Question

在这个查询中，我必须列出一对玩家ID和玩家名称的球员，他们为同一支球队效力。如果一名球员为3支球队效力，则另一支球员必须参加完全相同的3支球队。不能少，不多了。如果两名球员目前不参加任何球队，他们也应该被包括在内。查询应该返回（playerID1，playername1，playerID2，playerName2）而没有重复，例如如果玩家1信息在玩家2之前出现，则不应该有另一个玩家2信息在玩家1之前出现的元组。

例如，如果玩家A为洋基队和红袜队队员比赛，而且队员队员为洋基队队员，红袜队队员和道奇队队员队效力，我就不应该参加比赛。他们都必须为洋基队和红袜队效力，而不是其他人。现在，如果玩家为同一个团队玩游戏，此查询会找到答案。

 player(playerID: integer, playerName: string)
 team(teamID: integer, teamName: string, sport: string)
 plays(playerID: integer, teamID: integer)

现在我的查询是

SELECT p1.playerID, p1.playerName, p2.playerID, p2.playerName
FROM player p1, player p2, plays
WHERE p1.teamID = p2.teamID AND teamID in.....

我坚持在此之后如何接近它。有关如何解决此问题的任何提示。谢谢你的时间。

Answer 1

我认为最简单的方法是将团队连接在一起，然后加入结果。 Postgres提供函数string_agg()来聚合字符串：

select p1.playerId, p1.playerName, p2.playerId, p2.playerName
from (select p.playerId, string_agg(cast(p.TeamId as varchar(255)), ',' order by TeamId) as teams,
             pp.PlayerName
      from plays p join
           players pp
           on p.playerId = pp.playerId
      group by p.playerId
     ) p1 join
     (select p.playerId, string_agg(cast(p.TeamId as varchar(255)), ',' order by TeamId) as teams,
             pp.PlayerName
      from plays p join
           players pp
           on p.playerId = pp.playerId
      group by p.playerId
     ) p2
     on p1.playerid < p2.playerid and p1.teams = p2.teams;

编辑：

您可以在不string_agg的情况下执行此操作。我们的想法是从所有可能的玩家组合列表开始。

然后，使用left outer join加入第一个玩家的团队。并使用full outer join并在团队和驱动程序名称上进行匹配，加入团队中的第二个团队。您需要驱动程序表的原因是为了确保id / name不会在完全外连接中丢失：

select driver.playerid1, driver.playerid2
from (select p1.playerId as playerId1, p1.playerName as playerName1,
             p2.playerId as playerId2, p1.playerName as playerName2
      from players p1 cross join
           players p2
      where p1.playerId < p2.playerId
     ) driver left outer join
     plays p1
     on p1.playerId = driver.playerId full outer join
     plays p2
     on p2.playerId = driver.playerId and
        p2.teamid = p1.teamid
group by driver.playerid1, driver.playerid2
having count(p1.playerid) = count(*) and
       count(p2.playerid) = count(*);

这会加入团队ID上的两个玩家（订购时只需考虑一对）。然后，当两个玩家的所有行都具有非NULL团队值时，它会说匹配。对于等效的having子句，这可能更清楚：

having sum(case when p1.playerid is null then 1 else 0 end) = 0 and
       sum(case when p2.playerid is null then 1 else 0 end) = 0;

当两名球员的球队不匹配时，全外联接将产生NULL值。因此，没有NULL值表示所有团队都匹配。

Answer 2

这是my answer改编自previous question你的。{/ p>

使用三角形连接获取所有玩家的独特组合：

SELECT p1.playerID, p1.playerName, p2.playerID, p2.playerName
FROM player p1
INNER JOIN player p2 ON p1.playerID < p2.playerID

从第一个玩家中减去第二个玩家的团队集，并检查结果中是否没有行：

NOT EXISTS (
    SELECT teamID
    FROM plays
    WHERE playerID = p1.playerID

    EXCEPT

    SELECT teamID
    FROM plays
    WHERE playerID = p2.playerID
)

交换集合，减去并再次检查：

NOT EXISTS (
    SELECT teamID
    FROM plays
    WHERE playerID = p2.playerID

    EXCEPT

    SELECT teamID
    FROM plays
    WHERE playerID = p1.playerID
)

最后，在步骤1中将两个条件应用于三角形连接的结果。

SELECT p1.playerID, p1.playerName, p2.playerID, p2.playerName
FROM player p1
INNER JOIN player p2 ON p1.playerID < p2.playerID
WHERE
    NOT EXISTS (
        SELECT teamID
        FROM plays
        WHERE playerID = p1.playerID

        EXCEPT

        SELECT teamID
        FROM plays
        WHERE playerID = p2.playerID
    )
AND
    NOT EXISTS (
        SELECT teamID
        FROM plays
        WHERE playerID = p2.playerID

        EXCEPT

        SELECT teamID
        FROM plays
        WHERE playerID = p1.playerID
    )
;

涉及多个表的复杂SQL查询

2 个答案: