Lecture8:Data Manipulation in SQL Advanced SQL queries

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8Lecture8:Data Manipulation in SQLAdvanced SQL queries

Prepared by L. Nouf Almujally

Ref. Chapter5

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The Process of Database Design

Real World Domain

Conceptual model (ERD)

Relational Data Model

Create schema

(DDL)

Load Data(DML)

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Sample Data in Customer Table

custNo custName custSt custCity age

1 C1 Olaya St Jeddah 20

2 C2 Mains St Riyadh 30

3 C3 Mains Rd Riyadh 25

4 C4 Mains Rd Dammam

5 C5 Mains Rd Riyadh

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Sample Data in Product Table

prodNo prodNam

eprodDes price

100 P0 Food 100

101 P1 healthy food 100

102 P2 200

103 P3 self_raising flour,80%wheat

300

104 P4 network 80x 3004

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Sample Data in Orders Table

ordNo ordDate custNo prodNo quantity

1 01-jan-2003 1 100 2

2 02-jan-2003 1 101 1

3 01-jan-2003 2 102 1

4 01-jan-2003 3 100 2

5 03-jan-2003 1 101 1

6 06-mar-2003 2 100 10

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Dept Number Dept Name Location Mail NumberD1 Computer Science Bundoora 39D2 Information Science Bendigo 30D3 Physics Bundoora 37D4 Chemistry Bendigo 35

DEPARTMENT

EMPLOYEE

Employee No. First Name Last Name Dept Number SalaryE1 Mandy Smith D1 50000E2 Daniel Hodges D2 45000E3 Shaskia Ramanthan D2 58000E4 Graham Burke D1 44000E5 Annie Nguyen D1 60000

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O_Id OrderDate OrderPrice Customer

1 2008/11/12 1000 Nora

2 2008/10/23 1600 Sara

3 2008/09/02 700 Nora

4 2008/09/03 300 Nora

5 2008/08/30 2000 Yara

6 2008/10/04 100 Sara

Table orders ( Example 3)

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JOIN

• Often two or more tables are needed at the same time to find all required data

• These tables must be "joined" together

• The formal JOIN basically,• it computes a new table from those to be joined, • the new table contains data in the matching rows of the

individual tables.

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Types of JOIN

• Different SQL JOINs• types of JOIN:.

• JOIN: Return rows when there is at least one match in both tables ( INNER JOIN is the same as JOIN)

• LEFT JOIN: Return all rows from the left table, even if there are no matches in the right table

• RIGHT JOIN: Return all rows from the right table, even if there are no matches in the left table

• Full Outer Joins : retains rows that area unmatched in both the tables.

NOTE: In all the above outer joins, the displayed unmatched columns are filled with NULLS.

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Types of JOIN

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SQL Examples of Joins ( 1)

• Simple JoinSELECT E.firstName, E.lastName, D.deptName FROM EMPLOYEE E, DEPARTMENT D

WHERE E.deptNumber = D.deptNumber;



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E.Firstname E.lastname D.deptNameMandy Smith Computer ScienceDaniel Hodges Information ScienceShaskia Ramanthan Information ScienceGraham Burke Computer ScienceAnnie Nguyen Computer Science

Employee No. First Name Last Name Dept Number Dept Name Location Mail Number SalaryE1 Mandy Smith D1 Computer Science Bundoora 39 50000E2 Daniel Hodges D2 Information Science Bendigo 30 45000E3 Shaskia Ramanthan D2 Physics Bundoora 37 58000E4 Graham Burke D1 Computer Science Bundoora 39 44000E5 Annie Nguyen D1 Computer Science Bundoora 39 60000

This is the result from the matching

This is the final result:

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SQL Examples of Joins ( 2 )• Joining more than two tablesSELECT E.firstName, E.lastName, P.projTitle

FROM EMPLOYEE E, WORKS_ON W, PROJECT P WHERE E.employeeNo = W.employeeNo AND W.projNo = P.projNo;

Employee No. First Name Last Name Dept Number SalaryE1 Mandy Smith D1 50000E4 Graham Burke D1 44000E5 Annie Nguyen D1 60000E2 Daniel Hodges D2 45000E3 Shaskia Ramanthan D2 58000

Employee No. ProjNoE1 1E4 1E5 2E2 3E3 1

ProjNo Project Title1 Project A2 Project B3 Project C

EMPLOYEE

WORKS_ON

PROJECT

E.Firstname E.lastname P.projtitle

Mandy Smith Project A

Graham Burke Project A

Annie Nguyen Project B

Daniel Hodges Project C

Shaskia Ramanthan Project A

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SQL Examples of Joins ( 3 )

• List customers (by customer number, name and address) who have ordered the product 100.

SELECT c.custNo, custName, custSt, custCity FROM customer c, orders o WHERE c.custNo=o.custNo AND prodNo=100;

custNo

custName

custSt custCity age







1 01-jan-2003 1 100 2

2 02-jan-2003 1 101 1

3 01-jan-2003 2 102 1

4 01-jan-2003 3 100 2

5 03-jan-2003 1 101 1

6 06-mar-2003 2 100 10

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SQL Examples of Joins ( 4 )

• Find the total price of the products ordered by customer 1. SELECT sum(price*quantity) FROM orders, product WHERE orders.prodNo = product.prodNo AND custNo = 1;

prodNo prodName

prodDes price

100 P0 Food 100


102 P2 200


300

104 P4 network 80x 300


1 01-jan-2003 1 100 2

2 02-jan-2003 1 101 1

3 01-jan-2003 2 102 1

4 01-jan-2003 3 100 2

5 03-jan-2003 1 101 1

6 06-mar-2003 2 100 10

sum(price*quantity)

400

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Outer Joins in Oracle SQL

• Put an (+) on the potentially deficient side, ie the side where nulls may be added

• The (+) operator is placed in the join condition next to the table that is allowed to have NULL values.

