Question

我有一个基本的表结果（id，winner_id，loser_id）。 winner_id和loser_id都引用“玩家”表中的相同列。我可以通过1个GROUP BY查询来检索总赢数，并且可以检索非常类似的一个来检索总损失数。

select winner_id, count(*) as total_wins
from results group by winner_id order by total_wins desc

select loser_id, count(*) as total_losses
from results group by loser_id order by total_losses desc

现在我正在尝试执行一个基本上将这两个查询组合在一起并返回按ID分组的total_results的查询。有人能指出我正确的方向吗？

Answer 1

如果您确实返回了播放器的ID，那么查询似乎更有意义。否则，您只需订购一定数量的数字，但不能参考他们对应的数字：

select player_id, sum(total) total from (
  select winner_id player_id, count(*) total from results group by winner_id
  union all
  select loser_id, count(*) from results group by loser_id
) s
group by player_id
order by total desc

Answer 2

您可以在每个子查询的结果中包含id并将它们连接在一起。然后你可以减去两个计数：

SELECT id, total_wins - total_losses AS total_results
FROM  (
    SELECT winner_id AS id, count(*) AS total_wins
    FROM   results
    GROUP  BY winner_id
    ) AS w
JOIN (
    SELECT loser_id AS id, count(*) AS total_losses
    FROM   results
    GROUP  BY loser_id
    ) AS l USING (id)
ORDER  BY total_results DESC;

请注意，[INNER] JOIN会从两个子查询中未显示的结果中删除任何id。

~~您可以使用OUTER JOIN来保留此类行，并且必须添加COALESCE来处理NULL值：~~

SELECT id, COALESCE(total_wins, 0) - COALESCE(total_losses, 0) AS total_results FROM ( SELECT winner_id AS id, count(*) AS total_wins FROM results GROUP BY winner_id ) AS w FULL OUTER JOIN ( SELECT loser_id AS id, count(*) AS total_losses FROM results GROUP BY loser_id ) AS l USING (id) ORDER BY total_results DESC;

<击>

我刚刚意识到MySQL doesn't support FULL OUTER JOIN 使用@Mosty supplied的UNION ALL方法或者使用LEFT JOIN将两个子查询连接到包含所有ID的表。

如何通过查询对2组的结果进行求和

2 个答案: