Php 对不同类别中的每个队员的所有分数求和

我需要为每个玩家计算所有分数（他的核心）

现在，我可以求player1或player2的和。例如，在第7类中，玩家1是汤姆，而在第5类中，玩家2也是汤姆。结果应为Tom=2+10=12分

SELECT SUM(subpoints) AS hiscore, player1_id, player2_id, FROM (
    SELECT COUNT(current_record)*2 AS subpoints, player1_id, player2_id FROM db WHERE category_id IN (7,8) GROUP BY player1_id
       UNION ALL
    SELECT COUNT(current_record)*10 AS subpoints, player1_id, player2_id FROM db WHERE category_id IN (1,2,3,4,5,6) GROUP BY player1_id
) AS hi
GROUP BY player1_id

样本表：

 category_id | player1 | player2 | subpoints |
-------------+---------+---------+-----------+
           7 |   Tom   |  Mike   |     2     |
           5 |  Peter  |  Tom    |    10     |
----------------------------------------------

最终结果应为：

  Player | hiscore |
  -------+---------+
  Tom    |   12    |
  Mike   |    2    |
  Peter  |   10    |

您可以先取消填充列，然后将其聚合：

select player, sum(subpoints) hiscore
from
(
  select category_id, player1 player, subpoints
  from db
  union all
  select category_id, player2 player, subpoints
  from db
) d
group by player;

看

如果将原始查询集成到此中，则代码将类似于：

select player, sum(subpoints) hiscore
from
(
  SELECT COUNT(current_record)*2 AS subpoints, player1_id as player
  FROM db 
  WHERE category_id IN (7,8) 
  GROUP BY player1_id
  UNION ALL
  SELECT COUNT(current_record)*10 AS subpoints, player2_id as player
  FROM db 
  WHERE category_id IN (1,2,3,4,5,6) 
  GROUP BY player2_id
) d
group by player

---

好的，这很有效。我把“player2\u id作为player1\u id”搞错了。