slide - Information & Database Systems

Report
CSED421
Database Systems Lab
Constraints
Group functions
Connect to mysql server
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Connect to linux server
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brynn.postech.ac.kr
Id : student
pw : student
Connect to sql
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Type in terminal : mysql -u [hemos ID] –p
Pw : student id
Introduction
Integrity Constraints(ICs)
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Condition that must be true for any instance of databases
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Specified when schema is defined
Checked when relations are modified
5 types of constraints
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NOT NULL
UNIQUE
PRIMARY KEY
FOREIGN KEY
CHECK
NOT NULL Constraints
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Prohibits a database value from being null.
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null : either unknown or not applicable
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To satisfy a NOT NULL constraint,
every row in the table must contain a value for the column.
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Create table with NOT NULL constrained attribute
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Give NOT NULL constraint to existing table
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CREATE TABLE Students (
name CHAR(20) NOT NULL, … … …);
ALTER TALBE Students MODIFY name CHAR(20) NOT NULL;
Remove NOT NULL constraint
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ALTER TALBE Students MODIFY name CHAR(20);
UNIQUE Constraints
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Prohibits multiple rows from having the same value in the same column
or combination of columns, but allows some values to be null
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Create table with UNIQUE constrained attribute
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Give UNIQUE constraint to existing table
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CREATE TABLE Students (
login CHAR(10) UNIQUE, … … …);
ALTER TABLE Students ADD UNIQUE (login);
ALTER TABLE Students MODIFY login CHAR(10) UNIQUE;
Remove UNIQUE constraint
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ALTER TABLE Students DROP INDEX login;
ALTER TABLE Students MODIFY login CHAR(10);
PRIMARY KEY Constraints
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Prohibits multiple rows from having the same value in the same column or
combination of columns, and prohibits values from being null
NOT NULL constraint + UNIQUE constraint
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Create table with PRIMARY KEY
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… … …);
Give PRIMARY KEY constraint to existing attribute
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CREATE TABLE Students (
sid CHAR(20) PRIMARY KEY,
CREATE TABLE Students (
sid CHAR(20),… … …
PRIMARY KEY (sid));
ALTER TABLE Students ADD PRIMARY KEY (sid);
ALTER TABLE Students MODIFY sid CHAR(20) PRIMARY KEY;
Remove PRIMARY KEY constraint
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ALTER TABLE Students DROP PRIMARY KEY;
ALTER TABLE Students MODIFY sid CHAR(20)
FOREIGN KEY Constraints
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Values in one table must appear in another table
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Create table with FOREIGN KEY
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Give FOREIGN KEY constraint to existing attribute
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ALTER TABLE Enrolled ADD FOREIGN KEY (sid) REFERENCES Students
(sid);
Remove FOREIGN KEY constraint to existing attribute
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CREATE TABLE Enrolled (
sid
CHAR(20),
FOREIGN KEY (sid) REFERENCES Students (sid));
CREATE TABLE Enrolled (
sid CHAR(20) REFERENCES Students (sid));
ALTER TABLE Enrolled DROP FOREIGN KEY constraint_name;
Confirm whether two columns are linked
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SHOW CREATE TABLE Enrolled;
SELECT * FROM information_schema.KEY_COLUMN_USAGE;
FOREIGN KEY Constraints
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Referential actions
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What if referenced table is deleted or updated, 5 different actions take place
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CASCADE
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NO ACTION
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integrity check is done after trying to alter the table
RESTRICT
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changes from the parent table and automatically adjust the matching rows in the
child table
Rejects the delete or update operation for the parent table
SET DEFAULT, SET NULL
FOREIGN KEY (sid) REFERENCES Students (sid)
ON UPDATE cascade ON DELETE restrict
CHECK Constraints
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Requires a value in the database to comply with a specified condition
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Create table with CHECK constraint
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Give CHECK constraint to existing attribute
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ALTER TABLE Students ADD CHECK (age > 0);
Change CHECK constraint
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CREATE TABLE Students (
… … … , age INTEGER CHECK (age > 0));
ALTER TABLE Students MODIFY age CHECK (age > 0);
In MySQL, use TRIGGER instead
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“The CHECK clause is parsed but ignored by all storage engines.”
