Day07 operation

The first 1 topic :

1. Create table celebrity , works , summsry

create table celebrity(
sid int primary key,
sname varchar(50),
sage int,
ssex varchar(5)
)

create table works(
wid int primary key,
wwork varchar(50),
wdynasty varchar(20)
)

create table summary(
sid int,
wid int,
sassess varchar(255)
)

2. Add data to the table

insert into celebrity values
(1,' Li Bai ',34,' male '),
(2,' Du Fu ',24,' male '),
(3,' Bai Juyi ',31,' male '),
(4,' Li shangyin ',40,' Woman '),
(5,' Su shi ',26,' male '),
(6,' Xin qiji ',22,' male ')


insert into works values
(1,' Drink in ',' Tang Dynasty '),
(2,' Intones difficult ',' Tang Dynasty '),
(3,' Rain in the night sent to the North ',' Tang Dynasty '),
(4,' In the Quiet Night ',' Tang Dynasty '),
(5,' Wang Yue ',' Tang Dynasty '),
(6,' Qiantang lake spring line ',' Tang Dynasty '),
(7,' Nostalgia for Chibi ',' Tang Dynasty '),
(8,' Shu Shui tunes ',' Tang Dynasty ')


insert into summary values 
(1,1,' I'm born to be useful '),
(1,2,' The danger is high !'),
(6,8,' When is the moon '),
(3,6,' The flowers are getting more and more attractive '),
(4,3,' But when it rains at night '),
(5,7,' the mighty river flows eastward '),
(2,5,' The hills are small '),
(1,4,' look at the bright moon ')

3.(1). Look up the author whose gender is male in the celebrity table

select sname from celebrity where ssex=' male '

 

(2) Query the author of Jingyesi

select celebrity.sname from summary  inner join celebrity on summary.sid=celebrity.sid inner join works  on summary.wid=works.wid where works.wwork=' In the Quiet Night '

(3) Inquire about the ages of Li Bai and Du Fu

select sname,sage from celebrity where sname in(' Li Bai ',' Du Fu ')

 

(4) Check the number of men and women in the celebrity table

select count(*),ssex from celebrity group by ssex

 

(5)  Query the name of the work written by Bai Juyi and the corresponding famous sentences

select works.wwork,summary.sassess from summary  inner join celebrity on summary.sid=celebrity.sid inner join works  on summary.wid=works.wid where celebrity.sname=' Bai Juyi '

(6) The query age is 25 To 30 The number between the ages

select count(*) from celebrity where sage between 25 and 30

 

(7)  Query the two youngest data information in the celebrity list

select * from celebrity order by sage limit 0,2

(8)  Query the name of Li Bai's work , Famous sentences and age

select works.wwork,summary.sassess,celebrity.sage from summary  inner join celebrity on summary.sid=celebrity.sid inner join works  on summary.wid=works.wid where celebrity.sname=' Li Bai '

(9) Change the Dynasty whose work is Wangyue to the Northern Song Dynasty

update works set wdynasty=' Northern song dynasty ' where wwork=' Wang Yue '

 

 (10) Add an author Wang Wei in the celebrity list , Age 25, Gender male

insert into celebrity values(7,' Wang wei ',25,' male ')

The first 2 topic :

1. Create table product , user , orders 

create table product(
pid int primary key,
pname varchar(50),
pprice int
)
create table user(
uid int primary key,
uname varchar(50),
uage int,
usex char(5)
)
create table orders(
opid int,
ouid int,
onumber int
)

2. Add data

insert into product values
(1,' mobile phone ',2300),
(2,' The computer ',5600),
(3,' camera ',1200),
(4,' Projector ',2500)
insert into user values
(1,' Li San ',20,' male '),
(2,' Zhang Si ',23,' male '),
(3,' Zhao Wu ',25,' male '),
(4,' Sun Liu ',18,' Woman '),
(5,' The Monkey King ',24,' male ')
insert into orders values
(1,1,123123),
(1,2,112233),
(2,1,234567),
(2,5,787878),
(3,4,343421),
(3,5,909090),
(4,2,212112),
(4,1,343421)

3.(1) Query the user surnamed sun  

select uname from user where uname like ' Grandchildren %'

  

 (2) Query the oldest 2 Users

select uname from user order by uage desc limit 0,2

 (3) Check the goods bought by Li San

select product.pname from orders inner join product on orders.opid=product.pid inner join user on orders.ouid=user.uid where user.uname=' Li San '

  (4) Query the user name of the purchased computer

select user.uname from orders inner join product on orders.opid=product.pid inner join user on orders.ouid=user.uid where product.pname=' The computer '

   (5) The query order is 909090 Corresponding user name and commodity name

select user.uname,product.pname from orders inner join product on orders.opid=product.pid inner join user on orders.ouid=user.uid where orders.onumber=909090

  

 (6) Query the first two data of the highest price in the commodity table  

select * from product order by pprice desc limit 0,2

The first 3 topic

1. Create a student information table S--student , Course selection information table SC--studentcourse , Course information sheet C--course

create table student(
sno int primary key,
sname varchar(50),
age int,
sex varchar(5)
)
create table course(
cno int primary key,
cname varchar(50),
cteacher varchar(50)
)
create table studentcourse(
sno int,
cno int,
scgrade int
)

2. Add data

insert into student values
(1,' Zhang San ',18,' male '),
(2,' Li Si ',20,' Woman '),
(3,' Wang Wu ',23,' male '),
(4,' Zhao Liu ',22,' Woman ')
insert into course values
(1,'java',' He Hao '),
(2,'php',' Mei Jun Li '),
(3,'android',' Wang Chao ')
insert into studentcourse values
(1,1,50),
(2,1,66),
(3,1,90),
(1,2,55),
(2,2,68),
(3,2,61),
(4,2,44),
(1,3,90),
(2,3,78),
(3,3,22),
(4,3,55)

3.(1) Inquire about SC The information in the table corresponds to the girls in the courses taught by Mr. He Hao

select * from studentcourse sc inner join student on sc.sno=student.sno 
inner join course on sc.cno=course.cno where course.cteacher=' He Hao ' and student.sex=' Woman '

 

(2) Find out the information of all students who have not taken any teacher's courses

select * from studentcourse sc inner join student on sc.sno=student.sno 
inner join course on sc.cno=course.cno where not course.cteacher=' He Hao '

(3) List courses that have failed ( The score is less than 60) Of students

select student.sname from studentcourse sc inner join student on sc.sno=student.sno 
inner join course on sc.cno=course.cno where sc.scgrade<60 

 

The first 4 topic

1. Create student table student , The teacher table teacher

create table student(
id int primary key,
name varchar(20),
score int
)
create table teacher(
id int primary key,
name varchar(30),
class varchar(20),
classroom varchar(20),
student_id int
)

2. Add data

insert into student values
(1,' Zhang San ',80),
(2,' Li Si ',56),
(3,' Wang Wu ',72),
(4,' Zhao Liu ',30),
(5,' Liu Qi ',66)
insert into teacher values
(1,' Liuliangyu ',' test ','2102A',1),
(2,' Liuliangyu ',' test ','2011A',2),
(3,' Fan Qingxiao ','android','2012A',3),
(4,' Sun Wenlong ','PHP','2101A',4),
(5,' Xu Tengsheng ','JAVA','2013B',5)

3.(1) Inquire the names of the students and the corresponding subjects brought by teacher Liu Liangyu

select student.name,teacher.class from teacher inner join student on teacher.student_id=student.id where teacher.name=' Liuliangyu '

(2) Inquire about the students led by teacher Liu Liangyu , The name of the student who passed the grade

select student.name from teacher inner join student on teacher.student_id=student.id where teacher.name=' Liuliangyu ' and student.score>=60

(3) Query teacher Zhao Liu's teacher name and the corresponding subject name

select teacher.name,teacher.class from teacher inner join student on teacher.student_id=student.id where student.name=' Zhao Liu '