我有一个包含许多列的调查表，但我关注的是这两个，survey_date和over_rating。我不确定是否可以在单个查询中完成。我正在使用sql server 2012.这是我的示例数据。
select survey_date, overall_rating from survey
survey_date overall_rating
2017-01-06 15:09:51.940 6
2017-02-06 14:18:18.620 4
2017-05-07 16:03:12.037 7
2017-05-23 10:41:30.357 7
2017-05-23 10:41:30.357 5
2017-05-24 12:05:25.217 8
2017-06-01 09:03:47.727 7
2017-06-05 09:01:07.283 9
2017-06-05 09:28:12.597 6
2017-06-15 09:47:29.407 7
2017-07-06 12:10:50.003 2
2017-07-06 13:45:52.997 7
2017-08-06 14:00:35.403 5
2017-08-09 12:21:17.367 8
我需要计算每个等级1-10的发生次数，并将其总结。示例6月15日，评级10有1，评级9有10，... 这是结果表：
Month 10 9 8 7 6 5 4 3 2 1 Avg Score Total Total >=6 CSI
June'15 1 10 20 3 0 0 0 0 0 0 8 34 34 100%
July'15 1 16 14 0 0 0 0 0 1 0 9 32 31 99%
August'15 7 6 6 0 0 0 0 0 0 0 9 19 19 100%
September'15 0 2 2 0 0 0 0 0 0 0 9 4 4 100%
November'15 0 1 2 0 0 0 0 0 0 0 8 3 3 100%
December'15 0 7 3 4 2 0 0 0 0 0 8 16 16 100%
我尝试了这个查询，但部分错误，因为每个评级都有重复的月份：
select si.yr, si.mn,
case when si.overall_rating = 10 then count(si.overall_rating) else 0 end as '10',
case when si.overall_rating = 9 then count(si.overall_rating) else 0 end as '9',
case when si.overall_rating = 8 then count(si.overall_rating) else 0 end as '8',
case when si.overall_rating = 7 then count(si.overall_rating) else 0 end as '7',
case when si.overall_rating = 6 then count(si.overall_rating) else 0 end as '6',
case when si.overall_rating = 5 then count(si.overall_rating) else 0 end as '5',
case when si.overall_rating = 4 then count(si.overall_rating) else 0 end as '4',
case when si.overall_rating = 3 then count(si.overall_rating) else 0 end as '3',
case when si.overall_rating = 2 then count(si.overall_rating) else 0 end as '2',
case when si.overall_rating = 1 then count(si.overall_rating) else 0 end as '1',
sum(si.overall_rating) as month_count
from
(select YEAR(s.survey_date) yr, MONTH(s.survey_date) mn, s.overall_rating
from survey s where s.status='Submitted' and s.survey_date >= '2017-01-01' AND s.survey_date <= '2017-12-31'
group by YEAR(s.survey_date), MONTH(s.survey_date), s.overall_rating) si group by si.yr, si.mn, si.overall_rating;
结果：
yr mm 10 9 8 7 6 5 4 3 2 1 total
2017 1 0 0 0 0 1 0 0 0 0 0 6
2017 2 0 0 0 0 0 0 1 0 0 0 4
2017 5 0 0 0 0 0 1 0 0 0 0 5
2017 5 0 0 0 1 0 0 0 0 0 0 7
2017 5 0 0 1 0 0 0 0 0 0 0 8
2017 6 0 0 0 0 1 0 0 0 0 0 6
2017 6 0 0 0 1 0 0 0 0 0 0 7
2017 6 0 1 0 0 0 0 0 0 0 0 9
2017 7 0 0 0 0 0 0 0 0 1 0 2
2017 7 0 0 0 1 0 0 0 0 0 0 7
2017 8 0 0 0 0 0 1 0 0 0 0 5
2017 8 0 0 1 0 0 0 0 0 0 0 8
如您所见，5和6重复不同的评级。如果有人能告诉我是否可以在一个查询中完成。感谢
答案 0 :(得分：1)
我想我明白你要在这里实现的目标，你会很高兴知道你并不遥远。您在使用条件聚合时有正确的想法，但是您需要在聚合中包装条件case表达式，而不是相反。要执行条件count，您只需返回条件匹配的1和不匹配的0，然后sum结果。
这样做可以使group by保持简洁：
declare @t table(survey_date datetime,overall_rating int);
insert into @t values ('2017-01-06 15:09:51.940',6),('2017-02-06 14:18:18.620',4),('2017-05-07 16:03:12.037',7),('2017-05-23 10:41:30.357',7),('2017-05-23 10:41:30.357',5),('2017-05-24 12:05:25.217',8),('2017-06-01 09:03:47.727',7),('2017-06-05 09:01:07.283',9),('2017-06-05 09:28:12.597',6),('2017-06-15 09:47:29.407',7),('2017-07-06 12:10:50.003',2),('2017-07-06 13:45:52.997',7),('2017-08-06 14:00:35.403',5),('2017-08-09 12:21:17.367',8);
