SQL - 分区根据列值重新启动
SQL - Partition restarted based on a column value
我需要创建一个新列，在每个 Customer_ID 的列 Repeated Call 的每个 0 值处重新启动：
+-------------+---------+----------------------+---------------+
| 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 |
+-------------+---------+----------------------+---------------+
进入这样的事情：
+-------------+---------+----------------------+---------------+-------------+
| 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！
您可以使用 SUM() window 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
参见 demo（对于 MySql 但它是标准的 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 window 具有 ROWS UNBOUNDED PRECEDING 的分析函数 COUNT 计算重复呼叫列（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
我需要创建一个新列，在每个 Customer_ID 的列 Repeated Call 的每个 0 值处重新启动：
+-------------+---------+----------------------+---------------+
| 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 |
+-------------+---------+----------------------+---------------+
进入这样的事情：
+-------------+---------+----------------------+---------------+-------------+
| 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！
您可以使用 SUM() window 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
参见 demo（对于 MySql 但它是标准的 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 window 具有 ROWS UNBOUNDED PRECEDING 的分析函数 COUNT 计算重复呼叫列（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