Question

美国大学的数据库题目 大神帮我看看 绝对给高分

Answer 1

1，该博物馆是唯一有印象派风格的绘画作品
作为一项权利选择博物馆，收集了参加绘画AS B，上A.Painting ID = B.Painting的ID的不B.Style（选择绘画风格从为C c.Style >'印象派'组者C.Style）和B.Style ='印象派'的A.Museum组
2，馆内没有印象派风格的绘画作品。
博物馆，收集作为一项权利选择AS B A.Painting ID = B.Painting编号，在那里B.Style >'印象派'组
3 A.Museum，博物馆包含了参加绘画至少在19世纪的各种风格的绘画作品之一
然后再考虑这有点困难，

Answer 2

1、该馆只存有impressionist风格的画作
Select Museum from Collection AS A right out join Painting AS B on A.Painting ID = B.Painting id where B.Style not in ( Select Style from Painting as C where c.Style <> 'impressionist' group by C.Style) and B.Style = 'impressionist' group by A.Museum
2、该馆没有impressionist风格的画作
Select Museum from Collection AS A right out join Painting AS B on A.Painting ID = B.Painting id where B.Style <> 'impressionist' group by A.Museum
3、该馆至少含有1张19世纪各种风格的画作
这个有点难，再考虑考虑

Answer 3

1、只存有impressionist风格画作的馆
Select Museum from Collection AS A right out join Painting AS B on A.Painting ID = B.Painting id where B.Style not in ( Select Style from Painting as C where c.Style <> 'impressionist' group by C.Style) and B.Style = 'impressionist' group by A.Museum
2、没有impressionist风格画作的馆
Select Museum from Collection AS A right out join Painting AS B on A.Painting ID = B.Painting id where B.Style <> 'impressionist' group by A.Museum
3、至少含有1张19世纪各种风格画作的馆
select C.Museum from (select distinct A.Museum,B.Style from Collection as A inner join Painting as B on A.PaintingId=B.PaintingId where B.Century='19th') as C group by C.Museum having count(C.Style)=(select count(D.Style) from (select distinct Style from Painting) as D)
这个有点复杂，所以解释一下我的思路：
先求出含有19世纪画作的Museum与Style的所有组合C：(select distinct A.Museum,B.Style from Collection as A inner join Painting as B on A.PaintingId=B.PaintingId where B.Century='19th')
再对C用求出各Museum的Style数量count(C.Style)，
同时求出所有Style的总数：(select count(D.Style) from (select distinct Style from Painting) as D)
最后在having语句中设置条件：风格数量等于风格总数的馆，即为所需列出的馆
以上语句可能有细微错误，仅提供思路给你参考，希望对你有用，并得到你的奖励。谢谢！

Answer 4

妈妈呀，一个字都不认识为什么不来点中国的呢