对于一个项目,我需要为小型奥运会创建一个数据库.我的ER图如图所示……
链接到更好的图片:
https://i.imgur.com/xgfurWO.png?1
我需要创建一个查询,其中只包含竞争多个竞争对手的竞争对手(只有两个竞争对手),并列出名称,事件,地点和结果.
对于竞争对手参加两项活动,下面的查询可以产生70(竞争对手的#)记录,计数为2 …
SELECT c.firstname, c.lastname, COUNT(r.competitorid)
FROM COMPETITOR c LEFT OUTER JOIN ( REGISTRATION r LEFT OUTER JOIN EVENT e
ON r.eventid = e.eventid ) ON r.competitorid = c.competitorid
GROUP BY c.firstname, c.lastname
ORDER BY c.firstname;
-----------------------------------------
NORRIS HOLMWOOD 1
OCTAVIO MARTINEZ 1
ORFEO SILVA 2
etc...
在查询中包含事件名称后,它会生成72个结果,竞争对手在两个列出的事件中竞争两次,但COUNT()对于所有事件都是1.
SELECT c.firstname, c.lastname, e.eventname, COUNT(r.competitorid)
FROM COMPETITOR c LEFT OUTER JOIN ( REGISTRATION r LEFT OUTER JOIN EVENT e
ON r.eventid = e.eventid ) ON r.competitorid = c.competitorid
GROUP BY c.firstname, c.lastname, e.eventname
ORDER BY c.firstname;
----------------------------------------------------------------
NORRIS HOLMWOOD 100 METER BUTTERFLY 1
OCTAVIO MARTINEZ FLOOR EXERCISE 1
ORFEO SILVA 100 METER BUTTERFLY 1
ORFEO SILVA 400 METER INDIV MEDLEY 1
PONCIO ASIS POMMEL HORSE 1
PONCIO BARROS LONG JUMP 1
然后这就是事情真的搞砸了.当我将结果表与事件表连接时,输出与72条记录保持相同,但计数完全错误.
SELECT c.firstname, c.lastname, e.eventname, COUNT(r.competitorid)
FROM COMPETITOR c LEFT OUTER JOIN ( REGISTRATION r LEFT OUTER JOIN (
EVENT e right OUTER JOIN RESULT rs ON e.eventid = rs.eventid)
ON r.eventid = e.eventid ) ON r.competitorid = c.competitorid
GROUP BY c.firstname, c.lastname, e.eventname
ORDER BY c.firstname;
-----------------------------------------------------------------------
NIKOLAI MIKHAILOV POMMEL HORSE 6
NOEMI PELAEZ BALANCE BEAM 7
NORRIS HOLMWOOD 100 METER BUTTERFLY 6
OCTAVIO MARTINEZ FLOOR EXERCISE 6
ORFEO SILVA 100 METER BUTTERFLY 6
ORFEO SILVA 400 METER INDIV MEDLEY 6
PONCIO ASIS POMMEL HORSE 6
我的问题是我做错了什么?连接似乎至少部分地工作我想要它们,问题是乱搞count().数字越大并不一定重要,问题是由于某种原因,计数似乎相当随机.
最佳答案 考虑:(结果也特定于竞争对手,因此您错过了加入标准.)
此外,右连接意味着您希望结果中的所有记录和其他表中的所有记录都是LEFT加入竞争对手.由于结果在逻辑上基于竞争对手,我猜测正确的连接是不正确的.
因此,在计数1上出现双重问题,缺失的连接标准导致计数膨胀,而正确的连接可能也会导致问题,但由于结果取决于竞争对手,因此您可能看不到任何问题.
如果你想要竞争对手注册的事件数量,你需要分别得出那个数量[绝对拼写错误太可笑了,不能取出]然后加入(有其他方法,但我通常更喜欢这种方法,因为它似乎是DB不可知的)
SELECT c.firstname, c.lastname, e.eventname, CE.TotalNumberOFEvents
FROM COMPETITOR c
LEFT OUTER JOIN REGISTRATION r
ON r.competitorid = c.competitorid
LEFT OUTER JOIN EVENT e
ON r.eventid = e.eventid
LEFT OUTER JOIN RESULT rs --Not sure why this is a right join... on your qry.
ON e.eventid = rs.eventid
AND Rs.CompetitorID = C.CompetitorID ---ADDED this and removed the ()'s putting join critier under the joins.
LEFT JOIN (SELECT count(RegistrationID) as TotalNumberOfEvents, CompetitorID
FROM Registration) as CE --added this to get the # of events a competitor is in (independant of the event name listed)
ON Ce.CompetitorID = C.CompetitorID
GROUP BY c.firstname, c.lastname, e.eventname
ORDER BY c.firstname;