穷人的SQL枢纽。 列出每个用户问题为列，答案一行 穷人的SQL枢纽。 列出每个用户问题为列，

当前查询 ：

SELECT order_id AS OrderNumber, ordName, ordLastName, question, answer 
FROM cart_survey 
JOIN orders 
    ON cart_survey.order_id=orders.ordID 
JOIN survey_answers 
    ON survey_answers.id=cart_survey.answer_id 
JOIN survey_questions 
    ON survey_questions.id=cart_survey.question_id

结果 ：

OrderNumber ordName ordLastName question                        answer
8591        Larry   Marshburn   Type of Surgery:                Colostomy  
8591        Larry   Marshburn   Month of Surgery:               2
8591        Larry   Marshburn   Year of surgery:                2010
8591        Larry   Marshburn   Current Ostomy System Brand:    ConvaTec  
8591        Larry   Marshburn   Degree of Satisfaction:         Somewhat Satisfied  
8593        Melvin  Belcher     Type of Surgery:                Urostomy
8593        Melvin  Belcher     Month of Surgery:               9
8593        Melvin  Belcher     Year of surgery:                2010
8593        Melvin  Belcher     Current Ostomy System Brand:    ConvaTec  
8593        Melvin  Belcher     Degree of Satisfaction:         Very Satisfied  

如何正确查询表来拉动的结果会是这样？ 名字和姓氏在单行和问题为每列列和答案。

期望的结果

OrderNumber ordName ordLastName "Type of Surgery" "Month of Surgery" "Year of Surgery" etc.
8591        Larry   Marshbourn   Colostomy         2                  2010
8593        Melvin  Belcher      Urostomy          9                  2010

Answer 1:

张贴的答案工作，但笨拙和缓慢的。 你可以做我称之为平行聚集：

SELECT
     ID,
     SUM(case when question_id = 1 then 1 else 0 end) as sum1,
     SUM(case when question_id = 2 then 1 else 0 end) as sum2,
     SUM(case when question_id = 3 then 1 else 0 end) as sum3
GROUP BY ID

这将做一个传过来的表，而不是三个，是非常短的。 这是不是一个完整的步行通过，但你一定能适应的概念，您的需求。

Answer 2:

这就是所谓的支点，其中行的信息来确定列的列表。 这种查询的需要，如果在查询完全完成动态计算的SQL，通常更适合于客户端的格式，而不是（许多工具称之为枢纽或交叉表查询，SSRS称之为矩阵查询）。

Answer 3:

这就是所谓的PIVOT有两种方法与静态版或动态版本执行此操作。

静态版本是当你硬编码值成为列：

SELECT *
FROM
(
  SELECT order_id AS OrderNumber, ordName, ordLastName, question, answer 
  FROM cart_survey 
  JOIN orders 
      ON cart_survey.order_id=orders.ordID 
  JOIN survey_answers 
      ON survey_answers.id=cart_survey.answer_id 
  JOIN survey_questions 
      ON survey_questions.id=cart_survey.question_id
) x
pivot
(
  min(answer)
  for question in ([Type of Surgery:], [Month of Surgery:], 
                [Year of surgery:], [Current Ostomy System Brand:], 
                [Degree of Satisfaction:])
) p

动态透视，获取在运行时的列的列表：

DECLARE @cols AS NVARCHAR(MAX),
    @query  AS NVARCHAR(MAX)

select @cols = STUFF((SELECT distinct ', ' + QUOTENAME(question)
                    from survey_questions
            FOR XML PATH(''), TYPE
            ).value('.', 'NVARCHAR(MAX)') 
        ,1,1,'')


set @query 
      = 'SELECT OrderNumber, ordname, orderLastName,' + @cols + ' from 
         (
            SELECT order_id AS OrderNumber, ordName, ordLastName, question, answer 
            FROM cart_survey 
            JOIN orders 
                ON cart_survey.order_id=orders.ordID 
            JOIN survey_answers 
                ON survey_answers.id=cart_survey.answer_id 
            JOIN survey_questions 
                ON survey_questions.id=cart_survey.question_id
         ) x
         pivot 
         (
            min(answer)
            for question in (' + @cols + ')
         ) p '

execute(@query)

Answer 4:

这是MySQL版本

SELECT o . * , 
q1.answer AS  'Type of Surgery:',
q2.answer AS  'Month of Surgery:',
q3.answer AS  'Year of Surgery:',
q4.answer AS  'Current Brand:',
q5.answer AS  'Degree of Satisfaction:'
FROM (
    SELECT DISTINCT ordID, ordName, ordLastName
    FROM orders
)o
LEFT JOIN (
    SELECT cs.order_id, a.answer
    FROM cart_survey cs
    LEFT JOIN survey_answers a ON cs.answer_id = a.id
    WHERE cs.question_id =18
)q1 ON o.ordID = q1.order_id
LEFT JOIN (
    SELECT cs.order_id, a.answer
    FROM cart_survey cs
    LEFT JOIN survey_answers a ON cs.answer_id = a.id
    WHERE cs.question_id =19
)q2 ON o.ordID = q2.order_id
LEFT JOIN (
    SELECT cs.order_id, a.answer
    FROM cart_survey cs
    LEFT JOIN survey_answers a ON cs.answer_id = a.id
    WHERE cs.question_id =20
)q3 ON o.ordID = q3.order_id
LEFT JOIN (
    SELECT cs.order_id, a.answer
    FROM cart_survey cs
    LEFT JOIN survey_answers a ON cs.answer_id = a.id
    WHERE cs.question_id =21
)q4 ON o.ordID = q4.order_id
LEFT JOIN (
    SELECT cs.order_id, a.answer
    FROM cart_survey cs
    LEFT JOIN survey_answers a ON cs.answer_id = a.id
    WHERE cs.question_id =22
)q5 ON o.ordID = q5.order_id

Answer 5:

这是MSSQL版

select o.*, q1.[Type of Surgery:], q2.[Month of Surgery:], q3.[Year of surgery:]
    , q4.[Current Ostomy System Brand:]
    , q5.[Degree of Satisfaction with the fit and comfort of your Current Ostomy System:]
from (
    select distinct ordID, ordName + ' ' + ordLastName as [name] from dbo.Orders
) o
left join (
    select *, a.[Answer] as [Type of Surgery:] from cart_survey cs
    left join dbo.survey_answers a on cs.answer_id = a.id
    where cs.question_id = 1
) q1 on o.ordID = q1.[order_id]
left join (
    select *, a.[Answer] as [Month of Surgery:] from cart_survey cs
    left join dbo.survey_answers a on cs.answer_id = a.id
    where cs.question_id = 2
) q2 on o.ordID = q2.[order_id]
left join (
    select *, a.[Answer] as [Year of surgery:] from cart_survey cs
    left join dbo.survey_answers a on cs.answer_id = a.id
    where cs.question_id = 3
) q3 on o.ordID = q3.[order_id]
left join (
    select *, a.[Answer] as [Current Brand:] from cart_survey cs
    left join dbo.survey_answers a on cs.answer_id = a.id
    where cs.question_id = 4
) q4 on o.ordID = q4.[order_id]
left join (
    select *, a.[Answer] as [Degree of Satisfaction:] from cart_survey cs
    left join dbo.survey_answers a on cs.answer_id = a.id
    where cs.question_id = 5
) q5 on o.ordID = q5.[order_id]

文章来源: Poor Man's SQL Pivot. List Questions as Columns and Answers per User in one row