Oracle SQL：计数和求和，例如4个小时？

Question

我有一个关于已经解决的 Q (Oracle SQL: Counting in 4h?) 的后续问题。

如果数据样本中还有另一列，金额列，我想按照解决方案中的建议进行操作

select t.*, 
from (select t.*, lag(dt, 6) over (partition by employeeId order by dt) as ord6dt
      from t
     ) t
where dt - ord6dt < 4.0/24

，有一个额外的条件和金额列> 10？

所以查询应该只列出那些在 4 小时内有 6 个订单并且 sum(order_value)>10 的人。

Answer 1

您刚刚回顾 6 行的解决方案（这可能不是满足 4 小时前的第一个订单）。因此，添加另一个分析以保持该解决方案：

select *
from (select t1.*, lag(dt, 6) over (partition by employee order by dt) as ord6dt,
             sum(order_value) over (partition by employee order by dt rows 6 preceding) val
      from emptest t1
     ) 
where dt - ord6dt < 4.0/24
  and val >= 10;

将再次只计算 6 行（如果我们在过去四个小时内有超过 6 个订单，这可能会错过一些真正的匹配）。

我们可以改为建模：

select *
  from (select *
          from emptest
        model
        dimension by (employee, dt)
        measures (  order_value, order_id, 0 order_sum, 0 number_of_orders)
        rules (
          order_sum[any,any] = sum(order_value)[cv(employee), dt between cv(dt)-(4/24) and cv(dt)],
          number_of_orders[any,any] = count(*)[cv(employee), dt between cv(dt)-(4/24) and cv(dt)]

        ))
 where order_sum > 10
   and number_of_orders >= 6
order by employee, dt;

它会在 4 或更多时查找 number_of_orders，并且还会检查 4 小时内的计数。

小提琴：http://sqlfiddle.com/#!4/3045d/1