1. CSED421 Database Systems Lab Constraints Group functions

2.  Connect to linux server  brynn.postech.ac.kr  Id : student  pw : student  Connect to sql  Type in terminal : mysql -u [hemos ID] –p  Pw : student id Connect to mysql server

3. Introduction

4.  Condition that must be true for any instance of databases  Specified when schema is defined  Checked when relations are modified  5 types of constraints  NOT NULL  UNIQUE  PRIMARY KEY  FOREIGN KEY  CHECK Integrity Constraints(ICs)

5.  Prohibits a database value from being null.  null : either unknown or not applicable  To satisfy a NOT NULL constraint, every row in the table must contain a value for the column.  Create table with NOT NULL constrained attribute  CREATE TABLE Students ( name CHAR(20) NOT NULL, … … …);  Give NOT NULL constraint to existing table  ALTER TALBE Students MODIFY name CHAR(20) NOT NULL;  Remove NOT NULL constraint  ALTER TALBE Students MODIFY name CHAR(20); NOT NULL Constraints

6.  Prohibits multiple rows from having the same value in the same column or combination of columns, but allows some values to be null  Create table with UNIQUE constrained attribute  CREATE TABLE Students ( login CHAR(10) UNIQUE, … … …);  Give UNIQUE constraint to existing table  ALTER TABLE Students ADD UNIQUE (login);  ALTER TABLE Students MODIFY login CHAR(10) UNIQUE;  Remove UNIQUE constraint  ALTER TABLE Students DROP INDEX login;  ALTER TABLE Students MODIFY login CHAR(10); UNIQUE Constraints

7.  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  Create table with PRIMARY KEY  CREATE TABLE Students ( sid CHAR(20) PRIMARY KEY, … … …);  CREATE TABLE Students ( sid CHAR(20),… … … PRIMARY KEY (sid));  Give PRIMARY KEY constraint to existing attribute  ALTER TABLE Students ADD PRIMARY KEY (sid);  ALTER TABLE Students MODIFY sid CHAR(20) PRIMARY KEY;  Remove PRIMARY KEY constraint  ALTER TABLE Students DROP PRIMARY KEY;  ALTER TABLE Students MODIFY sid CHAR(20) PRIMARY KEY Constraints

8.  Values in one table must appear in another table  Create table with FOREIGN KEY  CREATE TABLE Enrolled ( sid CHAR(20), FOREIGN KEY (sid) REFERENCES Students (sid));  CREATE TABLE Enrolled ( sid CHAR(20) REFERENCES Students (sid));  Give FOREIGN KEY constraint to existing attribute  ALTER TABLE Enrolled ADD FOREIGN KEY (sid) REFERENCES Students (sid);  Remove FOREIGN KEY constraint to existing attribute  ALTER TABLE Enrolled DROP FOREIGN KEY constraint_name;  Confirm whether two columns are linked  SHOW CREATE TABLE Enrolled;  SELECT * FROM information_schema.KEY_COLUMN_USAGE; FOREIGN KEY Constraints

9.  Referential actions  What if referenced table is deleted or updated, 5 different actions take place  CASCADE  changes from the parent table and automatically adjust the matching rows in the child table  NO ACTION  integrity check is done after trying to alter the table  RESTRICT  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 FOREIGN KEY Constraints

10.  Requires a value in the database to comply with a specified condition  Create table with CHECK constraint  CREATE TABLE Students ( … … …, age INTEGER CHECK (age > 0));  Give CHECK constraint to existing attribute  ALTER TABLE Students ADD CHECK (age > 0);  Change CHECK constraint  ALTER TABLE Students MODIFY age CHECK (age > 0);  In MySQL, use TRIGGER instead  “The CHECK clause is parsed but ignored by all storage engines.” CHECK Constraints

11. Page 11  1. 다음의 IC 를 만족하는 두 테이블을 생성하라  Table Customer  id varchar(20)  ame varchar(20)  pw varchar(10)  age integer  Address varchar(20)  Table Orders  customer_id varchar(20)  customer_addr varchar(20)  amout integer Example of ICs  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

12.  CREATE TABLE Customer( id VARCHAR(20) PRIMARY KEY, name VARCHAR(20) NOT NULL, pw VARCHAR(10) age INTEGER CHECK(age>0), addressVARCHAR(20) );  CREATE TABLE Orders( customer_id VARCHAR(20) REFERENCES Customer(id), customer_addr VARCHAR(20), amountINTEGER, ); Example of ICs

13. Group function  GROUP BY  Sort the data with distinct value for data of specified columns  Usage form of GROUP BY  Select column from table [wherecondition] [GROUP BYcolumn[, column2, …]] [order bycolumn [ASC|DESC]

14.  Table DevelopTeam  Select job Select job,salary from DevelopTeamfrom DevelopTeam group by jobgroup by job,salary GROUP BY

15. Page 15  Group by clause is usually used with aggregate function(min, max, count, sum, avg)  Find the job and average salary of each jobs  Select job, avg(salary) from DevelopTeam group by job;  Find the job and largest salary of each jobs  Select job, max(salary) from DevelopTeam group by job; GROUP BY

16. Page 16  Giving condition on data is applied with group by clause  Usage form of HAVING clause  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  Select job, avg(salary) from DevelopTeam group by job having avg(salary)>350; HAVING clause

17. Page 17  Select job, avg(salary) from DevelopTeam where salary>350 ⋯⋯ ① group by job ⋯⋯ ② having avg(salary)>350; ⋯⋯ ③  Note :  Where is applied before grouping  Having is applied after grouping → aggregate function can be used only with having clause Difference between WHERE and HAVING ① ② ③

18. Page 18  각 직업별 연봉이 300 이상인 사람수를 검색하시오  Select job, count(*) as ‘num of person’ from DevelopTeam where salary>=300 group by job;  각 직업별 연봉의 최소값이 400 이상인 직업을 검색하시오  Select job, min(salary) From DevelopTeam Group by job Having min(salary)>=400; Example

19. Page 19  1. 다음의 IC 를 만족하는 두 테이블을 생성하라  Table Course  cName varchar(20)  language varchar(20)  room varchar(30)  Table Enrolled  cName varchar(20)  sName varchar(20)  gpa float  department varchar(20)  midterm int  final int Practice  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

20. Page 20  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); Insert data into table

21. Page 21  2. Enrolled 테이블에서 각 과목별로 몇 명의 수강생이 있는지를 검색 하시오.  결과는 과목명과 수강생 수를 출력  3. Enrolled 테이블에서 각 과목별 학점이 4.0 이상인 학생수를 검색하 시오  결과는 과목명, 학생 수 (column 명을 numStu 로 표현 ) 를 출력 Practice

22. Page 22  4. Enrolled 테이블에서 수강생이 4 명 이상인 과목의 중간고사 평균을 구하시오.  결과는 과목명, 수강생 수와 중간고사 평균을 출력  5. Enrolled 테이블에서 CSE 학생들의 각 과목별 기말고사의 최고점을 검색하시오.  결과는 과목명과 점수를 출력.  단, 최고점이 90 점 이상일 때만 출력 Practice

23.  6. Enrolled 테이블에서 과목별, 학과별 중간고사, 기말고사 평균을 검 색하시오.  결과는 과목명, 학과명과 중간고사, 기말고사 평균 출력 Practice