穷人的 SQL 枢轴。将问题列为列并在一行中列出每个用户的答案

Question

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 问题答案 8591 Larry Marshburn 手术类型：结肠造口术 8591 Larry Marshburn 手术月：2 8591 Larry Marshburn 手术年份：2010 8591 Larry Marshburn 当前造口系统品牌：ConvaTec 8591 Larry Marshburn 满意程度：有点满意 8593 Melvin Belcher 手术类型：泌尿造口术 8593 Melvin Belcher 手术月：9 8593 Melvin Belcher 手术年份：2010 8593 Melvin Belcher 当前造口系统品牌：ConvaTec 8593 Melvin Belcher 满意程度：非常满意

如何正确查询表以提取如下所示的结果？ 姓名和姓氏在一行，列的问题和每列的答案。

期望的结果

OrderNumber ordName ordLastName “手术类型” “手术月份” “手术年份”等。 8591 拉里·马什伯恩结肠造口术 2 2010 8593 Melvin Belcher 泌尿造口术 9 2010

Answer 1

发布的答案有效，但笨拙且缓慢。你可以做我所说的并行聚合：

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

这将在桌子上进行一次而不是三次，并且非常短。这不是一个完整的演练，但您肯定可以根据自己的需要调整这个概念。

Answer 2

这称为数据透视表，其中行中的信息用于确定列列表。如果完全在查询中完成，这种查询需要动态计算的 SQL，并且通常更适合客户端格式化（许多工具将其称为数据透视表或交叉表查询，SSRS 将其称为矩阵查询）。

Answer 3

这被称为PIVOT，有两种方法可以使用静态版本或动态版本执行此操作。

静态版本是当您将值硬编码为列时：

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

动态枢轴，在运行时获取列的列表：

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 版本

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]