Question

我有一个数据库，其中包含玩过的游戏及其得分。现在，我想运行一个查询，计算每个团队的总进球数超过x个（主场和客场总和）的所有结果。

数据库结构：

+----------+----------+---------------+---------------+
| HomeTeam | AwayTeam | HomeTeamGoals | AwayTeamGoals |
+----------+----------+---------------+---------------+
| team1    | team2    |             3 |             1 |
| team3    | team4    |             1 |             2 |
| team1    | team3    |             4 |             4 |
| team4    | team2    |             0 |             1 |
+----------+----------+---------------+---------------+

想要的结果：

+-------+----------+--------------------+
| team  | played   |  games > 3.5 goals |
+-------+----------+--------------------+
| team1 |        2 |                  2 |
| team2 |        2 |                  1 |
| team3 |        2 |                  1 |
| team4 |        2 |                  0 |
+-------+----------+--------------------+

我正在看这样的东西：

SELECT *, COUNT(*) AS total FROM games WHERE homeTeamGoals + awayTeamGoals > 3.5 GROUP BY homeTeam

但这仅适用于主场比赛，因此我需要一种将客场比赛也添加到其中的方法。

Answer 1

分别以HomeTeam和AwayTeam的形式获得团队的结果。
使用UNION ALL
将结果用作Derived Table。

派生表

Group by组成一个团队，以获取结果。

尝试以下操作（ SQL Fiddle DEMO ）：

SELECT derived_t.team, 
       SUM(derived_t.played) AS played, 
       SUM(derived_t.games)  AS 'games > 3.5 goals'
FROM 
(
  SELECT t1.HomeTeam as team, 
         COUNT(*) AS played, 
         SUM( IF(t1.HomeTeamGoals + t1.AwayTeamGoals > 3.5, 1, 0) ) AS games  
  FROM your_table AS t1 
  GROUP BY t1.HomeTeam

  UNION ALL 

  SELECT t1.AwayTeam as team, 
         COUNT(*) AS played, 
         SUM( IF(t1.HomeTeamGoals + t1.AwayTeamGoals > 3.5, 1, 0) ) AS games  
  FROM your_table AS t1 
  GROUP BY t1.AwayTeam
) AS derived_t
GROUP BY derived_t.team

php / mysql-计数和分组

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