如何在动态查询中将行值连接到列名

Question

我正在开发一个允许配置问题和答案的应用程序。目前最多可以有20个答案，但可能更少。

我的结构如下：

问题

+----+--------+--------------+-------------+
| ID | FormId | QuestionText | AnswerField |
+----+--------+--------------+-------------+
|  1 |      1 | Name         | Answer01    |
|  2 |      1 | Address      | Answer02    |
|  3 |      1 | Phone        | Answer03    |
|  4 |      1 | Email        | Answer04    |
|  5 |      2 | First Name   | Answer01    |
|  6 |      2 | Surname      | Answer02    |
+----+--------+--------------+-------------+

答案

+----+--------+----------+------------+--------------+--------------+----------------+----------+----------+
| ID | FormId | RecordId |  Answer01  |   Answer02   |   Answer03   |    Answer04    | Answer05 | Answer06 |
+----+--------+----------+------------+--------------+--------------+----------------+----------+----------+
|  1 |      1 |        1 | Bob Smith  | Bobs Address | 01234 111222 | [e-mail]  | Null     | Null     |
|  2 |      1 |        2 | Joe Bloggs | Joes Address | 04321 333444 | [e-mail] | Null     | Null     |
|  3 |      2 |        3 | David      | Jones        | Null         | Null           | Null     |          |
+----+--------+----------+------------+--------------+--------------+----------------+----------+----------+

因此，在Questions表中AnswerField Answer01映射到Answers表中的Answer01列

我想做的就是得到一个看起来像这样的结果集：

对于表格ID 1和记录ID 1：

+--------------+---------------+
| QuestionText |    Answer     |
+--------------+---------------+
| Name         | Bob Smith     |
| Address      | Bobs Address  |
| Phone        | 01234 111222  |
| Email        | [e-mail] |
+--------------+---------------+

然后对于表单ID 2和记录ID 3：

+--------------+---------+
| QuestionText | Answer  |
+--------------+---------+
| First Name   | David   |
| Surname      | Jones   |
+--------------+---------+

我尝试使用数据透视表：

SELECT QuestionText, Answer01, Answer02, Answer03, Answer04
FROM (
    SELECT DISTINCT Q.AnswerField, Q.QuestionText, Q.ID, A.Answer01, A.Answer02, A.Answer03, A.Answer04
    FROM Questions Q
    INNER JOIN Answers A ON A.FormId= Q.FormId
    WHERE A.ID = 17
) 
AS src
PIVOT (MAX(question_id) FOR Answer IN(answer_01, answer_02, answer_03, answer_04)) AS pvt

但这在所有列中都重复了答案：

+--------------+-----------+--------------+--------------+---------------+
| QuestionText | Answer01  |   Answer02   |   Answer03   |   Answer04    |
+--------------+-----------+--------------+--------------+---------------+
| Name         | Bob smith | Bobs Address | 01234 111222 | [e-mail] |
| Address      | Bob smith | Bobs Address | 01234 111222 | [e-mail] |
| Phone        | Bob smith | Bobs Address | 01234 111222 | [e-mail] |
| Email        | Bob smith | Bobs Address | 01234 111222 | [e-mail] |
+--------------+-----------+--------------+--------------+---------------+

这显然是不对的。

任何人都可以建议在SQL Server存储过程中如何做到这一点吗？

Answer 1

首先，您的Answers桌子经过精心设计。该表未规范化，当您要返回数据时会给您带来麻烦。如果可能，您需要重组该表。

如果您不能重新设计表格，那么您将不得不取消对答案表的透视，以便能够轻松地将答案与问题结合起来。

UNPIVOT将获取您的列并将其转换为行。不可更改的代码将是：

select formid, RecordId, answer, answercol
from answers a
unpivot
(
  answer
  for answerCol in ([Answer01], [Answer02], [Answer03], 
                    [Answer04], [Answer05], [Answer06])
) unpiv;

请参阅带有演示的SQL Fiddle。结果如下：

| FORMID | RECORDID |         ANSWER | ANSWERCOL |
--------------------------------------------------
|      1 |        1 |      Bob Smith |  Answer01 |
|      1 |        1 |   Bobs Address |  Answer02 |
|      1 |        1 |   01234 111222 |  Answer03 |
|      1 |        1 |  [e-mail] |  Answer04 |

数据放入行后，您可以加入问题表以返回所需的结果：

select q.questiontext, d.answer
from questions q
inner join
(
  select formid, RecordId, answer, answercol
  from answers a
  unpivot
  (
    answer
    for answerCol in ([Answer01], [Answer02], [Answer03], 
                      [Answer04], [Answer05], [Answer06])
  ) unpiv
) d
  on q.AnswerField = d.answercol
  and q.formid = d.formid
where d.recordid = 1;

请参阅带有演示的SQL Fiddle。结果如下：

| QUESTIONTEXT |        ANSWER |
--------------------------------
|         Name |     Bob Smith |
|      Address |  Bobs Address |
|        Phone |  01234 111222 |
|        Email | [e-mail] |