Question

我正在寻找一种方法来编写一个查询来比较多个mysql子查询的结果，并返回每个查询中的用户。

我有一个包含玩家幻想足球统计数据。为了简化，在这种情况下，我使用了3列：播放器，点，年。

我希望运行一个查询，返回2010年和2011年排名前50位（基于积分）的玩家列表。

我在搜索子查询，在一张桌子上进行连接等方面做了大量的搜索，但是我仍然不知道如何处理它。

Answer 1

您可以这样做：

SELECT  a.player

FROM    (SELECT player FROM players WHERE Year = 2010 ORDER BY points DESC LIMIT 50) a

        JOIN

        (SELECT player FROM players WHERE Year = 2011 ORDER BY points DESC LIMIT 50) b
        ON a.player = b.player

Answer 2

这是一个例子。我假设您根据积分总和计算top50，并且每年每位玩家都有几个参赛作品。


select y2010.player 
from (
  select player, sum from (
    select st1.player player, sum(st1.points) sum from stats st1 where st1.year = 2010 group by st1.player order by sum desc
  ) t1 limit 50 offset 0
) y2010, (
  select player, sum from (
    select st1.player player, sum(st1.points) sum from stats st1 where st1.year = 2011 group by st1.player order by sum desc
  ) t1 limit 50 offset 0
) y2011
where y2010.player = y2011.player

Answer 3

您可以使用UNION ALL，这会在两年内获得Top 50并将它们放在相同的结果集中，不需要加入：

(
    select player, year, points 
    from players
    where year = 2010
    order by points desc
    limit 50
)
union all
(
    select player, year, points 
    from players
    where year = 2011
    order by points desc
    limit 50
);

Answer 4

您是否想要：

，这有点模棱两可

2010年进入前50名的所有球员，以及2011年进入前50名的所有球员：

SELECT *
FROM   scores
WHERE  year = 2010
   AND points >= (SELECT MIN(points) FROM (
                   SELECT   points
                   FROM     scores
                   WHERE    year = 2010
                   ORDER BY points DESC
                   LIMIT    50
                 ) t)
UNION ALL
SELECT *
FROM   scores
WHERE  year = 2011
   AND points >= (SELECT MIN(points) FROM (
                   SELECT   points
                   FROM     scores
                   WHERE    year = 2011
                   ORDER BY points DESC
                   LIMIT    50
                 ) t)

所有在 2010年和2011年都进入前50名的玩家，在这种情况下，您需要进一步对结果进行分组：

SELECT player FROM ( -- query as above ) t GROUP BY player HAVING COUNT(DISTINCT year) = 2

Mysql返回子查询中的匹配用户

4 个答案: