Tsql 可怜的人';s的SQL Pivot。以列形式列出问题,并在一行中列出每个用户的答案
当前查询:Tsql 可怜的人';s的SQL Pivot。以列形式列出问题,并在一行中列出每个用户的答案,tsql,Tsql,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
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
订单号ordName ordLastName问题解答
8591 Larry Marshburn手术类型:结肠造口术
8591拉里·马什烧伤手术月数:2
8591拉里·马什伯恩手术年份:2010年
8591 Larry Marshburn当前造口系统品牌:ConvaTec
8591 Larry Marshburn满意度:有点满意
8593 Melvin-Belcher手术类型:尿造口术
8593梅尔文·贝尔彻手术月:9
8593梅尔文·贝尔彻手术年份:2010年
8593 Melvin Belcher当前造口系统品牌:ConvaTec
8593 Melvin Belcher满意度:非常满意
我如何正确地查询这些表,以获取如下所示的结果?
姓名和姓氏在一行上,列的问题和每列的答案
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
订单号ordName ordLastName“手术类型”“手术月份”“手术年份”等。
8591 Larry Marshbourn结肠造口术2010年2月
8593 Melvin Belcher泌尿造口术2010年9月
这称为透视,其中行中的信息用于确定列列表。这种查询如果完全在查询中完成,则需要动态计算SQL,通常更适合客户端格式化(许多工具称之为透视或交叉表查询,SSRS称之为矩阵查询)。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]
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
这是MySQL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]
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
贴出的答案有效,但笨拙而缓慢。您可以执行我称之为并行聚合的操作:
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
这将在桌子上做一次传球,而不是三次,而且非常短。这不是一个完整的演练,但您肯定可以根据您的需要调整该概念。这称为
透视图
使用静态版本或动态版本执行此操作有两种方法
静态版本是指将值硬编码为列:
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,获取运行时的列列表:
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)
您需要执行
透视
/交叉选项卡
查询。如果列是动态的,那么您需要动态SQL。好的,我必须研究一下,发现问题也不会改变。这也是我最喜欢的方法,但我实际上称之为“穷人的轴心”!它实际上相当于SQLServerPivot子句。