使用NOT EXISTS选择所有成员

时间:2015-03-29 08:26:27

标签: mysql where exists

我正在尝试使用MySQL为库数据库编写SQL查询。 有两个表shelf(studentnumber, booknumber)booklist(booknumber,booktitle,language)。书单表条目中有4种不同的语言,即italian, spanish, hungarian, german

我想了解已阅读所选语言所有图书的studentnumbers

表格的示例数据:

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表时,您会看到有2号学生的学生阅读所有德语书籍,而有学生3号的学生阅读所有匈牙利语书籍。 但是没有学生读过意大利语或西班牙语的所有书籍。

代码的最后一部分将如下所示,但我无法构建第一部分,可能包括NOT EXISTS SELECT booknumber FROM booklist WHERE language ='italian';

1 个答案:

答案 0 :(得分:0)

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选择语句语言和计数(*),它将为您提供更多信息(他们已阅读的语言以及书籍数量),即

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

输出

2, german, 2
3, hungarian, 3

<强>更新

要显示所有以指定语言阅读书籍的学生(如评论中所要求的),只需添加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