我需要计算周数。它工作正常,但我不知道是 [34]、[35]、[36]、[37]、[38]、[39] 的星期几。weekno 仅在运行时知道。例如,一年中它总共包含 52 周。我在@Bluefeet 的帮助下得到了这个查询,他在同一个stackoverflow 上发布了这个查询。如果我在 between 字段中更改起始日期和截止日期,它将不起作用。因为我在下面的查询中硬编码了 weekno。无论一年中中间字段的日期如何,请给我解决方案。
SET DATEFIRST 1
SELECT case when InstanceType is not null then InstanceType else 'Sum' End InstanceType ,
sum([34]) AS FirstWeek,
sum([35]) AS SecondWeek,
sum([36]) AS ThirdWeek,
sum([37]) AS FourthWeek,
sum([38]) AS FifthWeek,
sum([39]) AS SixthWeek,
max(InstanceDescription) AS InstanceDescription
FROM
(
SELECT [SPGI01_INSTANCE_TYPE_C] AS InstanceType,
[34], [35], [36], [37], [38], [39], InstanceDescription
FROM
(
SELECT I01.[SPGI01_INSTANCE_TYPE_C],
DatePart(wk, I01.[SPGI01_CREATE_S]) WeekNo,
DATEADD(DAY, 7 -DATEPART(WEEKDAY,I01.[SPGI01_CREATE_S]), I01.[SPGI01_CREATE_S]) WeekEnd,
J03.SPGJ03_MSG_TRANSLN_X InstanceDescription
FROM [SUPER-G].[dbo].[CSPGI01_ASN_ACCURACY] I01
INNER JOIN [SUPER-G].[dbo].[CSPGI50_VALID_INSTANCE_TYPE] I50
ON I50.[SPGI50_INSTANCE_TYPE_C] = I01.[SPGI01_INSTANCE_TYPE_C]
LEFT JOIN CSPGJ02_MSG_OBJ J02
ON I50.SPGJ02_MSG_K = J02.SPGJ02_MSG_K
LEFT JOIN CSPGJ03_MSG_TRANSLN J03
ON J02.SPGJ02_MSG_K = J03.SPGJ02_MSG_K
where I50.[SPGA04_RATING_ELEMENT_D] = 1
and I01.[SPGI01_EXCEPTIONED_F] = 'N'
and I01.[SPGI01_DISPUTED_F] != 'Y'
AND J03.[SPGJ03_LOCALE_C] = 'en_US'
and I01.[SPGA02_BUSINESS_TYPE_C] = 'PROD'
and I01.[SPGA03_REGION_C] = 'EU'
and I01.[SPGI01_SUB_BUSINESS_TYPE_C] = 'PRD'
and I01.[SPGI01_CREATE_S] between '10-08-2012 00:00:00.000' AND '11-18-2012 23:59:59.000'
) x
pivot
(
count(WeekEnd)
FOR weekno IN ([34], [35], [36], [37], [38], [39])
) p
) x1
GROUP BY InstanceType WITH ROLLUP