在考虑唯一推销员姓名计数的同时，查找随着日期增加的保险销售金额的累积总和

我有下表

+------------+----------+------------+
| SaleDate   | SaleName | SaleAmount |
+------------+----------+------------+
| 2021-03-01 | Jack     |         40 |
| 2021-03-02 | Mark     |         60 |
| 2021-03-03 | Sam      |         30 |
| 2021-03-03 | Mark     |         70 |
| 2021-03-02 | Sam      |        100 |
| 2021-03-01 | Jack     |         70 |
+------------+----------+------------+

我想找到按日期累计的 SalesAmount，同时还保持表中销售人员姓名的数量不断增加。

+------------+-----------------+-----------------+
| SaleDate   | Count(SaleName) | Sum(SaleAmount) |
+------------+-----------------+-----------------+
| 2021-03-01 | 1               |              110|      
| 2021-03-02 | 3               |              270|
| 2021-03-03 | 3               |              370|
+------------+-----------------+-----------------+

所以基本上任务是消除可能每天或以累积方式发生的重复名称？

您可以使用窗口函数来获取销售人员首次出现的日期。然后使用聚合和累积和：

select saledate,
       sum(sum(case when seqnum = 1 then 1 else 0 end)) over (order by saledate),
       sum(sum(amount)) over (order by saledate) as running_amount
from (select t.*,
             row_number() over (partition by salename order by saledate) as seqnum
      from t
     ) t
group by saledate
order by saledate;