Question

我从SQL开始，当我想运行一些查询时遇到问题。数据库有这两个表：

我想检索“id_equip”27和58的球队名称，以及两者之间的中场比赛的百分比。

到目前为止，我已经完成了：

SELECT   j.id_eq_local , 
         Avg(res_local) 
FROM     juga AS j 
WHERE    ( 
                  j.id_eq_local = 58 
         OR       j.id_eq_local = 27) 
AND      ( 
                  res_local > res_visitant 
         OR       res_local < res_visitant) 
AND      ( 
                  j.id_eq_visitant = 58 
         OR       j.id_eq_visitant = 27) 
GROUP BY (j.id_eq_local, j.id_eq_visitant);

但它没有给出结果：

巴塞罗那 - 57

皇马 - 28

这意味着：巴塞罗那赢得了57％的比赛，皇家马德里获得了28％的胜利，其余的则以平局结束。

数据库的简要说明：

表JUGA：

id_jornada =本周的身份
id_temp =本赛季的身份
id_lliga =联赛的ID（在这种情况下必须是“ESP”）
id_eq_local =本地团队的ID
id_eq_visitant =来宾团队的身份
res_local =本地团队成绩
res_vistitant =来宾团队成绩

表设备：

id_equip =团队的身份
nom_equip =团队名称

一些数据样本：

我想要的是什么：

提前谢谢。

Answer 1

可以在聚合函数上使用postgre的filter - 说明符，它将聚合函数考虑的行限制为满足特定条件的行。因此，我们可以比较同一查询中不同行集的计数：

select 'Real Madrid' as equip,
  100*(count(*) filter (where (id_eq_local > id_eq_visitant and res_local > res_visitant) OR (id_eq_local < id_eq_visitant and res_local < res_visitant))) / count(*) as percentage
from juga
where id_eq_visitant IN (27,58)
  and id_eq_local IN (27,58)

UNION 

select 'Barcelona' as equip,
  100*(count(*) filter (where (id_eq_local < id_eq_visitant and res_local > res_visitant) OR (id_eq_local > id_eq_visitant and res_local < res_visitant))) / count(*) as percentage
from juga
where id_eq_visitant IN (27,58)
  and id_eq_local IN (27,58)

Answer 2

我认为这样做会。

${propA${propB}}

Answer 3

我认为这就是你要找的东西：

select (local.local_wins + guest.guest_wins) / (local.cnt + guest.cnt) as Team1,
       (local.guest_wins + guest.local_wins) / (local.cnt + guest.cnt) as Team2
from 
    (select sum(case when res_local > res_visitant then 1 else 0 end) as local_wins, sum(case when res_visitant > res_local then 1 else 0 end) guest_wins, count(*) as cnt
     from JUGA where id_local = 53 and id_guest = 17) local,

    (select sum(case when res_local > res_visitant then 1 else 0 end) as local_wins, sum(case when res_visitant > res_local then 1 else 0 end) as guest_wins, count(*) as cnt 
     from JUGA where id_local = 17 and id_guest = 53) guest

具有2个表的SQL查询百分比

3 个答案: