sql - 如果一周中超过 x 天则返回

Question

我有一个表，其中包含每天和每个客户的交易列表。我需要查找 6 个月内周日交易次数超过 x 的客户/交易日期。

请注意，每个客户每天可能有超过 1 笔交易，但只要他们在周日有 1 笔交易，那么该周日就计入 6 个月期间的周日计数。

这是我到目前为止的代码。我使用 sum(transactionvalue) 作为将一天可能的多个交易合并为 1 条记录的方法:

select customernumber,sum(transactionvalue),date from transactions  
where date between '2015-01-01' and '2015-06-01'
and datename(weekday, date) = 'Sunday'
group by customernumber,date
having count(date) >= x

但是，当我更改计数值(即“x”变大)时，给定客户的记录会变小。如果客户在该时间段内有 7 个星期日，那么无论 x 是 1 还是 7，我都希望返回 7 条记录。只有当 x 大于 7 时，才不会返回该客户的所有交易。

这里是一些示例数据:

+-----------------+------------+--------------------+
| Customer Number | Date       | Transaction Amount |
+-----------------+------------+--------------------+
| 1               | 17/05/2015 | 11.00              |
| 2               | 17/05/2015 | 21.00              |
| 2               | 17/05/2015 | 22.00              |
| 3               | 17/05/2015 | 31.00              |
| 3               | 17/05/2015 | 32.00              |
| 3               | 17/05/2015 | 33.00              |
| 1               | 24/05/2015 | 11.00              |
| 2               | 24/05/2015 | 21.00              |
| 3               | 24/05/2015 | 31.00              |
| 2               | 31/05/2015 | 21.00              |
+-----------------+------------+--------------------+

在此示例中，如果 x = 1，我希望返回以下内容:

+-----------------+------------+--------------------+
| Customer Number | Date       | Transaction Amount |
+-----------------+------------+--------------------+
| 1               | 17/05/2015 | 11.00              |
| 2               | 17/05/2015 | 43.00              |
| 3               | 17/05/2015 | 96.00              |
| 1               | 24/05/2015 | 11.00              |
| 2               | 24/05/2015 | 21.00              |
| 3               | 24/05/2015 | 31.00              |
| 2               | 31/05/2015 | 21.00              |
+-----------------+------------+--------------------+

但是如果 x = 3 则返回:

+-----------------+------------+--------------------+
| Customer Number | Date       | Transaction Amount |
+-----------------+------------+--------------------+
| 2               | 17/05/2015 | 43.00              |
| 2               | 24/05/2015 | 21.00              |
| 2               | 31/05/2015 | 21.00              |
+-----------------+------------+--------------------+

谢谢

Answer 1

试试这个:

select customernumber, date, sum(transactionvalue) as transaction_amt
from transactions   
where date >= '2015-01-01' and date < '2015-07-01'
    and datename(weekday, date) = 'Sunday'
and customernumber in (
  select customernumber
  from transactions  
  where date >= '2015-01-01' and date < '2015-07-01'
      and datename(weekday, date) = 'Sunday'
  group by customernumber
  having count(distinct date) >= 3
)    
group by customernumber, date

<强> SQL Fiddle