[英]Complicated SQL Query involving multiple tables
在這個查詢中,我必須列出一對玩家ID和玩家名稱的球員,他們為同一支球隊效力。如果一名球員為3支球隊效力,則另一支球員必須參加完全相同的3支球隊。 不能少,不多了。 如果兩名球員目前不參加任何球隊,他們也應該被包括在內。 查詢應該返回(playerID1,playername1,playerID2,playerName2)而沒有重復,例如如果玩家1信息在玩家2之前出現,則不應該有另一個玩家2信息在玩家1之前出現的元組。
例如,如果玩家A為洋基隊和紅襪隊隊員比賽,而隊員隊員為洋基隊隊員,紅襪隊隊員和道奇隊隊員隊效力,我就不應該參加比賽。 他們都必須為洋基隊和紅襪隊效力,而不是其他人。 現在,如果玩家為同一個團隊玩游戲,此查詢會找到答案。
player(playerID: integer, playerName: string)
team(teamID: integer, teamName: string, sport: string)
plays(playerID: integer, teamID: integer)
現在我的查詢是
SELECT p1.playerID, p1.playerName, p2.playerID, p2.playerName
FROM player p1, player p2, plays
WHERE p1.teamID = p2.teamID AND teamID in.....
在此之后,我被困在如何接近它。 有關如何解決此問題的任何提示。 謝謝你的時間。
我認為最簡單的方法是將團隊連接在一起,然后加入結果。 Postgres提供函數string_agg()
來聚合字符串:
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;
編輯:
你可以在沒有string_agg
情況下做到這一點。 我們的想法是從所有可能的玩家組合列表開始。
然后,使用left outer join
第一個玩家的團隊。 並通過在團隊和驅動程序名稱上使用full outer join
和匹配來full outer join
第二個團隊。 您需要驅動程序表的原因是為了確保id / name不會在完全外連接中丟失:
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(*);
這會加入團隊ID上的兩個玩家(有序,所以一對只被考慮一次)。 然后,當兩個玩家的所有行都具有非NULL團隊值時,它會說匹配。 對於等效的having
子句,這可能更清楚:
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;
當兩個玩家擁有不匹配的球隊時,完整的外部聯接將產生NULL
值。 因此,沒有NULL
值意味着所有團隊都匹配。
使用三角形連接獲取所有玩家的獨特組合:
SELECT p1.playerID, p1.playerName, p2.playerID, p2.playerName FROM player p1 INNER JOIN player p2 ON p1.playerID < p2.playerID
從第一個玩家中減去第二個玩家的團隊集,並檢查結果中是否沒有行:
NOT EXISTS ( SELECT teamID FROM plays WHERE playerID = p1.playerID EXCEPT SELECT teamID FROM plays WHERE playerID = p2.playerID )
交換集合,減去並再次檢查:
NOT EXISTS ( SELECT teamID FROM plays WHERE playerID = p2.playerID EXCEPT SELECT teamID FROM plays WHERE playerID = p1.playerID )
最后,將兩個條件應用於步驟1中三角形連接的結果。
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 ) ;
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