select dateadd(m,datediff(m,0,survey_date),0) as [Month]
,sum(case when overall_rating = 10 then 1 else 0 end) as [10]
,sum(case when overall_rating = 9 then 1 else 0 end) as [9]
,sum(case when overall_rating = 8 then 1 else 0 end) as [8]
,sum(case when overall_rating = 7 then 1 else 0 end) as [7]
,sum(case when overall_rating = 6 then 1 else 0 end) as [6]
,sum(case when overall_rating = 5 then 1 else 0 end) as [5]
,sum(case when overall_rating = 4 then 1 else 0 end) as [4]
,sum(case when overall_rating = 3 then 1 else 0 end) as [3]
,sum(case when overall_rating = 2 then 1 else 0 end) as [2]
,sum(case when overall_rating = 1 then 1 else 0 end) as [1]
,count(overall_rating) as ScoresReturned
,sum(overall_rating) as TotalScore
,avg(cast(overall_rating as decimal(10,0))) as Average
from @t
group by dateadd(m,datediff(m,0,survey_date),0)
order by [Month];
输出：
+-------------------------+----+---+---+---+---+---+---+---+---+---+----------------+------------+----------+
| Month | 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | ScoresReturned | TotalScore | Average |
+-------------------------+----+---+---+---+---+---+---+---+---+---+----------------+------------+----------+
| 2017-01-01 00:00:00.000 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 6 | 6.000000 |
| 2017-02-01 00:00:00.000 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 4 | 4.000000 |
| 2017-05-01 00:00:00.000 | 0 | 0 | 1 | 2 | 0 | 1 | 0 | 0 | 0 | 0 | 4 | 27 | 6.750000 |
| 2017-06-01 00:00:00.000 | 0 | 1 | 0 | 2 | 1 | 0 | 0 | 0 | 0 | 0 | 4 | 29 | 7.250000 |
| 2017-07-01 00:00:00.000 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 2 | 9 | 4.500000 |
| 2017-08-01 00:00:00.000 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 2 | 13 | 6.500000 |
+-------------------------+----+---+---+---+---+---+---+---+---+---+----------------+------------+----------+
答案 1 :(得分：0)
select si.yr, si.mn,
case when si.overall_rating = 10 then count(si.overall_rating) else 0 end as '10',
case when si.overall_rating = 9 then count(si.overall_rating) else 0 end as '9',
case when si.overall_rating = 8 then count(si.overall_rating) else 0 end as '8',
case when si.overall_rating = 7 then count(si.overall_rating) else 0 end as '7',
case when si.overall_rating = 6 then count(si.overall_rating) else 0 end as '6',
case when si.overall_rating = 5 then count(si.overall_rating) else 0 end as '5',
case when si.overall_rating = 4 then count(si.overall_rating) else 0 end as '4',
case when si.overall_rating = 3 then count(si.overall_rating) else 0 end as '3',
case when si.overall_rating = 2 then count(si.overall_rating) else 0 end as '2',
case when si.overall_rating = 1 then count(si.overall_rating) else 0 end as '1',
sum(si.overall_rating) as month_count
from
(select YEAR(s.survey_date) yr, MONTH(s.survey_date) mn, s.overall_rating
from survey s where s.status='Submitted' and s.survey_date >= '2017-01-01' AND s.survey_date <= '2017-12-31'
group by YEAR(s.survey_date), MONTH(s.survey_date), s.overall_rating) si group by si.yr, si.mn;
您必须删除si.overall_rating
Group By
答案 2 :(得分：0)
Oh! I posted late, anyway you can try this otherwise.
DATA:
IF ( OBJECT_ID('tempdb..#temptable') IS NOT NULL )
BEGIN
DROP TABLE #temptable
END
CREATE TABLE #temptable
(
survey_date DATETIME ,
overall_rating NUMERIC(22,2)
)
INSERT INTO #temptable
( survey_date, overall_rating )
VALUES ( '2017-01-06 15:09:51.940', 6 ),
( '2017-02-06 14:18:18.620', 4 ),
( '2017-05-07 16:03:12.037', 7 ),
( '2017-05-23 10:41:30.357', 7 ),
( '2017-05-23 10:41:30.357', 5 ),
( '2017-05-24 12:05:25.217', 8 ),
( '2017-06-01 09:03:47.727', 7 ),
( '2017-06-05 09:01:07.283', 9 ),
( '2017-06-05 09:28:12.597', 6 ),
( '2017-06-15 09:47:29.407', 7 ),
( '2017-07-06 12:10:50.003', 2 ),
( '2017-07-06 13:45:52.997', 7 ),
( '2017-08-06 14:00:35.403', 5 ),
( '2017-08-09 12:21:17.367', 8 )
QUERY:
;
WITH CTE
AS ( SELECT DATENAME(month, survey_date) + ' '''
+ RIGHT(CAST(YEAR(survey_date) AS NVARCHAR(4)),
2) AS [Month] ,
ISNULL([1], 0) [1] ,
ISNULL([2], 0) [2] ,
ISNULL([3], 0) [3] ,
ISNULL([4], 0) [4] ,
ISNULL([5], 0) [5] ,
ISNULL([6], 0) [6] ,
ISNULL([7], 0) [7] ,
ISNULL([8], 0) [8] ,
ISNULL([9], 0) [9] ,
ISNULL([10], 0) [10],
Total,
Average
FROM ( SELECT survey_date ,
COUNT(overall_rating) overall_rating,
CAST(SUM(overall_rating) AS INT) Total,
AVG(overall_rating) Average
FROM ( SELECT DATEADD(MONTH,
DATEDIFF(MONTH,
0, survey_date),
0) survey_date ,
overall_rating
FROM #temptable
) T
GROUP BY t.survey_date
) PVT PIVOT ( SUM(overall_rating) FOR overall_rating IN ( [1],
[2], [3], [4],
[5], [6], [7],
[8], [9], [10] ) ) P
)
SELECT [Month] ,
ISNULL([1], 0) [1] ,
ISNULL([2], 0) [2] ,
ISNULL([3], 0) [3] ,
ISNULL([4], 0) [4] ,
ISNULL([5], 0) [5] ,
ISNULL([6], 0) [6] ,
ISNULL([7], 0) [7] ,
ISNULL([8], 0) [8] ,
ISNULL([9], 0) [9] ,
ISNULL([10], 0) [10],
Total,
Average
FROM CTE
RESULT:
Month 1 2 3 4 5 6 7 8 9 10 Total Average
----------------------- ------ ---- ----- ----- ----- ----- ---- ---- ----- ----- ----------- -------------
January '17 1 0 0 0 0 0 0 0 0 0 6 6.000000
February '17 1 0 0 0 0 0 0 0 0 0 4 4.000000
May '17 0 0 0 4 0 0 0 0 0 0 27 6.750000
June '17 0 0 0 4 0 0 0 0 0 0 29 7.250000
July '17 0 2 0 0 0 0 0 0 0 0 9 4.500000
August '17 0 2 0 0 0 0 0 0 0 0 13 6.500000
(6 row(s) affected)