SQL:如何使多个连接到表的同一列，而无需覆盖结果

Question

我有篮球比赛的桌子和一支像这样的篮球队：

MATCHES:
ID   |   HOME_TEAM_ID  |  AWAY_TEAM_ID | SCORE_HOME | SCORE_AWAY
----------------------------------------------------------------
1    |       20        |       21      |     80     |    110
2    |       12        |       10      |     96     |     90

TEAMS:
ID   |   NAME
-------------------------
20   |   BULLS
21   |   KNICKS

给出了一个匹配ID，我想检索得分和团队名称。我该如何加入从团队表中检索两个团队名称？

我尝试过：

SELECT *
FROM matches AS m
JOIN teams AS t1 ON t.id = m.home_team_id
JOIN teams AS t2 ON ti.id = m.away_team_id
WHERE m.id = 1

...但是，这里的第二个加入语句的结果似乎覆盖了第一个的结果，所以我只得到一个名称：

[id] => 1
[score_home] => 80
[score_away] => 110
[name] => KNICKS

我也尝试了：

SELECT *
FROM matches AS m
JOIN teams AS t ON (t.id = m.home_team_id OR t.id = m.away_team_id)
WHERE m.id = 1

...返回两个结果：

[id] => 1
[score_home] => 80
[score_away] => 110
[name] => BULLS

和

[id] => 1
[score_home] => 80
[score_away] => 110
[name] => KNICKS

我想进行一个查询，以返回类似的东西

[id] => 1
[score_home] => 80
[score_away] => 110
[name_home_team] => BULLS
[name_home_team] => KNICKS

可能吗？

Answer 1

SELECT
    Matches.ID,
    Matches.Score_Home,
    Matches.Score_Away,
    HomeTeam.Name Home_Team_Name,
    AwayTeam.Name Away_Team_Name
FROM
    Matches
    INNER JOIN Teams HomeTeam ON Matches.Home_Team_ID = HomeTeam.ID
    INNER JOIN Teams AwayTeam ON Matches.Away_Team_ID = AwayTeam.ID

Answer 2

您只需要名称列名

SELECT m.ID,
       m.SCORE_HOME,
       m.SCORE_AWAY,
       t1.NAME as name_home_team,
       t2.NAME as name_home_team
FROM MATCHES AS m
JOIN teams AS t1 ON t1.id = m.home_team_id
JOIN teams AS t2 ON t2.id = m.away_team_id
WHERE m.id = 1

Answer 3

   select  m.ID,
           (select NAME from TEAM where id=m.HOME_TEAM_ID) HOME_TEAM_NAME,
           m.SCORE_HOME,
           (select NAME from TEAM where id=m.AWAY_TEAM_ID) AWAY_TEAM_NAME,    
           m.SCORE_AWAY 
    from 
    MATCHES m
    where m.ID=1

Answer 4

未测试，应起作用：

select x.id, x.score_home, x.score_away, y.name as home_team, z.name as away_team
from matches x,
  (select name from teams where id = x.home_team_id)y,
  (select name from teams where id = x.away_team_id)z
where x.id = 1;