structured query language agenda: - sql commands for db definition - sql dml commands - views and...

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Structured Query Language Agenda: - SQL commands for DB definition - SQL DML commands - VIEWS and security control using SQL

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Structured Query Language

Agenda:

- SQL commands for DB definition

- SQL DML commands

- VIEWS and security control using SQL

CREATE TABLE command

- Table name

- Description of the attributes:

-- Attribute name

-- Data type

-- Constraints on the attribute values, including:

--- NULL or NOT NULL

--- Domain of attribute

--- Referential constraints on the attribute

- Primary key attributes

- Foreign key attributes

CREATE TABLE, example 1

CREATE TABLE Book (

isbn VARCHAR(15) NOT NULL,

title VARCHAR(200) NOT NULL,

catalog_no VARCHAR(15) NOT NULL,

copy_no INT NOT NULL DEFAULT 1,

keywords CHAR(100) NULL,

purchase_date DATE NULL,

PRIMARY KEY CLUSTERED(catalog_no, copy_no))

Attribute nameTable name

Data type Domain constraint

Default

PK composed of two attributes

CREATE TABLE, example 2

CREATE TABLE Borrows ( catalog_num VARCHAR(15) NOT NULL, copy_num INT NOT NULL, issue_date DATE NOT NULL, person_id CHAR(8) NOT NULL,

PRIMARY KEY CLUSTERED(catalog_num, copy_num, person_id, issue_date),

CONSTRAINT fk_borrows_book FOREIGN KEY(catalog_num, copy_num) REFERENCES Books(catalog_no, copy_no),

CONSTRAINT fk_borrows_person FOREIGN KEY(person_id) REFERENCES Person( id) )R

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CREATE TABLE, example 3

CREATE TABLE Person (

lname VARCHAR(35) NOT NULL,

fnames VARCHAR(50) NOT NULL,

email VARCHAR(60) NOT NULL UNIQUE

CHECK ( email LIKE ‘%@%’),

id CHAR(8) NOT NULL,

phone CHAR(12) NULL,

PRIMARY KEY (id) )

Domain constraint

Key (but not PK)

DROP TABLE command

DROP TABLE Person;

DROP TABLE Person CASCADE;

Step 1. Referential constraints are deleted

Step 2. All data in the table is deleted

Step 3. The definition of the table is deleted from the DB.

Problem: Borrows( person_id) REFERENCES Person( id)

ALTER TABLE command

- Add a new column in a table

- Delete a column from a table

- Add/Delete a constraint specified on a table

Example 1

ALTER TABLE Person ADD fines FLOAT;

Personlname fnames email id phone

Personlname fnames email id phone fines

Initial design

ALTER-ed design

ALTER TABLE command…

Example 2: add a new attributeALTER TABLE Book ADD category VARCHAR(10) NOT NULL DEFAULT “normal” CHECK (category in (“normal”, “reserve”, “media”));

Example 3: drop a constraintALTER TABLE Borrows DROP CONSTRIANT fk_borrows_person;

Example 4: add a constraintALTER TABLE Borrows ADD CONSTRAINT fk_borrows_person FOREIGN KEY(person_id) REFERENCES Person( id);

Q: What happens to data entered in a table BEFORE adding a constraint?

INSERT INTO command, populating a DB

Example 1:

INSERT INTO PersonVALUES ( ‘Bush’, ‘George W.’, ‘[email protected]’, ‘09112001’, NULL, 0);

Person

lname fnames email id phone fines

Example 2:

INSERT INTO BookVALUES ( ‘0321122267’, ‘Fundamentals of Database Systems’, ‘QA76.9.D3’, 1, ‘Databases’, ‘2004-09-25’);

Book

isbn title catalog_no copy_no keywords Purchase_date

DELETE FROM command

- to delete one or more rows of data from a table

Example 1:

DELETE FROM Person WHERE id= ‘09112001’;

Example 2:

DELETE FROM Person WHERE lname=’Bush’;

Example 3:

DELETE FROM Borrows WHERE 1

DELETE FROM command …

Example 4:

DELETE FROM Borrows

WHERE person_id IN (“09112001”, “55554444”, “12345678”);

Example 5:

DELETE FROM Borrows

WHERE person_id IN ( SELECT id

FROM Person

WHERE lname= ‘Bush’);

UPDATE command

- Modify the value of one or more cells in a table

Example 1:

