我有以下表格:
match
id | rival_team
----------------------
| 1 | chelsea fc
| 2 | real madrid
player
ID | name | last_name |
---------------------------------
| 1 | John | Doe |
| 2 | Peter | Williams |
called_up_players
match_id | player_id | substitution_id |
---------------------------|-------------------|
| 1 | 1 | 1 |
| 1 | 2 | NULL |
substitution
| id | match_id | substitute_player_id | replaced_player_id |
---------------------------|----------------------|--------------------|
| 1 | 1 | 1 | 2 |
我有以下SQL语句
SELECT called_up_players.match_id, match.rival_team, player.name,
player.last_name, substitution.id
FROM called_up_players, substitution, player, match
WHERE called_up_players.substitution_id = substitution.substitute_player_id
AND player.id = called_up_players.substitution_id;
以及以下输出:
match_id| rival_team | name | last_name | substitution_id
--------+------------+----------+-----------+-----------
1 | chelsea fc | John | Doe | 1
1 | real madrid| John | Doe | 1
(2 rows)
但是,我希望输出像
match_id| rival_team | name | last_name | substitution_id
--------+------------+----------+-----=-----+----------------
1 | chelsea fc | John | Doe | 1
2 | real madrid| John | Doe | NULL
显示John Doe被召集到名单的所有比赛。无论玩家参与替换,我都想拥有名称和substitution_id
的列我想我可以使用JOINS实现这一点,但我不知道如何连接表,所以我可以有类似于上面的输出。我尝试了许多声明,但我得到了错误。
答案 0 :(得分:2)
您正在加入没有连接条件的匹配,因此这将成为交叉连接而不是内连接。
这是为什么不同的连接语法被认为是最佳实践的一个示例:
FROM called_up_players
INNER JOIN substitution ON called_up_players.substitution_id = substitution.substitute_player_id
INNER JOIN player ON player.id = called_up_players.substitution_id
INNER JOIN match ON ???
答案 1 :(得分:1)
感谢Learning2Code和他的回答我更了解连接,最后,我想出了一个答案来获得我想要的输出。
我没有使用INNER JOIN,而是使用LEFT JOIN,以便从左边的表中检索所有记录,如果右边的表中没有数据,则该字段为空
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输出:
SELECT match_id, rival_team, name, last_name, substitution_id
FROM match
LEFT JOIN called_up_players ON called_up_players.match_id = match.id
LEFT JOIN player ON player.id = called_up_player.player_id
LEFT JOIN substitution ON called_up_players.substitution_id = substitution.substitution_id
WHERE player.name = 'John' AND player.last_name = 'Doe';