Question

我正在整理一个新项目，我需要做的一件事情就是得到假足球（美国）比赛的结果。我在php和mysql中使用CodeIgniter作为我的框架。我的dbfiddle如下：

https://www.db-fiddle.com/f/5Q7QezddpNEGabaia2w5FQ/0

现在，如您所见，每个团队都有一个home_team_id和away_team_id条目。我想做的是查询每个队只得到一个结果（无论他们是在家还是不在）。如果不需要，我宁愿不以编程方式处理此数据。

结果应类似于以下内容：

SELECT week, home_team_id, away_team_id, 
    my_score, their_score, 
    a.team_name as home_team, b.team_name as away_team 
FROM nfl_user_matchups nm 
LEFT JOIN user_teams a ON nm.home_team_id = a.user_teams_id 
LEFT JOIN user_teams b ON nm.away_team_id = b.user_teams_id 
WHERE week = 1

预期的最终数据将是这样（my_score将是home_team_id得分，而他们的得分将是away_team_id得分）：

+------+--------------+--------------+----------+-------------+
| week | home_team_id | away_team_id | my_score | their_score |
+------+--------------+--------------+----------+-------------+
| 1    | 3            | 9            | 112      | 144         |
+------+--------------+--------------+----------+-------------+
| 1    | 7            | 2            | 85       | 96          |
+------+--------------+--------------+----------+-------------+
| 1    | 1            | 6            | 111      | 114         |
+------+--------------+--------------+----------+-------------+
| 1    | 4            | 5            | 99       | 125         |
+------+--------------+--------------+----------+-------------+
| 1    | 8            | 10           | 140      | 122         |
+------+--------------+--------------+----------+-------------+

Answer 1

您的问题陈述基本上意味着一对竞争的团队，即(1,2)，应与(2,1)相同。处理此类要求的一种方法是确保对于这两个团队之间的任何比赛，我们确保它们的顺序相同。因此，基本上，我们从两个团队中获得Least()团队ID值，并始终将其放在第一个索引中；并且Greatest()的值始终位于第二（最后）索引处。值得注意的是，Greatest()与Max()不同。 Max()函数可计算一列的最大值；而Greatest()通常用于比较行中的值。

现在，我们可以简单地在这对上GROUP BY并根据需要计算聚合。在确定家庭/离家分数时，您将需要使用CASE .. WHEN表达式来确定特定行的家庭ID最少，还是离家ID：

SELECT 
  week, 
  LEAST(home_team_id, away_team_id) AS home_id, 
  GREATEST(home_team_id, away_team_id) AS away_id, 
  MAX(CASE 
      WHEN LEAST(home_team_id, away_team_id) = home_team_id 
      THEN my_score 
      ELSE their_score
      END) AS home_score,
  MAX(CASE 
      WHEN GREATEST(home_team_id, away_team_id) = away_team_id 
      THEN their_score 
      ELSE my_score
      END) AS away_score 
FROM nfl_user_matchups 
WHERE week = 1
GROUP BY
  week, 
  home_id, 
  away_id;

View on DB Fiddle

每个团队每周只有一次“足球比赛”结果

1 个答案: