Question

我有一个由以下列组成的表：

  id      name    value
serial    text    text

我需要移除id值最小且name = 'SUPPORT_EMAIL'的行。我怎样才能以最佳方式做到这一点？我试过了：

DELETE FROM propertie 
WHERE name = 'SUPPORT_EMAIL' 
HAVING id = MIN(id) 
GROUP BY name ;

和

DELETE FROM propertie 
WHERE name = 'SUPPORT_EMAIL' 
GROUP BY name 
HAVING id = MIN(id)  ;

但它没有用。当然，我可以写

DELETE FROM propertie 
WHERE id = (SELECT id 
        FROM propertie 
        WHERE name = 'SUPPORT_EMAIL' and id = (SELECT MIN(id) 
                            FROM propertie 
                            WHERE name = 'SUPPORT_EMAIL')
)

但对我来说似乎很奇怪。

Answer 1

你在那里非常接近，但可以通过以下方式简化你的工作：

DELETE FROM propertie 
WHERE id = (SELECT MIN(id) 
            FROM   propertie 
            WHERE  name = 'SUPPORT_EMAIL')

你也可以：

DELETE FROM propertie 
WHERE id = (SELECT id
            FROM   propertie 
            WHERE  name = 'SUPPORT_EMAIL'
            ORDER BY id
            LIMIT  1)

我不确定这是否是合法的PostgreSQL语法：

DELETE FROM   propertie 
WHERE  name = 'SUPPORT_EMAIL'
ORDER BY id
LIMIT  1

Answer 2

为了在删除查询中使用聚合函数，我们需要使用如下的子查询。我觉得这是优化的方式：

DELETE FROM attendence 
WHERE
  id IN 
     (select min(id) from attendence where name = 'SUPPORT_EMAIL' );

Answer 3

您的问题不清楚，但是如果有多个重复行，并且您想要删除所有重复但最后一个（ID最高的那个），您可以使用此查询：

DELETE FROM propertie p1
USING propertie p2
WHERE
  p1.name=p2.name
  AND p1.id<p2.id;

或者如果你想删除重复的SUPPORT_EMAILs：

DELETE FROM propertie p1
USING propertie p2
WHERE
  p1.name='SUPPORT_EMAIL'
  AND p1.name=p2.name
  AND p1.id<p2.id;

请参阅示例小提琴here。

如何删除最小值的副本？

3 个答案: