MySQL的选择数

Question

PlayerLeague

• league_id

• player_id

• player_position

PlayerStat

• player_id

• 得分

我有两个表，我想更新PlayerLeague，设置player_position相等行数：

SELECT * FROM PlayerStat s JOIN PlayerLeague l 
ON s.player_id=l.player_id WHERE l.league_id=3 ORDER BY score DESC; 

我试着在我的选择中使用@i，但与连接ORDER BY被忽略。

UPDATE

所以，现在我也这样与它的工作，但我不知道它的正常方式。 它将每小时运行三个联赛40-500名球员。

SET @i=0; 
UPDATE PlayerLeague ll set ll.player_position = (SELECT position FROM (SELECT 
s.player_id, @i:=@i+1 as position 
FROM PlayerStat s 
where s.player_id in 
    (SELECT player_id from PlayerLeague l WHERE l.league_id = 3) 
    ORDER BY score DESC) AS t WHERE t.player_id=ll.player_id); 

Answer 1

满足这个小可爱怪物：

UPDATE PlayerLeague AS l 
    JOIN (
     SELECT p.player_id, @i := @i + 1 AS player_position 
     FROM (
      SELECT ss.player_id 
      FROM PlayerStat AS ss 
       JOIN PlayerLeague AS ll 
        ON ss.player_id=ll.player_id 
      WHERE ll.league_id=3 
      ORDER BY ss.score DESC 
     ) AS p, (SELECT @i := 0) AS tmp 
    ) AS s ON s.player_id=l.player_id 
SET l.player_position = s.player_position 

因此，这里发生了什么？

• 最内层查询按照他们的得分顺序选择联盟中的所有玩家。
• 它包含在添加行数的查询中。
• 最后，外层UPDATE可以满足您的需求：将行数添加到PlayerLeague表中。

Answer 2

尝试

SELECT * 
    FROM PlayerStat s 
where s.player_id in 
     (select player_id from PlayerLeague l WHERE l.league_id = 3) 
ORDER BY score DESC;