Question

我有一张表，其中列出了每天和每位客户的交易清单。我需要找到在6个月内的星期日发生超过x次交易的客户/交易日期。

请注意，每位客户每天可能有超过1笔交易，但只要他们在星期日甚至有1笔交易，那么该周日就会计入6个月期间的星期日计数。

这是我到目前为止的代码。我使用sum（transactionvalue）作为将一天中可能的多个事务组合成1个记录的方法：

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

然而，当我更改计数值，即'x'变大时，给定客户的记录变小。如果客户在该时间段内有7个星期日，那么我希望返回7个记录，无论x是1还是7.只有当x大于7时，才能返回该客户的所有交易。

以下是一些示例数据：

+-----------------+------------+--------------------+
| 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              |
+-----------------+------------+--------------------+

在这个例子中，我希望在x = 1时返回以下内容：

+-----------------+------------+--------------------+
| 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              |
+-----------------+------------+--------------------+

但如果x = 3：

，则会返回

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

由于

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

如果超过一周的x天，则返回

1 个答案: