mysql - Mysql 在子查询中返回匹配的用户

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。

我有一个包含球员的梦幻足球统计数据。为简化起见，在这种情况下，我使用了 3 列：玩家、积分、年份。

我希望运行一个查询，以返回 2010 年和 2011 年均进入前 50 名（基于积分）的球员列表。

我已经在使用子查询、在一张表上进行连接等方面进行了很多搜索，但我仍然对如何处理这个问题感到茫然。

Answer 1

你可以这样做：

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

您可以使用 a UNION ALL，这将Top 50在两年内为您提供并将它们放在相同的结果集中，无需加入：

(
    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 3

Here is an example. I assumed that you calculate top50 based on sum of points and you have several entries for each player in each year.

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 4

您是否想要：

• 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年均进入前 50 名的球员，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