对于每个国家/地区，查找已访问该国家/地区的船舶的平均重量

Question

尝试使用以下LEFT JOIN解决问题但在此之后陷入困境

SELECT COUNTRY FROM VISIT LEFT JOIN S ON S.CRUISENUM=VISIT.CRUISENUM (SELECT SHIP.SHIPNUM , WEIGHT , CRUISENUM FROM CRUISE LEFT JOIN SHIP ON CRUISE.SHIPNUM = SHIP.SHIPNUM) AS S

Answer 1

您可以在同一查询中进行两次连接，无需创建子查询。

     SELECT V.COUNTRY, AVG(S.WEIGHT)
       FROM visit V
  LEFT JOIN cruise C ON ( C.CRUISENUM = V.CRUISENUM )
  LEFT JOIN ship   S ON ( S.SHIPNUM = C.SHIPNUM )
   GROUP BY V.COUNTRY

Answer 2

除非确实需要，否则请避免使用子查询。您通过JOIN进入了正确的解决方案。 （但是你不需要左连接，因为如果ship记录不存在，那么你不想计算任何东西。）

SELECT COUNTRY, AVG(weight)
FROM VISIT
JOIN CRUISE USING (cruisenum)
JOIN SHIP USING (shipnum)
GROUP BY COUNTRY

Answer 3

按总重量计算的最大构建者

select builder from ship where max(weight) in
(select SUM(WEIGHT) from ship group by builder)

列出至少访问过两个不同国家/地区的巡航主管的名字。我尝试过此代码，但它对我来说也不起作用

 SELECT DISTINCT C1.DIRECTOR
FROM CRUISE C1 , CRUISE C2
WHERE C1.DIRECTOR =C2.DIRECTOR AND C1.SHIPNUM = C2.SHIPNUM