SQL - 根据列值重新启动分区答案

【问题标题】：SQL - Partition restarted based on a column valueSQL - 根据列值重新启动分区
【发布时间】：2020-08-18 19:07:32
【问题描述】：

我需要创建一个新列，该列在每个Repeated Call 的每个Repeated Call 的每个0 值处重新启动Customer_ID：

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

变成这样：

+-------------+---------+----------------------+---------------+-------------+
| 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！

【问题讨论】：

标签： sql window-functions rank partition

【解决方案1】：

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

请参阅demo（适用于 MySql，但它是标准 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           |

【讨论】：

【解决方案2】：

您可以使用窗口分析函数 COUNT 和 ROWS UNBOUNDED PRECEDING 计算列重复呼叫（针对每个客户）中的每个 0 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

【讨论】：