UPDATE Borrows SET issue_date=CURRENT_DATE( )WHERE person_id=’09112001’;

Borrows

catalog_num copy_num issue_date person_id

Example 2:

UPDATE Person SET fines= fines*2.0 WHERE id=’09112001’;

Function provided by SQL

SQL allows the use of arithmetic expressions

Person

lname fnames email id phone fines

SELECT command

- Output required information from one or more tables

For the following examples (unless stated otherwise):

- Use the EMPLOYEE-DEPARTMENT-PROJECTS database

- Assume the initial data in the tables as provided earlier

SELECT command (1)

Example 1:Report the birth date and address of employee named "John Smith"

OUTPUT

BDate Address

9-Jan-55 731 Fonden

SELECT BDate, AddressFROM EMPLOYEEWHERE Fname = ‘John’ AND

Lname = ‘Smith’;

SELECT command (1a)

Example 1a:Report the SSN of Employees who spend more than 15 hourson some project.

SELECT DISTINCT ESSNFROM WORKS_ONWHERE Hours > 15;

SELECT ESSNFROM WORKS_ONWHERE Hours > 15;

OUTPUT

ESSN

123456789

666884444

453453453

999887777

987987987

OUTPUT

ESSN

123456789

666884444

453453453

453453453

999887777

987987987

987987987

SELECT command (2)

Example 2:Report the Name and address of employees workingin the “Research” department.

OUTPUT

Fname Lname Address

John Smith 731 Fonden

Franklin Wong 638 Voss

Ramesh Narayan 975 Fire Oak

Joyce English 5631 Rice

SELECT Fname, Lname, AddressFROM EMPLOYEE, DEPARTMENTWHERE Dname = ‘Research’ AND

Dnumber = Dno

SELECT command (3)

Example 3: For each project located in Stafford,list the project number, the controlling department, andthe department manager's last name and address.

SELECT Pnumber, Dnum, Lname, AddressFROM PROJECT, DEPARTMENT, EMPLOYEEWHERE Dnum = Dnumber AND

MgrSSN = SSN ANDPlocation = ‘Stafford’;

OUTPUT

Pnumber Dnum Lname Address

10 4 Wallace 291 Berry

30 4 Wallace 291 Berry

SELECT command (4): Alias and Dot-notation

Example 4: For each employee, give the last name,and the last name of his/her supervisor.

SELECT E.Lname, S.LnameFROM EMPLOYEE AS E, EMPLOYEE AS SWHERE E.SuperSSN = S.SSN

OUTPUT

E.Lname S.Lname

Smith Wong

Wong Borg

Zeleya Wallace

Wallace Borg

Narayan Wong

English Wong

Jabbar Wallace

SELECT command (5, 6): no WHERE

Example 5: Print SSN of all employees.

SELECT SSNFROM EMPLOYEE

OUTPUT

SSN

123456789

333445555

999887777

987654321

666884444

453453453

987987987

888665555

SELECT SSN, DnameFROM EMPLOYEE, DEPARTMENT

How many rows in output ?

SELECT command (7): wildcard

Example 7: Show the EMPLOYEE table

SELECT * FROM EMPLOYEE

SELECT DEPT_LOCATION.*, DEPARTMENT.DnameFROM DEPT_LOCATION, DEPARTMENTWHERE DEPT_LOCATION.Dnumber = DEPARTMENT.Dnumber

Example 7a: Report Department information including locations

OUTPUT

Dnumber Dlocation Dname

1 Houston Headquarters

4 Stafford Administration

5 Bellaire Research

5 Sugarland Research

5 Houston Research

SELECT command (8): UNION

Example 8: List all projects which either use employee "Wong",or are controlled by a department managed by "Wong".

(SELECT Pname FROM PROJECT, WORKS_ON, EMPLOYEE WHERE Pnumber = PNo AND ESSN = SSN AND LName = 'Wong' ) UNION(SELECT Pname FROM PROJECT, DEPARTMENT, EMPLOYEE WHERE DNum = Dnumber AND SSN = MgrSSN AND LName = 'Wong');

First sub-query Second sub-query OUTPUT

Pname

ProductY

ProductZ

Computerization

Reorganisation

Pname

ProductX

ProductY

ProductZ

OUTPUT

PName

ProductX

ProductY

ProductZ

Computerization

Reorganisation

SELECT command (9): nested queries

Example 9: Report the name and address of all employees working in the 'Research' department.

