ID POSITION EXPERIENCE SALARY 1 top 90 1500 2 bottom 100 1500 3 top 90 750 4 left 90 1000 5 right 100 1300 6 top 90 1500 7 left 80 2000 8 top 80 1000 9 bottom 100 2000 10 left 100 2000
所以,这是我的表 SERVICE,其中(如我们所见)最大经验是 100。 我需要编写一个查询来查找通过位置(左、右、上、下)组成的每个组中经验中出现 100 的次数。
所以我写道:-
select position,count(*) from service group by position having experience=(select max(experience) from service);
预期输出:-
POSITION COUNT(*) bottom 2 left 1 right 1 top 0
但是, 它给了我一个错误:-“不是 GROUP BY 表达式”
我的逻辑是,首先我将其分为几组,然后使用having子句来计算每组中经验等于最大值的元组。经验。
P粉7187309562024-03-31 14:00:05
一种方法是使用带有子查询的左连接,它仅返回最大值。需要使用 case 来返回具有任意最大值的组。
SELECT s.position, sum(case when max_experience is null then 0 else 1 end ) as max_count FROM service s LEFT JOIN ( select max(experience) as max_experience from service ) as s1 ON s.experience = s1.max_experience group by s.position order by max_count desc ;
为了更容易理解,运行下面的查询,您将发现服务表中除值 100 之外的每一行中 max_experience 均为空。简单来说,您只需要计算组中值为 100 和 0 的行还没有达到最大经验值。
SELECT s.*,s1.* FROM service s LEFT JOIN (select max(experience) as max_experience from service ) as s1 ON s.experience = s1.max_experience ;
编辑。答案在 Oracle 中也有效,但需要删除子查询后面的关键字 as
SELECT s.position, sum(case when max_experience is null then 0 else 1 end ) as max_count FROM service s LEFT JOIN ( select max(experience) as max_experience from service ) s1 ON s.experience = s1.max_experience group by s.position order by max_count desc ;
P粉9365685332024-03-31 09:12:10
使用求和
:
select position, sum(experience = 100) from tbl group by position