Question

我有两张桌子。这是关于国际象棋比赛和它的球员。

在第一张桌子中，我有关于玩家的详细信息。在第二个表格中，我有关于游戏的详细信息（玩家1，玩家2），（游戏类型）。

两个表链接在一起。（两个参考，玩家1与第一个表的id链接，玩家2再次与第一个表的id链接）

我的任务是进行审讯。

我必须显示，玩过大多数游戏的玩家。

问题在于我将这个播放器放在两个表中。到目前为止，我设法显示了一个玩家在一个列中玩的游戏的最大数量：

select Player1,count(Player1) as number_ofgames
group by Player1 
having count(Player 1) in 
(select max(count(Player 1)) from games group by Player1);

Answer 1

我不确定您的表格结构因此假设您的player表格包含列player_id以及game表格player1_id和player2_id此查询可能符合您对Oracle 12c的要求。

select
    p.player_id,
    count(*) as games_played

from
    player p join game g on p.player_id = g.player1_id or p.player_id = g.player2_id

group by
    p.player_id

order by
    2 desc

fetch first 1 rows with ties;

如果多行与第N行的值匹配，则使用WITH TIES子句可能会导致返回更多行。

有关详细信息，请参阅http://www.oracle-base.com/articles/12c/row-limiting-clause-for-top-n-queries-12cr1.php#top-n。

在MySQL中使用LIMIT代替：

select
    p.player_id,
    count(*) as games_played

from
    player p join game g on p.player_id = g.player1_id or p.player_id = g.player2_id

group by
    p.player_id

order by
    2 desc

limit 1;

在此处阅读更多内容：http://dev.mysql.com/doc/refman/5.0/en/select.html。

Answer 2

如果您使用的是11G R2，则可以使用listagg功能。如果没有，很容易找到不同的方式

   With distrib as
                (select O as dabegin, 17 as daend, 'under 18' as category from dual
                union all 
                select 18 as dabegin ,25 as daend, 'between 18-25' as category from dual
                union all 
                select 26 as dabegin ,30 as daend,  'between 26-30' as category from dual
                 )
                , games as
                (select
                    p.player_id,
                    count(*) as games_played

    from
        player p join game g on p.player_id = g.player1_id or p.player_id = g.player2_id

group by
    p.player_id)
select max(games_played), listagg(player_id,',') WITHIN GROUP (ORDER BY player_id) as 'winner(s)', category
from distrib d, games g, players p
where p.player_id=g.player_id
and g. age between d.dabegin and daend
group by g.category

SQL-Oracle Interrogation

2 个答案: