如何解决如何在sql中按最近六周日的每个星期分组?
我对您发布的数据做了一些假设。
首先,您发布的所有值均以年份为单位,2011
但最终的结束日期与列标题不对应2011
,它们是的Sunday
值,2012
因此我更改了数据。也是的最后一个条目Early
ASN 8/15/2011 12:00
,我相信应该是一个Late ASN
条目,否则总和要匹配。
要获得结果,您希望应用该PIVOT
功能。此功能使您可以汇总值,然后将其转换为列。
SET DATEFirsT 1 -- set this so the start of the week is Sunday
select InstanceType,
sum([39]) as Sep_23,
sum([38]) as Sep_16,
sum([37]) as Sep_09,
sum([36]) as Sep_02,
sum([35]) as Aug_26,
sum([34]) as Aug_19
from
(
select SPGI01_INSTANCE_TYPE_C as InstanceType,
[39], [38], [37], [36], [35], [34]
from
(
select SPGI01_INSTANCE_TYPE_C,
DatePart(wk, SPGI01_CREATE_S) WeekNo,
DATEADD(DAY, 7 -DATEPART(WEEKDAY,SPGI01_CREATE_S),SPGI01_CREATE_S) WeekEnd
from table1
) x
pivot
(
count(WeekEnd)
for weekno in ([39], [38], [37], [36], [35], [34])
) p
) x1
group by InstanceType with rollup
解决方法
目前正在研究报告。我需要的是样品台,
Instance Type Sep-23 Sep-16 Sep-09 Sep-02 Aug-26 Aug-19
-------------------------------------------------------------------------
Early ASN 4 2 4 1 1 2
Late ASN 2 1 5 3 1 1
Sum 6 3 9 4 2 3
但是实际表是
SPGI01_INSTANCE_TYPE_C SPGI01_CREATE_S
--------------------------------------------------------------
Early ASN 9/17/2012 12:00:00.000
Early ASN 9/18/2012 10:06:11.000
Early ASN 9/19/2012 8:00:04.000
Early ASN 9/20/2012 3:00:05.000
Early ASN 9/10/2012 12:00:07.000
Early ASN 9/11/2012 12:00:32.000
Early ASN 9/3/2012 12:00:17.000
Early ASN 9/4/2012 10:06:00.000
Early ASN 9/5/2012 8:00:00.000
Early ASN 9/6/2012 3:00:00.000
Early ASN 8/31/2012 12:00:00.000
Early ASN 8/26/2012 12:00:00.000
Early ASN 8/14/2012 12:00:00.000
Early ASN 8/15/2012 12:00:00.000
Late ASN 9/17/2012 12:00:00.000
Late ASN 9/18/2012 10:06:00.000
Late ASN 9/11/2012 12:00:00.000
Late ASN 9/3/2012 12:00:00.000
Late ASN 9/4/2012 10:06:00.000
Late ASN 9/5/2012 8:00:00.000
Late ASN 9/6/2012 3:00:00.000
Late ASN 9/6/2012 2:00:00.000
Late ASN 8/31/2012 12:00:00.000
Late ASN 8/31/2012 12:00:00.000
Late ASN 8/31/2012 12:00:00.000
Early ASN 8/15/2012 12:00:00.000
我需要对“
SPGI01_INSTANCE_TYPE_C”列进行分组,并将每周的星期日分组,直到最近六周的星期日分组。在这里,我粘贴了两个示例表,一个是我想要的表,另一个是我拥有的表。给我解决方案。
我的查询是
SELECT distinct I01.[SPGI01_INSTANCE_TYPE_C],count (I01.[SPGI01_INSTANCE_TYPE_C])
FROM [SUPER-G].[dbo].[CSPGI01_ASN_ACCURACY] I01,[SUPER-G].[dbo].[CSPGI50_VALID_INSTANCE_TYPE] I50
where
I01.[SPGA02_BUSINESS_TYPE_C] = 'prod'
and
I01.[SPGA03_REGION_C] in( 'ap','na','sa','eu')
and
I01.[SPGI01_SUB_BUSINESS_TYPE_C] = 'PRD'
and
(I01.[SPGI01_CREATE_S] between '2012-01-01 12:00:00.000' AND DATEADD(day,7,'2012-01-15 00:00:00.000'))
and
I01.[SPGI01_EXCEPTIONED_F] = 'N'
and
I01.[SPGI01_DISPUTED_F] != 'Y'
and
I50.[SPGI50_INSTANCE_TYPE_C] = I01.[SPGI01_INSTANCE_TYPE_C]
and
I50.[SPGA04_RATING_ELEMENT_D] = 1
group by I01.[SPGI01_INSTANCE_TYPE_C]
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。