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  • sql语句之连表查询

    学习链接:https://blog.csdn.net/qq_43119297/article/details/82559687

    1:创建book表,并插入数据

    CREATE TABLE book(
             bookID VARCHAR(20) PRIMARY KEY ,
              bookName VARCHAR(50) DEFAULT NULL
       );
    
    INSERT INTO book VALUES
    (1,'java基础'),
    (2,'javaweb'),
    (3,'JDBC'),
    (4,'HTML高级');
    

    2:创建loan表

    CREATE TABLE loan(
             uid INT,
             bookID VARCHAR(32),
             lnum INT,
             PRIMARY KEY(uid,bookID)
           );
    
    INSERT INTO loan VALUES
    (1,1,8),
    (2,4,2),
    (3,3,3),
    (4,2,7),
    (5,1,1),
    (2,3,10),
    (2,2,3),
    (3,1,5);
    

    3.创建user表,并插入数据

     CREATE TABLE USER(
        uid INT AUTO_INCREMENT PRIMARY KEY,
        username VARCHAR(20) UNIQUE,
        age INT,
        sex VARCHAR(2)
     );
    
    INSERT INTO USER VALUES 
    (NULL,'周瑜',22,'男'),
    (NULL,'小乔',18,'女'),
    (NULL,'甄宓',23,'女'),
    (NULL,'曹操',30,'男'),
    (NULL,'貂蝉',26,'女');  
    

      

    4,问题

    – 1.查询年龄小于 25的学生姓名、学生年龄
    
    – 2.查询年龄在18-22之间(包含18和22)的学生信息
    
    –3. 统计学生表中男女的数量分别是多少
    
    – 4.查询学生的总人数、平均年龄、最小年龄
    
    – 5.查询借阅过图书的每个学生ID借阅的总次数
    
    – 6.查询图书名称包含”java”的图书数量
    
    –7. 查询借阅总次数大于50的学生学号
    
    –8. 查询借阅了’HTML高级’的学生学号
    
    –9. 查询没有借阅过图书的学生学号、学生姓名
    
    – 10.查询借阅过’javaWeb’图书的学生学号、借阅次数
    
    – 11.查询年龄比”周瑜”大的所有学生姓名、学生年龄
    
    – 12.查询年龄最大的学生ID、学生姓名,学生年龄
    
    –13. 查询借阅了图书的学生姓名、借阅的不同图书总数、借阅所有图书总次
    
    –14. 查询被借阅的每本图书的图书名称、借阅总次数
    
    – 15.查询借阅次数最多的图书ID和借阅的总次数
    --------------------- 
    

    5,答案

    1:
    SELECT username,age
    FROM USER
    WHERE age<25;
    
    2:
    SELECT username,age
    FROM USER
    WHERE age BETWEEN 18 AND 22;
    
    3:
    SELECT sex,COUNT(username) 数量
    FROM USER
    GROUP BY sex;
    
    4:
    SELECT COUNT(uid) 总人数,AVG(age) 平均年龄,MIN(age) 最小年龄
    FROM USER;
    
    5:
    SELECT uid,COUNT(lnum)借阅次数
    FROM loan
    GROUP BY uid;
    
    6:
    SELECT SUM(b.`lnum`) 总数
    FROM book a,loan b
    AND bookname LIKE "java%";
    
    7:
    SELECT uid
    FROM loan
    GROUP BY uid HAVING SUM(lnum)>5;
    
    8:
     SELECT user.`uid`
     FROM book,USER,loan
     WHERE loan.`bookID`=book.`bookID` AND 
     loan.`uid` =user.`uid` AND
     book.`bookName`='HTML高级';
    
    9:
     SELECT user.`uid`,user.`username`
     FROM book,USER,loan
     WHERE loan.`bookID`=book.`bookID` AND 
     loan.`uid` =user.`uid` AND
     book.`bookName`NOT IN
     ('HTML高级','JDBC','javaweb','java基础');
    --------------------- 
    
    10:
    SELECT user.`uid`,SUM(loan.`lnum`) 次数
     FROM book,USER,loan
     WHERE loan.`bookID`=book.`bookID` AND 
     loan.`uid` =user.`uid` AND
     book.`bookName`='javaWeb'
     GROUP BY book.`bookName`;
    
    11:
     SELECT username,age
    WHERE username='周瑜') t
     WHERE a.`age`>t.b;
    
    12:
    SELECT uid, username,age
     FROM USER a , (SELECT MAX(age) b FROM USER ) t
     WHERE a.`age`=t.b;
    
    13:
     SELECT a.`username`,s.e 借阅的不同图书总数,s.b 借阅所有图书总次
     FROM USER a,(SELECT d.`uid` k,SUM(d.`lnum`) e ,
     COUNT(d.`bookID`) b FROM loan d GROUP BY uid ) s
     WHERE a.`uid`=s.k;
    
    
    14:
     SELECT  book.bookname 书名,n.v 借阅次数  
     FROM book,(SELECT bookid s,SUM(lnum) v FROM
      loan GROUP BY bookid)  n
     WHERE n.s=book.`bookID`;
    
    15:
    SELECT  bookID,MAX(n.a)  
    FROM (SELECT bookid,SUM(lnum) a 
    FROM loan GROUP BY bookid) n;
    

      

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  • 原文地址:https://www.cnblogs.com/zhuhaofeng/p/9782590.html
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