Question

我需要选择所有匹配并在指定的比赛中按团队进行分组。
问题是，在匹配中，我有2个外键到团队
1）host_id
2）guest_id

要选择两个团队，我选择比赛并加入：

JOIN
  teams host ON matches.host_id = host.id
JOIN
  teams guest ON matches.guest_id = guest.id

但我有问题如何将它们分组，以便对来自不同团队的匹配的所有统计数据进行合并。主人和客人都不会成倍增加因为如果我按照host.name和guest.name对它们进行分组，我会像Team一样获得Team1的SUM，而像Team1一样获得Team1的SUM。

顺便说一下。欢迎stackoverflow;）

Answer 1

创建一个子集合，用于获取所需的聚合，然后返回到主要语句。

(rest of your joins)
join
(select host.id, sum(score) totalscore,avg(gametime) average_gametime,sum(whateverelse)
from host
group by host.ID) a
on host.id = a.id

现在你可以调用a.totalscore，a.average_gametime或select子句中的任何其他内容

Answer 2

你似乎想要一个团队在家里和客人时的匹配信息：

select t.id, count(*), sum(...), ...
from teams t join matches m
where t.id = m.home or t.id=m.guest
group by (t.id)

您的团队ID以外的比赛信息现在也可以分开给家庭和客人。例如匹配列home_score＆amp; away_score。然后你需要用“或”来获取那些信息：

select t.id,
    case when when t.id = m.home then "home" when t.id = m.guest then "guest" end case) as role,
    sum(*) as games_played,
    sum(case when t.id = m.home then m.home_score when t.id = m.guest then m.home_guest end case) as total_score,
    case when t.id = m.home then m.home_penalties when t.id = m.guest then m.home_penalties end case) as total_penalties,
    ...

这可以用较少的case表达式编写，但可能更慢：

select id, role, count(*), sum(score), sum(penalty)
from teams t
join (select home as team, "home" as role, m.home_score as score, m.home_penalty as penalty, ..., match.* from matches)
    union
    select away as team, "away" as role, m.away_score as score, m.away_penalty as penalty, ..., match.* from matches) m
on t.id = m.team
group by (t.id)

为了避免这些并发症，您可以保持一个团队是在家还是离开（角色）加上其在表中每场比赛的结果，并且只保留另一个表中匹配的团队配对，然后定义与其home_ / away_列的匹配作为他们的观点。

同一个表和组的多个外键

2 个答案: