如何结合结果来获得阿森纳赢得所有可用六分的球队数量

Question

我有一个名为“home”的表，用于存储阿森纳足球俱乐部在 2003-04 赛季主场联赛中的表现，而“away”表则存储阿森纳在同一赛季客场的表现。

Table `home`:

Opponent           Goal_Scored      Goal_Conceded
Everton                2                 1
Aston Villa            2                 0
Portsmouth             1                 1
Newcastle United       3                 2
Chelsea                2                 1
Tottenham Hotspur      2                 1
Fulham                 0                 0
Blackburn Rovers       1                 0
Wolverhampton Wonderers 3                0
Middlesbrough          4                 1


Table `away`:

Opponent           Goal_Scored      Goal_Conceded
Everton                1                 2
Aston Villa            2                 2
Portsmouth             1                 2
Newcastle United       2                 3
Chelsea                2                 2
Tottenham Hotspur      2                 3
Fulham                 2                 0
Blackburn Rovers       1                 2
Wolverhampton Wonderers 3                2
Middlesbrough          1                 4

请注意，一支球队获胜获得 3 分，平局获得 1 分，失败获得 0 分。

我想确定阿森纳赢得所有可用六分的球队数量。

为此，我编写了部分代码：

CREATE TABLE home (
    Opponent varchar(60),
    Goal_Scored int(8),
    Goal_Conceded int(8)
);

CREATE TABLE away (
    Opponent varchar(60),
    Goal_Scored int(8),
    Goal_Conceded int(8)
);

INSERT INTO home (Opponent, Goal_Scored, Goal_Conceded)
VALUES 
("Everton",2,1),
("Aston Villa",2,0),
("Portsmouth",1,1),
("Newcastle United",3,2),
("Chelsea",2,1),
("Tottenham Hotspur",2,1),
("Fulham",0,0),
("Blackburn Rovers",1,0),
("Wolverhampton Wonderers",3,0),
("Middlesbrough",4,1);


INSERT INTO away (Opponent, Goal_Scored, Goal_Conceded)
VALUES
("Everton",1,2),
("Aston Villa",2,2),
("Portsmouth",1,2),
("Newcastle United",2,3),
("Chelsea",2,2),
("Tottenham Hotspur",2,3),
("Fulham",2,0),
("Blackburn Rovers",1,2),
("Wolverhampton Wonderers",3,2),
("Middlesbrough",1,4);

为了得到综合结果，我尝试了下面的代码，但没有产生预期的结果：

select a.opponent
from home a 
where a.goals_scored > a.goals_conceded
union all
select b.opponent
from away b 
where b.goals_scored > b.goals_conceded

然后我应该如何结合结果来获得阿森纳赢得所有可用六分的球队数量

Answer 1

没有必要根据您的要求检查积分。
条件应该是同一个对手Goal_Scored > Goal_Conceded为真：

SELECT Opponent
FROM (
  SELECT * FROM home WHERE Goal_Scored > Goal_Conceded
  UNION ALL
  SELECT * FROM away WHERE Goal_Scored > Goal_Conceded
) t
GROUP BY Opponent
HAVING COUNT(*) = 2 -- for 2 matches against the same opponent

如果您只想要团队的数量，那么使用EXISTS 聚合会更好：

SELECT COUNT(*) number_of_teams 
FROM home h
WHERE h.Goal_Scored > h.Goal_Conceded
  AND EXISTS (SELECT 1 FROM away a WHERE a.Opponent = h.Opponent AND a.Goal_Scored > a.Goal_Conceded) 

请参阅demo。

Answer 2

既然您已经有了解决方案，这里是加入解决方案：

select opponent
from home
join away using (opponent)
where home.goal_scored > home.goal_conceded
  and away.goal_scored > away.goal_conceded;

我们将每支球队的主客场记录加入其中，然后过滤掉阿森纳没有赢得两回合的所有记录。

Answer 3

如果我理解正确，你可以这样做：

select t.Opponent, sum(CASE WHEN goal_scored - goal_conceded > 0 THEN 3
WHEN goal_scored - goal_conceded = 0 THEN 1 ELSE 0 end ) points 
from 
( SELECT * FROM home a
  union all
  SELECT * FROM away a
) t
group by Opponent
having points = 6;

db小提琴here