如何使用光標刪除重複條目

Question

附上的圖像具有兩個表表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 

所需輸出：

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

問：

我要保持近期更新CUSTOMER_ID（使用表2更新timestamp列）在表1和使用遊標刪除表1中的所有其他條目。 表格圖像

有人可以請用光標解釋邏輯。

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

純SQL將是：

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) 

刪除之前單獨執行每個子查詢，以確保它是正確的。在第一個子查詢row_number()函數返回1爲最近更新的行，2 - 爲第二等。第二個子查詢選擇除最近（rn > 1）之外的所有ID，並且它是要刪除的ID。

Answer 3

使用rowid僞列。

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

凡列1，列2和欄3彌補每條記錄的標識鍵。您可能會列出所有列。