SELECT Fname, Lname, AddressFROM EMPLOYEEWHERE Dno IN ( SELECT Dnumber

FROM DEPARTMENT WHERE Dname = 'Research' )

Result of inner query:

Dnumber

5

OUTPUT

Fname Lname Address Dno

John Smith 731 Fonden 5

Franklin Wong 638 Voss 5

Ramesh Narayan 975 Fire Oak 5

Joyce English 5631 Rice 5

uncorrelated

SELECT command (10): nested queries

Example 10: Get the names of all employees who have a dependent with the same first name.

SELECT E.Fname, E.LnameFROM EMPLOYEE AS EWHERE E.SSN IN ( SELECT ESSN

FROM DEPENDENT WHERE ESSN = E.SSN

AND E.Fname = DependentName )

OUTPUT

E.Fname E.Lname

correlated

SELECT command (11): EXISTS operator

Example 11:Get names of employees who work for at least one project.

SELECT Fname, LnameFROM EMPLOYEEWHERE EXISTS ( SELECT *

FROM WORKS_ON WHERE SSN = ESSN )

OUTPUT

Fname Lname

John Smith

Franklin Wong

Alicia Zeleya

Ramesh Narayan

Joyce English

Ahmad Jabbar

James Borg

SELECT command (12): NOT EXISTSExample 12:Find names of employees who do not work for even one project.

SELECT Fname, LnameFROM EMPLOYEEWHERE NOT EXISTS ( SELECT *

FROM WORKS_ON WHERE SSN = ESSN )

OUTPUT

Fname Lname

Jennifer Wallace

SELECT command (13): matching CHAR(n) types

Example 13:Find names of all Employees who live on Fonden street.

SELECT LnameFROM EMPLOYEEWHERE Address LIKE ‘%Fonden%’;

matchingoperator

matcheszero or more chars

OUTPUT

Lname

Smith NOTES:(1) ‘_’ matches exactly one char(2) RLIKE operator allows REG_EXP

SELECT command (14): aggregates

Example 14:Get the minimum, maximum, average and total salariesfor employees of the Research department.

SELECT sum(Salary), max( Salary), min( Salary), avg( Salary)FROM EMPLOYEE, DEPARTMENTWHERE Dno = Dnumber AND Dname = 'Research'

OUTPUT

13300 4000 2500 3325

SELECT sum(Salary) AS Tot, max( Salary) AS Max, min( Salary) AS Min, avg( Salary) AS MeanFROM EMPLOYEE, DEPARTMENTWHERE Dno = Dnumber AND Dname = 'Research'

OUTPUT

Tot Max Min Mean

13300 4000 2500 3325

SELECT command (15): GROUP BY

Example 15:For departments other than Headquarters, get the Dno,the No. of employees in that department, and their average salary.

SELECT Dno, count(*) AS HeadCount, avg(Salary) AS MeanSalaryFROM EMPLOYEE, DEPARTMENTWHERE Dno = Dnumber AND Dname <> 'Headquarters'GROUP BY Dno;

OUTPUT

Dno HeadCount MeanSalary

5 4 3325

4 3 3100

SELECT command (16): GROUP BY .. HAVING ..

Example 16:For ‘Large’ departments other than Headquarters, get the Dno,the No. of employees in that department, and their average salary.

SELECT Dno, count(*) AS HeadCount, avg(Salary) AS MeanSalaryFROM EMPLOYEE, DEPARTMENTWHERE Dno = Dnumber AND Dname <> 'Headquarters'GROUP BY DnoHAVING HeadCount > 3;

OUTPUT

Dno HeadCount MeanSalary

5 4 3325

SELECT command (17): Mathematical operators

Example 17:Display the result of a 10% increase in Salary of employeeswhose Last name starts with "B".

SELECT Lname, 1.1 * Salary AS IncreasedSalaryFROM EMPLOYEEWHERE Lname LIKE 'B%'

OUTPUT

Lname IncreasedSalary

Borg 6050

SELECT command (18): sorting

Example 18:Report names and salaries of employees, in descending order by salary

SELECT Lname, SalaryFROM EMPLOYEEORDER BY Salary DESC OUTPUT

Lname Salary

Borg 5500

Wallace 4300

Wong 4000

Narayan 3800

Smith 3000

Zeleya 2500

English 2500

Jabbar 2500

VIEWS

VIEW:A virtual table derived from a set of existing tables.