Example of ICs
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1. 다음의 IC를 만족하는 두 테이블을 생성하라
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Table Customer
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id varchar(20)
ame varchar(20)
pw varchar(10)
age integer
Address varchar(20)
Table Orders
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customer_id varchar(20)
customer_addr varchar(20)
amout integer
 Constraints of Customer
 Id is unique and not null
 Name is not null
 Age must be bigger than 0
 Constraints of Orders
 Customer id references id of
customer table
Page 11
Example of ICs
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CREATE TABLE Customer(
id
VARCHAR(20) PRIMARY KEY,
name
VARCHAR(20) NOT NULL,
pw
VARCHAR(10)
age
INTEGER
CHECK(age>0),
address VARCHAR(20)
);
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CREATE TABLE Orders(
customer_id
customer_addr
amount
);
VARCHAR(20) REFERENCES Customer(id),
VARCHAR(20),
INTEGER,
Group function
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GROUP BY
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Sort the data with distinct value for data of specified columns
Usage form of GROUP BY
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Select
from
[where
[GROUP BY
[order by
column
table
condition]
column[, column2, …]]
column [ASC|DESC]
GROUP BY
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Table DevelopTeam
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Select job
from DevelopTeam
group by job
Select job,salary
from DevelopTeam
group by job,salary
GROUP BY
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Group by clause is usually used with aggregate function(min, max, count,
sum, avg)
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Find the job and average salary of each jobs
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Select job, avg(salary)
from DevelopTeam
group by job;
Find the job and largest salary of each jobs
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Select job, max(salary)
from DevelopTeam
group by job;
Page 15
HAVING clause
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Giving condition on data is applied with group by clause
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Usage form of HAVING clause
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SELECT
column1
FROM
table
[WHERE
condition]
[GROUP BY column2]
[HAVING
group_function_condition]
[ORDER BY column3 [ASC|DESC]]
Find the job and average salary of all job whose average salary is greater
than 350
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Select job, avg(salary) from DevelopTeam
group by job
having avg(salary)>350;
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Difference between WHERE and HAVING
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Select job, avg(salary) from DevelopTeam
where salary>350
group by job
having avg(salary)>350;
①
⋯⋯ ①
⋯⋯ ②
⋯⋯ ③
②
③
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Note :
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Where is applied before grouping
Having is applied after grouping
→aggregate function can be used only with having clause
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Example
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각 직업별 연봉이 300이상인 사람수를 검색하시오
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Select job, count(*) as ‘num of person’
from DevelopTeam
where salary>=300
group by job;
각 직업별 연봉의 최소값이 400이상인 직업을 검색하시오
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Select job, min(salary)
From DevelopTeam
Group by job
Having min(salary)>=400;
Page 18
Practice
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1. 다음의 IC를 만족하는 두 테이블을 생성하라
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Table Course
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cName varchar(20)
language varchar(20)
room varchar(30)
Table Enrolled
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cName varchar(20)
sName varchar(20)
gpa float
department varchar(20)
midterm int
final int
 Constraints of Course
 Course name is primary key
 Constraints of Enrolled
 Course name references
course’s course name
 Department is not null
 Midterm and final must lie in
0~100
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Insert data into table
<course>
<enrolled>
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insert into Course values('DB','english','PIRL 142'), ('AI','english','B4 101'), ('PL','korean','B2
102');
insert into Enrolled values('DB','a',3.3,'CSE',80,90), ('DB','b',4.0,'CSE',85,70),
('DB','c',3.9,'MGT',75,85), ('DB','d',3.1,'MGT',70,80), ('DB','e',4.1,'MTH',90,100),
('AI','a',3.3,'CSE',90,70), ('AI','g',3.3,'CSE',95,75), ('AI','h',3.2,'CSE',85,80),
('PL','a',3.3,'CSE',65,95), ('PL','e',4.1,'MTH',100,100), ('PL','k',3.4,'MTH',75,90),
('PL','i',2.7,'MGT',55,70);
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Practice
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2. Enrolled 테이블에서 각 과목별로 몇 명의 수강생이 있는지를 검색
하시오.
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결과는 과목명과 수강생 수를 출력
3. Enrolled 테이블에서 각 과목별 학점이 4.0이상인 학생수를 검색하
시오
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결과는 과목명, 학생 수(column명을 numStu로 표현)를 출력
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Practice
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4. Enrolled 테이블에서 수강생이 4명 이상인 과목의 중간고사 평균을
구하시오.
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결과는 과목명, 수강생 수와 중간고사 평균을 출력
5. Enrolled 테이블에서 CSE 학생들의 각 과목별 기말고사의 최고점을
검색하시오.
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결과는 과목명과 점수를 출력.
단, 최고점이 90점 이상일 때만 출력
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Practice
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6. Enrolled 테이블에서 과목별, 학과별 중간고사, 기말고사 평균을 검
색하시오.
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결과는 과목명, 학과명과
중간고사, 기말고사 평균 출력

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