问题描述
我需要创建一个新列，该列在每个Repeated Call的列Customer_ID的每0个值处重新开始：
+-------------+---------+----------------------+---------------+
| Customer_ID | Call_ID | Days Since Last Call | Repeated Call |
+-------------+---------+----------------------+---------------+
| 1 | 1 | Null | 0 |
| 1 | 2 | 45 | 0 |
| 1 | 3 | 0 | 1 |
| 1 | 4 | 0 | 1 |
| 1 | 5 | 0 | 1 |
| 1 | 6 | 48 | 0 |
| 1 | 7 | 1 | 1 |
| 2 | 8 | Null | 0 |
| 2 | 9 | 1 | 1 |
+-------------+---------+----------------------+---------------+
在这种情况下：
+-------------+---------+----------------------+---------------+-------------+
| Customer_ID | Call_ID | Days Since Last Call | Repeated Call | Order_Group |
+-------------+---------+----------------------+---------------+-------------+
| 1 | 1 | Null | 0 | 1 |
| 1 | 2 | 45 | 0 | 2 |
| 1 | 3 | 0 | 1 | 2 |
| 1 | 4 | 0 | 1 | 2 |
| 1 | 5 | 0 | 1 | 2 |
| 1 | 6 | 48 | 0 | 3 |
| 1 | 7 | 1 | 1 | 3 |
| 2 | 8 | Null | 0 | 1 |
| 2 | 9 | 1 | 1 | 1 |
+-------------+---------+----------------------+---------------+-------------+
感谢您的建议，谢谢！
解决方法
您可以使用SUM（）窗口函数：
select t.*,sum(case when Repeated_Call = 0 then 1 else 0 end)
over (partition by Customer_ID order by Call_Id) Order_Group
from tablename t
请参见demo（对于MySql，但这是标准SQL）。
结果：
| Customer_ID | Call_ID | Days Since Last Call | Repeated_Call | Order_Group |
| ----------- | ------- | -------------------- | ------------- | ----------- |
| 1 | 1 | | 0 | 1 |
| 1 | 2 | 45 | 0 | 2 |
| 1 | 3 | 0 | 1 | 2 |
| 1 | 4 | 0 | 1 | 2 |
| 1 | 5 | 0 | 1 | 2 |
| 1 | 6 | 48 | 0 | 3 |
| 1 | 7 | 1 | 1 | 3 |
| 2 | 8 | | 0 | 1 |
| 2 | 9 | 1 | 1 | 1 |
您可以使用窗口分析功能COUNT和ROWS UNBOUNDED PRECEDING来计算“重复呼叫”列中的（对于每个客户）每0个值：
SELECT *,COUNT(CASE WHEN Repeated Call=0 THEN 1 ELSE NULL END )OVER(PARTITION BY Customer_ID
ORDER BY Call_ID ROWS UNBOUNDED PRECEDING)Order_Gr FROM Table