Main uses of a view:

- Security (selective display of information to different users)

- Ease-of-use-- Explicit display of derived attributes-- Explicit display of related information from different tables-- Intermediate table can be used to simplify SQL query

CREATE VIEW command (1)

Example 1:Create a view showing the names of employees, which project they work on,and how many hours they spend on each project.

CREATE VIEW EMP_WORKS_ON AS SELECT Fname, Lname, Pname, Hours FROM EMPLOYEE, PROJECT, WORKS_ON WHERE SSN = ESSN AND Pno = Pnumber;

SELECT * FROM EMP_WORKS_ON

EMP_WORKS_ON

Fname Lname Pname Hours

John Smith ProductX 32.5

John Smith ProductY 7.5

Ramesh Narayan ProductZ 40

Joyce English ProductX 20

Joyce English ProductY 20

Franklin Wong ProductY 10

Franklin Wong ProductZ 10

Franklin Wong Computerization 10

Franklin Wong Reorganization 10

Alicia Zeleya Newbenefits 30

Alicia Zeleya Computerization 10

Ahmad Jabbar Computerization 35

Ahmad Jabbar Newbenefits 5

Ahmad Jabbar Newbenefits 20

Ahmad Jabbar Reorganization 15

James Borg Reorganization null

VIEWS: mechanism of storage/updates

1. Only definition of view is stored in memory

2. Data for a view is only generated when the query is processed

3. Update a view attribute data in the underlying table is updated

VIEWS: mechanism of storage/updates..

Example 2:What happens to employee hours if they work one-shift overtime?

UPDATE EMP_WORKS_ONSET Hours = Hours * 1.5

SELECT * FROM WORKS_ON

WORKS_ON

ESSN Pno Hours

123456789 1 48.75

123456789 2 11.25

666884444 3 60

453453453 1 30

453453453 2 30

333445555 2 15

333445555 3 15

333445555 10 15

333445555 20 15

999887777 30 45

999887777 10 15

987987987 10 52.5

987987987 30 7.5

987654321 30 30

987987987 20 22.5

888665555 20 null

VIEWS: mechanism of storage/updates…

Example 3:John Smith, currently working on ‘ProductX’ project,is reassigned to ‘ProductY’ project.

Example 3 (incorrect): UPDATE EMP_WORKS_ONSET Pname = 'ProductY'WHERE Lname = 'Smith' AND

Pname = 'ProductX'

Example 3 (correct):UPDATE WORKS_ONSET Pno = (SELECT Pnumber FROM PROJECTS WHERE Pname = 'ProductY')WHERE ESSN = ( SELECT SSN FROM EMPLOYEE WHERE Lname = 'Smith')

AND Pno = ( SELECT Pnumber FROM PROJECT WHERE Pname = 'ProductX');

Problem ?

VIEWS: using derived attributes

CREATE VIEW DEPT_INFO AS SELECT DName, count(*) AS NumEmps, sum( Salary) AS TotalSalary FROM DEPARTMENT, EMPLOYEE WHERE DNumber = DNo GROUP BY DName;

NOTE:

- you cannot UPDATE a computed attribute

VIEWS, GRANT: security control

Example 5:Allow user U1 to see/modify all Employee data except Salaries.

CREATE VIEW EMP_PERSONNEL ASSELECT Fname, Minit, Lname, SSN, BDate, Address, Sex, SuperSSN, DnoFROM EMPLOYEE;

GRANT SELECT, UPDATE ON EMP_PERSONNEL to U1;

GRANT SELECT, UPDATE ON EMP_PERSONNEL TO U1WITH GRANT OPTION;

OR

1

2

U1 > GRANT SELECT ON EMP_PERSONNEL TO U2;

GRANT, REVOKE commands

REVOKE SELECT ON EMP_PERSONNEL FROM U2;

Disallow U2 from seeing anything in EMP_PERSONNEL:

GRANT UPDATE ON EMPLOYEE( Salary) TO U3;

GRANT , REVOKE can refer to individual attributes

Allow U3 to change Salary in EMPLOYEE table:

Concluding remarks

GRANT on SELECT use VIEW or TABLE

GRANT on INSERT, DELETE, UPDATE use TABLE

Next topic: Indexes