更快选择Max SQL查询

Question

这是用于选择具有最大行数的配置类型的查询。 有没有第二种查询选择数据一样快的另一种方法呢？

select CONFIG_TYPE, MAX(COUNTING) FROM 
(select CONFIG_TYPE, COUNT(*) as COUNTING FROM NOTIFICATION_CONFIG GROUP BY CONFIG_TYPE)
WHERE COUNTING =
(select MAX(COUNTING) FROM 
(select COUNT(*) as COUNTING FROM NOTIFICATION_CONFIG GROUP BY CONFIG_TYPE)
)GROUP BY CONFIG_TYPE

第二个查询将在1个搜索中完成工作，而另一个查询将在2个搜索中完成工作。

select CONFIG_TYPE, COUNTING FROM
(select CONFIG_TYPE, COUNT(*) as COUNTING FROM NOTIFICATION_CONFIG
GROUP BY CONFIG_TYPE ORDER BY COUNTING DESC FETCH FIRST 1 ROW ONLY)

只是想知道是否还有另一种方法可以做到这一点。

Answer 1

使用窗口功能！

SELECT c.*
FROM (SELECT CONFIG_TYPE, COUNT(*) as COUNTING,
             RANK() OVER (ORDER BY COUNT(*) DESC) as seqnum
      FROM NOTIFICATION_CONFIG
      GROUP BY CONFIG_TYPE
     ) c
WHERE seqnum = 1;

这使用RANK() ，因此它将返回所有最大值（在平局的情况下）。 如果要在ROW_NUMBER()的情况下返回任意的最大值，请改用ROW_NUMBER() 。 或者，如果您不想要联系，则可以执行以下操作：

SELECT CONFIG_TYPE, COUNT(*) as COUNTING
FROM NOTIFICATION_CONFIG
GROUP BY CONFIG_TYPE
ORDER BY COUNT(*) DESC
FETCH FIRST 1 ROW ONLY;

Answer 2

这只是搞砸了

select CONFIG_TYPE, MAX(COUNTING) 
FROM  ( select CONFIG_TYPE, COUNT(*) as COUNTING 
        FROM NOTIFICATION_CONFIG 
        GROUP BY CONFIG_TYPE 
      )
WHERE COUNTING = ( select MAX(COUNTING) 
                   FROM ( select COUNT(*) as COUNTING 
                          FROM NOTIFICATION_CONFIG 
                          GROUP BY CONFIG_TYPE
                        )
                 )
GROUP BY CONFIG_TYPE