Mysql返回匹配用戶的子查詢

Question

我正在尋找一種方法來編寫一個查詢來比較多個mysql子查詢的結果，並返回每個查詢中的用戶。

我有一個包含幻想足球統計球員。爲了簡化，在這種情況下，我使用了3列：玩家，積分，年份。

我期待在2010年和2011年

運行（基於分）返回的球員誰在排名前50位完成了一個列表的查詢我已經做了與玩弄搜索的手子查詢，在一張桌子上進行連接等，但我仍然對如何解決這個問題感到迷茫。

Answer 1

你可以做這樣的事情：

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

下面是一個例子。我假設您根據積分總和來計算top50，並且每年有多個參賽者。

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

可以使用UNION ALL，這將讓你在這兩年的Top 50並把它們放在相同的結果集，沒有連接所需要：

(
    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

這是有些不明確您是否希望：

• 所有在2010年獲得前50名的選手以及2011年排名前50位的所有選手：

  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) 
  

• 誰在排名前50位中都 2010和2011年完成，在這種情況下，您需要進一步的組，結果所有的球員：

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