sql语句之连表查询

Posted 2021-01-12 36°艳阳天

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;