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我在 MySQL 实例中使用Lahman 棒球数据库。我想找到每年都在本垒打（HR）中名列前茅的球员。Batting 表的架构具有以下（相关部分）：

+-----------+----------------------+------+-----+---------+-------+
| Field     | Type                 | Null | Key | Default | Extra |
+-----------+----------------------+------+-----+---------+-------+
| playerID  | varchar(9)           | NO   | PRI |         |       |
| yearID    | smallint(4) unsigned | NO   | PRI | 0       |       |
| HR        | smallint(3) unsigned | YES  |     | NULL    |       |
+-----------+----------------------+------+-----+---------+-------+

对于每一年，每个玩家都有一个条目（每年数百到 12k，可以追溯到 1871 年）。一年内获得前 N 名击球手很容易：

SELECT playerID,yearID,HR
FROM Batting
WHERE yearID=2009
ORDER BY HR DESC LIMIT 3;
+-----------+--------+------+
| playerID  | yearID | HR   |
+-----------+--------+------+
| pujolal01 |   2009 |   47 |
| fieldpr01 |   2009 |   46 |
| howarry01 |   2009 |   45 |
+-----------+--------+------+

但我有兴趣找到每年的前三名。我找到了这样的解决方案，描述了如何从一个类别中选择顶部，我试图将它应用于我的问题，但最终得到一个永远不会返回的查询：

SELECT
    b.yearID, b.playerID, b.HR
FROM
    Batting AS b
LEFT JOIN
    Batting b2
    ON
    (b.yearID=b2.yearID AND b.HR <= b2.HR)
GROUP BY b.yearID HAVING COUNT(*) <= 3;

我哪里出错了？

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2 回答 2

3

像这样的东西应该工作：

SELECT b.playerID, b.yearID, b.HR
FROM Batting b 
WHERE HR >= (
    SELECT b2.HR 
    FROM Batting b2 
    WHERE b2.yearID=b1.yearID
    ORDER BY b2.HR DESC
    LIMIT 2, 1
)
ORDER BY b.yearID DESC, b.HR DESC;

解释：选择所有本垒打数 >= 数的行作为当年第三高的行。这不会断绝关系。因此，如果有多个击球手拥有相同数量的本垒打，他们都会出现。

结果是从最近一年开始排序的，按每年的排名进行子排序。

注意：LIMIT是从0开始的偏移量，所以2、1表示从第二行开始抓一行，即：第三行。

于 2010-06-17T17:06:18.583 回答
0

哇，随意。我碰巧在 Lahman Baseball DB 上使用关于模拟 Oracle 分析函数的文章进行了相同的查询（尽管是薪水）。这个版本的查询很简洁，但不是那么直观。

select *
from (

select 
    b.yearID as year,
    b.teamID as team,
    m.nameFirst as first,
    m.nameLast as last,
    find_in_set(b.HR, x.teamRank) as rank,
    b.HR as HR


from 
    Batting b
    inner join Master m on m.playerID = b.playerID
    inner join (select yearID, group_concat(distinct HR order by HR desc) as teamRank from Batting group by yearID) x on x.yearID = b.yearID

) x

where 
    rank <= 10 and rank > 0 

order by    
    year desc, rank

或者 2010 年每个团队的前 5 名人力资源总数...

select *
from (

select 
    b.yearID as year,
    b.teamID as team,
    m.nameFirst as first, 
    m.nameLast as last, 
    b.HR as HR,
    find_in_set(b.HR, x.teamRank) as rank

from 
    Batting b
    inner join Master m on m.playerID = b.playerID
    inner join (select teamID, group_concat(distinct HR order by HR desc) as teamRank from Batting where yearID = 2010 group by teamID) x on x.teamID = b.teamID
where 
    b.yearID = 2010
) x

where 
    rank <= 5 and rank > 0 

order by    
    team, rank

limit 12

显示这些结果...

year    team    first   last        HR  rank
2010    ARI Mark    Reynolds    32  1
2010    ARI Chris   Young       27  2
2010    ARI Kelly   Johnson     26  3
2010    ARI Adam    LaRoche     25  4
2010    ARI Justin  Upton       17  5
2010    ATL Brian   McCann      21  1
2010    ATL Jason   Heyward     18  2
2010    ATL Troy    Glaus       16  3
2010    ATL Martin  Prado       15  4
2010    ATL Eric    Hinske      11  5
2010    BAL Luke    Scott       27  1
2010    BAL Ty      Wigginton   22  2
于 2011-01-19T05:21:12.773 回答