如何使用光标删除重复条目

Question

附上的图像具有两个表表1和表2

表1：

+-------------------------------------------------+--+ 
| First_name Last_name Date_of_birth Customer_ID | | 
+-------------------------------------------------+--+ 
| --------- --------- ---------  ------- | | 
| Anmol  Sharma 12/SEP/1988  A101  | | 
| Anmol  Sharma 12/SEP/1988  A102  | | 
| Pushkar  Gupta  03/feb/1987  A103  | | 
| Pushkar  Gupta  03/Feb/1987  A104  | | 
+-------------------------------------------------+--+ 

表2：

First_name Last_name Date_of_birth Customer_ID Updated_timestamp 
---------  --------- -------------- ------------ --------------- 
Anmol   Sharma  12/SEP/1988  A101  23/Feb/2015 
Anmol   Sharma  12/SEP/1988  A102  24/FEb/2015 
Anmol   Sharma  12/SEP/1988  A101  25/Feb/2015 
Pushkar   Sharma  03/FEB/1987  A104  12/Jan/2015 

所需输出：

First_name Last_name Date_of_birth Customer_ID 
--------- --------- -------------- ------------ 
Anmol  Sharma  12/SEP/1988  A101 
Pushar  Sharma  03/FEB/1987  A104 

问：

我要保持近期更新CUSTOMER_ID（使用表2更新timestamp列）在表1和使用游标删除表1中的所有其他条目。 表格图像

有人可以请用光标解释逻辑。

Answer 1

DELETE FROM table1 WHERE Customer_ID NOT IN 
(SELECT Customer_ID from table2 
GROUP BY Customer_ID 
ORDER BY Updated_timestamp DESC LIMIT 1); 

Answer 2

纯SQL将是：

delete from table1 
where customer_id in (
     select customer_id 
      from (select customer_id, 
         row_number() over 
          (partition by last_name, first_name 
          order by updated_timestamp desc) rn 
        from table2) t 
      where rn > 1) 

删除之前单独执行每个子查询，以确保它是正确的。在第一个子查询row_number()函数返回1为最近更新的行，2 - 为第二等。第二个子查询选择除最近（rn > 1）之外的所有ID，并且它是要删除的ID。

Answer 3

使用rowid伪列。

DELETE FROM your_table WHERE rowid not in (SELECT MIN(rowid) FROM your_table GROUP BY column1, column2, column3);

凡列1，列2和栏3弥补每条记录的标识键。您可能会列出所有列。