我写了下面的查询,它给了我结果集 -
select
a.character_name,
b.planet_name,
sum(b.departure - b.arrival) AS screen_time
from characters a
inner join timetable b
on a.character_name = b.character_name
group by a.character_name, b.planet_name;
结果集:
Character_name | planet_name | Screen_time
C-3 PO Bespin 4
C-3 PO Hoth 2
C-3 PO Tatooine 4
Chewbacca Bespin 4
Chewbacca Endor 5
Chewbacca Hoth 2
Chewbacca Tatooine 4
现在,如何选择具有max(screen_time(的每个字符的planet_name和character_name。例如,对于 C-3 采购订单,要显示两行
C-3 PO | Bespin
C-3 PO | Tattoine
对于丘巴卡,显示一行
Chewbacca | Endor
我面临的问题是因为我无法将条件实现到中间表。
编辑(删除了我之前的答案(
您还可以使用 HAVING
子句中的相关子查询来实现相同的目的:
select
a.character_name,
b.planet_name,
sum(b.departure - b.arrival) as screen_time
from characters a
inner join timetable b
on a.character_name = b.character_name
group by a.character_name, b.planet_name
having sum(b.departure - b.arrival) = (
select
max(tmp.screen_time)
from (
select b.character_name, sum(b.departure - b.arrival) as screen_time
from timetable b
group by b.character_name, b.planet_name) tmp
where a.character_name = tmp.character_name
group by tmp.character_name);
鉴于您的问题,我不确定您是要检索screen_time
还是只是character_name
和planet_name
.如果只需要最后两列,请从主查询中删除sum(b.departure - b.arrival) as screen_time
:
select
a.character_name,
b.planet_name
from characters a
inner join timetable b
on a.character_name = b.character_name
group by a.character_name, b.planet_name
having sum(b.departure - b.arrival) = (
select
max(tmp.screen_time)
from (
select b.character_name, sum(b.departure - b.arrival) as screen_time
from timetable b
group by b.character_name, b.planet_name) tmp
where a.character_name = tmp.character_name
group by tmp.character_name);
PS:我之前的答案不起作用。对此感到抱歉。
select character_name, planet_name
from (
select
a.character_name,
b.planet_name,
sum(b.departure - b.arrival) AS screen_time
from characters a
inner join timetable b
on a.character_name = b.character_name
group by a.character_name, b.planet_name
) sq
where screen_time = (SELECT MAX(ssq.screen_time) FROM (
select
a.character_name,
b.planet_name,
sum(b.departure - b.arrival) AS screen_time
from characters a
inner join timetable b
on a.character_name = b.character_name
group by a.character_name, b.planet_name
) ssq WHERE ssq.character_name = sq.character_name
);
- 看到它在SQLfiddle中实时工作
但别担心,性能应该不会像最初看起来那么糟糕。MySQL通常足够聪明,不会执行两次相同的查询。
还有其他方法可以实现相同的目的。如果您愿意,您也可以尝试这些: 保持特定列的组最大值的行