select 所有成员使用 NOT EXISTS

Usage of NOT EXISTS to select ALL members

我正在尝试使用 MySQL 为图书馆数据库编写 SQL 查询。 有两个 tables shelf(studentnumber, booknumber)booklist(booknumber,booktitle,language)。书单 table 条目中有 4 种不同的语言,即 italian, spanish, hungarian, german.

我想学 studentnumbers 已阅读所选语言所有书籍的

tables 的示例数据:

create table shelf(studentnumber INT, booknumber INT); 

INSERT INTO shelf values(1,1),(1,2),(1,3),(2,1),(2,3)(2,4),(2,5),
                        (2,6),(3,6),(3,7)(3,8),(3,9); 

create table booklist(booknumber INT, booktitle VARCHAR(50), language VARCHAR(10); 

INSERT INTO booklist values(1, 'FirstBook', 'italian'),(2,'SecondBook', 'spanish'),
(3,'ThirdBook','italian'),(4,'FourthBook','german'),(5,'FifthBook','german'),
(6, 'SixthBook','spanish'),(7,'SeventhBook','hungarian'),(8,'EightBook','hungarian'),
(9,'NinthBoo‌​k','hungarian'),(10,'TenthBook','Spanish'),(11,'EleventhBook', 'italian');

示例输出: 当您查看书架和书单 tables 时,您会看到学生编号为 2 的学生阅读了所有德语书籍,学生编号为 3 的学生阅读了所有匈牙利语书籍。 但是没有学生读完所有意大利语或西班牙语的书。

代码的最后一部分如下所示,但我无法构建第一部分,可能会包含 NOT EXISTS SELECT booknumber FROM booklist WHERE language='italian';

这应该很适合你:

select studentnumber 
from shelf s1 join booklist b1 on s1.booknumber=b1.booknumber
group by studentNumber, language
having count(language) >= (
        select count(*) from booklist b2
        where b1.language=b2.language
        group by b2.language)

输出:

2
3

您还可以添加第一个 select 语句语言和计数 (*),这将为您提供更多信息(他们阅读的语言以及书籍的数量)即

select studentnumber,language,count(*) 
//rest of code

输出

2, german, 2
3, hungarian, 3

更新

要显示所有阅读过指定语言书籍的学生(按照评论中的要求),只需添加一个 WHERE 子句:

select studentnumber
from shelf s1 join booklist b1 on s1.booknumber=b1.booknumber
where language='german'
group by studentNumber, language
having count(language) >= (
        select count(*) from booklist b2
        where b1.language=b2.language
        group by b2.language)

输出:

2