Question

我正在使用MySQL Workbench 6.2 CE。

我有以下表格列：

Artist (artistId, name, dateOfBirth, dateOfDeath, homeCountry)
Picture (picId, name, artistId, value)
Museum (museumName, city, country)
Exhibited (picId, museumName, from, to)

我想要实现这个问题的答案：

The museums (with name and city) which exhibit or exhibited pictures
of **all** the artists born between 1920 and 1950.

到目前为止我已经尝试过查询：

SELECT museum.museumName, museum.city, COUNT(artist.artistId) AS C FROM museum

INNER JOIN exhibited ON exhibited.museumName = museum.museumName
INNER JOIN picture ON picture.picId = exhibited.picId
INNER JOIN artist ON artist.artistId = picture.artistId

WHERE dateOfBirth BETWEEN 2000 AND 2004
GROUP BY museum.museumName
HAVING COUNT(C) = (SELECT COUNT(*) FROM artist WHERE dateOfBirth BETWEEN 2000 AND 2004);

------------------------

SELECT museumName, city FROM museum WHERE museumName IN 
(SELECT museumName FROM exhibited WHERE picId IN
 (SELECT picId FROM Picture WHERE artistId IN 
  (
   SELECT artistId FROM artist WHERE dateOfBirth BETWEEN 2000 AND 2004
  )
 )
);

我还尝试了其他查询但无法提供我删除的查询，因为他们没有帮助。

哪些查询可以打印询问的数据？

Answer 1

您的第一个查询非常接近。而不是count(c)，你需要计算艺术家的数量，以及让年代变得正确：

HAVING COUNT(DISTINCT artistId) = (SELECT COUNT(*) FROM artist WHERE dateOfBirth BETWEEN 1920 AND 1950);

您还需要修复where子句。现在在博物馆展出的艺术家不是那么多2000年后出生的艺术家。

MySQL嵌套查询问题

1 个答案: