer 1

将多行连接到单个列的过程不是PIVOT，PIVOT是将数据行转换为列时的过程。由于您要创建以逗号分隔的值列表，因此代码不同。

您可以使用FOR XML PATH和STUFF创建列表：

select t1.pid,
  t1.name, 
  STUFF(
         (SELECT ', ' + t2.response
          FROM yourtable t2
          where t1.pid = t2.pid
          FOR XML PATH (''))
          , 1, 1, '')  AS response
from yourtable t1
group by t1.pid, t1.name;

请参阅SQL Fiddle with Demo。

或者您可以CROSS APPLY使用FOR XML PATH：

select DISTINCT t1.pid,
  t1.name, 
  response = left(t2.response, len(t2.response)-1) 
from yourtable t1
cross apply
(
  select t2.response + ', '
  from yourtable t2
  where t1.pid = t2.pid
  FOR XML PATH('')
) t2 (response);

见SQL Fiddle with Demo。两个版本都会得到以下结果：

| PID |               NAME |                                       RESPONSE |
|-----|--------------------|------------------------------------------------|
| 147 | Factory Influenced | Fish Paste, Contract Packer, Cereals, Biscuits |
| 154 | Factory Influenced |                               Werda, Worcester |
| 157 | Factory Influenced |                                    Safari Nuts |
| 158 | Factory Influenced |              Moirs, Oats, Noogy, Savoury Foods |

Microsoft sql server：与连接一起旋转行？

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