SQL行到列

Question

我有一个表,并希望将其行转置为列,类似于数据透视表但没有汇总.

例如,我有以下表格:

Question
--QuestionID
--QuestionText

Response
--ResponseID
--ResponseText
--QuestionID

基本上我希望能够创建一个动态表,如:

Question 1 Text | Question 2 Text | Question 3 Text
---------------------------------------------------
Response 1.1 Text | Response Text 1.2 | Response 1.3
Response 2.1 Text | Response Text 2.2 | Response 2.3
Response 3.1 Text | Response Text 3.2 | Response 3.3
Response 4.1 Text | Response Text 4.2 | Response 4.3

主要要求是我在设计时不知道问题文本是什么.

请有人帮忙 - 我把头发拉出来:oS

基本上,您可以保证在此方案中每个相应问题都会有响应.

Answer 1

SQL除非您知道设计时间中的列数(即问题),否则您无法使用(动态查询除外).

您应该以表格格式提取所需数据,然后在客户端处理它:

SELECT  *
FROM    Question
LEFT OUTER JOIN
        Response
ON      Response.QuestionId = Question.QuestionID

或者,可能是这个(在SQL Server 2005+,Oracle 8i+和PostgreSQL 8.4+):

SELECT  *
FROM    (
        SELECT  q.*, ROW_NUMBER() OVER (ORDER BY questionID) AS rn
        FROM    Question q
        ) q
LEFT OUTER JOIN
        (
        SELECT  r.*, ROW_NUMBER() OVER (PARTITION BY questionID ORDER BY ResponseID) AS rn
        FROM    Response r
        ) r
ON      r.QuestionId = q.QuestionID
        AND q.rn = r.rn
ORDER BY
        q.rn, q.QuestionID

后一个查询将以这种形式给出结果(如果您有4疑问):

rn      question      response
---          ---           ---
1     Question 1  Response 1.1
1     Question 2  Response 2.1
1     Question 3  Response 3.1
1     Question 4  Response 4.1
2     Question 1  Response 1.2
2     Question 2  Response 2.2
2     Question 3  NULL
2     Question 4  Response 4.2
3     Question 1  NULL
3     Question 2  NULL
3     Question 3  Response 3.3
3     Question 4  NULL

,这是它将以表格形式输出数据,并rn标记行号.

每当您rn在客户端上看到更改时,您只需关闭<tr>并打开新客户端.

您可以安全地为<td>每个结果集行添加一个,因为每个结果集都保证返回相同的数字或行rn

这是一个非常常见的问题.

SQL 只是不是一个正确的工具来返回动态列数的数据.

SQL 对集合进行操作,列布局是集合的隐式属性.

您应该在设计时定义要获取的集合的布局,就像定义变量的数据类型一样C.

C使用严格定义的变量,SQL使用严格定义的集合.

请注意,我并不是说这是最好的方法.这只是SQL工作方式.

更新:

在SQL Server,您可以将表格HTML从数据库中拉出来:

WITH    a AS
        (
        SELECT  a.*, ROW_NUMBER() OVER (PARTITION BY question_id ORDER BY id) AS rn
        FROM    answer a
        ),
        rows AS (
        SELECT  ROW_NUMBER() OVER (ORDER BY id) AS rn
        FROM    answer a
        WHERE   question_id =
                (
                SELECT  TOP 1 question_id
                FROM    answer a
                GROUP BY
                        question_id
                ORDER BY
                        COUNT(*) DESC
                )
        )
SELECT  (
        SELECT  COALESCE(a.value, '')
        FROM   question q
        LEFT JOIN
                a
        ON      a.rn = rows.rn
                AND a.question_id = q.id
        FOR XML PATH ('td'), TYPE
        ) AS tr
FROM    rows
FOR XML PATH(''), ROOT('table')

有关详细信息,请参阅我的博客中的此条目:

• 动态枢轴