ORACLE SQL 获取连续28天有销售的店铺

最近公司一SSRS报表需求改变,同比店铺的选取为连续28天有销售的店铺,思考良久,发现可以利用ROW_NUMBER()
OVER(PARTITION BY COLUMN01 ORDER BY
COLUMN02)来实现,ROW_NUMBER()从1开始,按照COLUMN01分组以及COLUMN02排序.

1:先看数据源,要求取出上面连续28天内销售金额大于0的XF_STORECODE :

2: 分解一下,由图中看出,只要取XF_STORECODE,COUNT(1)>28就可以了.

3: 完整代码:

4: 文本代码

 1 WITH STORE_COMP AS

 2  (SELECT T.XF_STORECODE, T.XF_TXDATE, SUM(T.XF_AMTSOLD) XF_AMTSOLD

 3     FROM XF_VIPITEMDM T, XF_ITEMMAS T1

 4    WHERE T.XF_PLU = T1.XF_PLU

 5      AND SUBSTR(T.XF_STORECODE, 3, 2) IN (‘NK‘, ‘SC‘, ‘NS‘, ‘MB‘)

 6      AND T1.XF_GROUP2 = ‘NIKE‘

 7      AND T1.XF_GROUP9 <> ‘特许装备‘

 8      AND T.XF_TXDATE >= TO_DATE(‘20140501‘, ‘yyyymmdd‘)

 9      AND T.XF_TXDATE <= TO_DATE(‘20140531‘, ‘yyyymmdd‘)

10    GROUP BY T.XF_STORECODE, T.XF_TXDATE

11   HAVING SUM(T.XF_AMTSOLD) > 0)

12 SELECT XF_STORECODE

13   FROM (SELECT XF_STORECODE,

14                XF_TXDATE,

15                ROW_NUMBER() OVER(PARTITION BY XF_STORECODE ORDER BY XF_TXDATE) ROW_NUM,

16                TRUNC(XF_TXDATE - ROW_NUMBER()

17                      OVER(PARTITION BY XF_STORECODE ORDER BY XF_TXDATE)) ROWDIFF

18           FROM STORE_COMP)

19  GROUP BY XF_STORECODE, ROWDIFF

20 HAVING COUNT (1) >= 28

5: 如有更好的方法,欢迎赐教,谢谢!

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