Question

假设我的SQL数据库中有三列，如

ID | NAME | PHONE
-----------------    
 1 | JEFF | 467
 2 | JEFF | 489
 3 | JOHN | 234
 4 | JACK | 323
 5 | JEFF | 378

我想编写一个SQL查询，删除检测到每NAME次出现的所有行。这意味着在运行SQL查询后，该表应如下所示：

ID | NAME | PHONE
-----------------
 1 | JEFF | 467
 2 | JOHN | 234
 3 | JACK | 323

非常感谢你！

非常感谢，我现在把它改成了

delete from product_list y
    where exists (select 1 from product_list y2 where y.model = y2.model and y2.linkid < y.linkid);

但我总是得到这个错误：

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'delete * from product_list y where exists (select 1 from product_list y2 whe' at line 3

提前致谢！

Answer 1

标准的SQL方法是：

delete from yourtable y
    where exists (select 1 from yourtable y2 where y.name = y2.name and y2.id < y.id);

即，删除存在具有相同名称和较低ID的记录的所有记录。

如果您只想返回行，请使用相同的想法：

select y.*
from yourtable y
where not exists (select 1 from yourtable y2 where y.name = y2.name and y2.id < y.id);

Answer 2

我认为这个查询应该有效。在此查询中，Employee是我的表名，Employee-id是密钥。

DELETE from Employee WHERE EmpID in(select EmpID from Employee GROUP BY EmpId HAVING COUNT(*) >1)

删除SQL中的重复列

2 个答案: