Question

我在表格中列出了日期，交易类型和价值。如果该客户有3个或更多连续的“现金”，我需要退还给定客户的所有交易。按日期排序时连续的事务。

因此，在下面的示例数据中，我想要返回客户1和3的所有交易（包括信用交易），因为这两个客户连续有3个或更多现金交易。客户2被忽略，因为即使他们有超过3次现金交易，他们也不会连续。

╔════════════╦════════════╦═══════════╦═══════╗
║  Customer  ║    Date    ║ TransType ║ Value ║
╠════════════╬════════════╬═══════════╬═══════╣
║ Customer 1 ║ 1/01/2015  ║ cash      ║ 23.00 ║
║ Customer 1 ║ 2/01/2015  ║ cash      ║ 24.00 ║
║ Customer 2 ║ 2/01/2015  ║ cash      ║ 28.00 ║
║ Customer 2 ║ 4/01/2015  ║ credit    ║ 29.00 ║
║ Customer 3 ║ 5/01/2015  ║ credit    ║ 27.00 ║
║ Customer 2 ║ 6/01/2015  ║ cash      ║ 23.00 ║
║ Customer 2 ║ 8/01/2015  ║ credit    ║ 24.00 ║
║ Customer 3 ║ 9/01/2015  ║ cash      ║ 28.00 ║
║ Customer 3 ║ 13/01/2015 ║ cash      ║ 29.00 ║
║ Customer 1 ║ 15/01/2015 ║ cash      ║ 25.00 ║
║ Customer 1 ║ 17/01/2015 ║ credit    ║ 26.00 ║
║ Customer 3 ║ 18/01/2015 ║ cash      ║ 23.00 ║
║ Customer 1 ║ 20/01/2015 ║ cash      ║ 27.00 ║
║ Customer 3 ║ 20/01/2015 ║ credit    ║ 24.00 ║
║ Customer 2 ║ 21/01/2015 ║ cash      ║ 25.00 ║
║ Customer 3 ║ 22/01/2015 ║ credit    ║ 25.00 ║
║ Customer 2 ║ 23/01/2015 ║ cash      ║ 26.00 ║
╚════════════╩════════════╩═══════════╩═══════╝

Answer 1

因此，为了让拥有至少三个连续现金交易的客户可以使用自我加入，并且每行连接前后行，并测试这三个是否为现金转换。

用作第一个公用表表达式的查询对客户分区的所有行进行编号，因此我们有一个合适的列来连接它们。然后在第二个公用表表达式中建立连接，并将其结果输入到最终查询中。查询可以缩短，但为了清晰起见，我把它留了一点。

with cte as (
    select *, r = row_number() over (partition by customer order by date) 
    from table1 -- this is your source table
), cte2 as (
    select t1.customer 
    from cte t1
    join cte t2 on t1.customer = t2.customer and (t1.r = t2.r-1 or t1.r = t2.r+1)
    where t1.transtype = 'cash' and t2.transtype = 'cash'
    group by t1.customer
    having count(*) >= 3
)

select * from Table1 -- this is your source table
where Customer in (select Customer from cte2)
order by customer, date;

使用您的示例数据，这将返回客户1和3的所有行。

Sample SQL Fiddle

Answer 2

您可以使用技巧来枚举＆＃34;现金＆＃34;交易。这个技巧是行数的差异，它非常有用：

select t.*
from (select t.*, count(*) over (partition by grp, customerid, transtype) as cnt
      from (select t.*,
                   (row_number() over (partition by customerid order by date) -
                    row_nubmer() over (partition by customerid, transtype order by date)
                   ) as grp
            from t
           ) t
      where transtype = 'cash'
     ) t
where cnt >= 3;

这将返回客户和开始日期。如果要返回实际事务，可以使用额外级别的窗口函数：

select customerid, min(date) as start_date, sum(value) as sumvalue
from (select t.*,
             (row_number() over (partition by customerid order by date) -
              row_nubmer() over (partition by customerid, transtype order by date)
             ) as grp
      from t
     ) t
where transtype = 'cash'
group by grp, transtype, customerid
having count(*) >= 3;

查找具有3个或更多具有相同值的连续记录的记录

2 个答案: