使用max()获取多个行

Question

我写了下面的查询，给出了结果集-

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，如何选择具有最大值( planet_name )的每个字符的character_name和screen_time。对于ex，对于C-3PO，需要显示两行

C-3 PO | Bespin
C-3 PO | Tattoine

对于Chewbacca，要显示一行

Chewbacca | Endor

我所面临的问题是因为我无法实现中间表的条件。

Answer 1

编辑(删除了我以前的答案)

您还可以通过HAVING子句中的关联子查询实现相同的目标：

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);

考虑到您的问题，我不确定您是想检索screen_time，还是只想检索character_name和planet_name。如果只想使用最后两列，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:我以前的答案没有用。真对不起。

Answer 2

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 
);

• 看到它在木琴中运行

但别担心，表演不应该像最初看上去那么糟糕。MySQL通常足够聪明，不会执行相同的查询两次。

实现这一目标还有其他方法。你也可以试试，如果你喜欢的话：保留某一列的分组最大值的行。