SQL查询联接无法正常工作

时间:2016-07-15 17:32:03

标签: mysql join oracle-sqldeveloper

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...... enter image description here 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(这只是两个竞争对手),并列出名称,事件,地点和结果。

下面的查询有效,因为它应该产生70个(竞争对手的#)记录,计数为2,竞争对手参加两个赛事......

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,它产生了72个结果,竞争对手参加了两次列出的两个事件,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。

1 个答案:

答案 0 :(得分:1)

考虑:(结果是特定于竞争对手,因此您错过了加入条件。) 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,缺少连接标准会导致计数膨胀,而正确加入可能也会导致问题,但由于结果取决于竞争对手,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]然后重新加入(还有其他方法,但我通常更喜欢这种方法它似乎是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;