sql server连续统计次数问题？

[xuminjun]

比如A表，字段B为日期，字段C为人，字段D为买东西。请问怎么写代码达到下面效果。
第一天A买东西，统计为一次，第二天A又买了，统计为两次。第三天A没有买，统计为0.第四天买了，统计为一次。只要断了就重新计数。

 

[xuminjun]

@幸福感奇差 谢谢大佬

[幸福感奇差]

做两层窗口函数：

1. 最内层判断时间连续性，连续标 0，不连续标 1
2. 对上述标记取 cumsum 作为分组依据
3. 对 2 中分组做行号

select *, sum(1) over(partition by name, cumsum order by date) rownum
from 
(select *, sum(tag) over(partition by name order by date) cumsum
from
(SELECT *, 
    case when date = lag(date) over(partition by name order by date) + 1
            then 0 else 1 end tag
FROM test) t ) t
```sql

[幸福感奇差]

[xuminjun]

@二月十六 谢谢版主

[二月十六]

--测试数据
if not object_id(N'Tempdb..#T') is null
    drop table #T
Go
Create table #T([B] Date,[C] nvarchar(21),[D] nvarchar(22))
Insert #T
select '2023-03-10',N'A',N'香蕉' union all
select '2023-03-11',N'A',N'香蕉' union all
select '2023-03-12',N'A',N'香蕉' union all
select '2023-03-14',N'A',N'香蕉' union ALL
select '2023-03-15',N'A',N'香蕉' union all
select '2023-03-01',N'B',N'香蕉' union all
select '2023-03-02',N'B',N'香蕉' union all
select '2023-03-15',N'B',N'香蕉'
Go
--测试数据结束
;WITH cte AS (
Select *,ROW_NUMBER()OVER(PARTITION BY C ORDER BY B) AS rn from #T
),cteb AS (
SELECT *,ROW_NUMBER()OVER(PARTITION BY C,DAY(B)-rn ORDER BY B)  AS rn1 FROM cte
)
SELECT B,C,cteb.D,rn1 AS 次数 FROM cteb 

[二月十六]

[二月十六]

如果楼主考虑每个日期都要有数的话，可以利用master.dbo.spt_values 获得连续的日期，然后left join实现

[炎同学要努力]

【字节SQL面试真题拆解，地狱级难度，保姆级教程—连续登录问题的求解
】https://www.bilibili.com/video/BV1mQ4y1r7Kh/?spm_id_from=333.337.search-card.all.click&vd_source=32daa2743a817af5135ab88d7123ce16
【统计sql中连续出现的次数】https://blog.csdn.net/qq_37056683/article/details/112008505?ops_request_misc=&request_id=&biz_id=102&utm_term=sql%20%E8%BF%9E%E7%BB%AD%E5%87%BA%E7%8E%B0%E6%AC%A1%E6%95%B0&utm_medium=distribute.pc_search_result.none-task-blog-2~all~sobaiduweb~default-2-112008505.142^v88^control_2,239^v2^insert_chatgpt&spm=1018.2226.3001.4187