• Example (Left Outer Join) : List all customers, and the products ordered if they have ordered some products.

SELECT c.custNo, o.prodNo, quantity FROM customer c, orders o WHERE c.custNo = o.custNo (+);

• Note: • a table may be outer joined with only one other table.• Which table column to use is important, eg, in above example, do not use

o.custNo in place of c.custNo in the SELECT list.

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1 )Inner Join SQL Example

SELECT Student_Name, Advisor_NameFROM Students, Advisors

WHERE Students.Advisor_ID= Advisors.Advisor_ID;

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2 )Left Outer Join SQL Example


WHERE Students.Advisor_ID= Advisors.Advisor_ID (+);

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3 )Right Outer Join SQL Example


WHERE Students.Advisor_ID(+)= Advisors.Advisor_ID ;

Student_Name Advisor_NameStudent_1 advisor 1Student_5 advisor 3Student_7 advisor 3Student_9 advisor 1

Student_10 advisor 3null Advisor 5

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4 )Full Outer Join SQL Example

SELECT Student_Name, Advisor_NameFROM Students , Advisors

WHERE Students.Advisor_ID (+) = Advisors.Advisor_ID (+) ;

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Nested Queries (1)

• Query results are tables, which can also be queried.

SELECT * FROM (SELECT prodNo, sum(quantity) AS sum FROM orders

GROUP BY prodNo);WHERE sum>10;

Equivalent to

SELECT prodNo, sum(quantity) as sum FROM orders GROUP BY prodNo HAVING sum(quantity)>10;

• The inner query is referred to as a subquery

prodNo sum100 14101 2102 1

prodNo sum100 14

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Nested Queries (2)

• If the query result is a single value, it can be treated as a value, and be compared with other values.

Example: Find products with price more than average

SELECT prodNo, price FROM product WHERE price > (SELECT AVG(price)

FROM product);

AVG(price)

200

prodNo price103 300104 300

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Subquery

• Subquery with equality:

SELECT firstName, lastNameFROM EMPLOYEEWHERE deptNumber =(SELECT deptNumber

FROM DEPARTMENT WHERE mailNumber = 39);

deptNumberD1

firstName lastNameMandy SmithGraham BurkeAnnie Nguyene 24

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Subquery

• Subquery with aggregate function:

SELECT firstName, lastName, salary FROM EMPLOYEE WHERE salary > (SELECT avg(salary) FROM EMPLOYEE);

avg(salary)

51400

firstName lastName salary

Shaskia Raman than 58000Annie Nguyene 60000

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Subquery

• Nested Subquery (use of IN):

SELECT firstName, lastNameFROM EMPLOYEEWHERE deptNumber IN (SELECT deptNumber

FROM DEPARTMENT WHERE location = ‘Bundoora’);

deptNumber

D1

D3firstName lastName

Mandy Smith

Graham Burke

Annie Nguyene



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Subquery

• List the products ordered by customers living in Riyadh.

SELECT prodNo FROM orders WHERE custNo IN (SELECT custNo

FROM customer WHERE custCity =‘Riyadh');

- This query is equivalent to

SELECT prodNo FROM orders o, customer c WHERE o.custNo =c.custNo AND custCity = ‘Riyadh';

custNo

2

3

5

prodNo

100

102

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Queries using EXISTS or NOT EXISTS

Queries using EXISTS • Designed for use only with subqueries• EXISTS return true if there exists at least one row in the result table

returned by the subquery, it is false if the subquery returns an empty result table.

• Syntax

SELECT column_name FROM table_name WHERE EXISTS|NOT EXISTS ( subquery );

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Queries using EXISTS or NOT EXISTS

• ExampleSELECT firstName, lastName

FROM EMPLOYEE E WHERE EXISTS (SELECT * FROM DEPARTMENT D

WHERE E.deptNumber = D.deptNumber AND D.location = ‘Bendigo’);



firstName lastNameDaniel HodgesShaskia Ramanthan

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Example . EXISTS

• Find all customers who have ordered some products.SELECT * FROM customer cWHERE exists (SELECT *

FROM orders o WHERE o.custNo =c.custNo);

• If the subquery is not empty, then the exists condition is true.








1 01-jan-2003 1 100 2

2 02-jan-2003 1 101 1

3 01-jan-2003 2 102 1

4 01-jan-2003 3 100 2

5 03-jan-2003 1 101 1

6 06-mar-2003 2 100 1031

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Example . NOT EXISTS

• Find all customers such that no order made by them has a quantity less than 2.

SELECT * FROM customer cWHERE NOT EXISTS (SELECT *

FROM orders o WHERE o.custNo = c.custNo

AND quantity <2);








1 01-jan-2003 1 100 2

2 02-jan-2003 1 101 1

3 01-jan-2003 2 102 1

4 01-jan-2003 3 100 2

5 03-jan-2003 1 101 1

6 06-mar-2003 2 100 10





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UNION

• The UNION operator is used to combine the result-set of two or more SELECT statements.

• Notice that each SELECT statement within the UNION must1. have the same number of columns. 2. The columns must also have similar data types. 3. the columns in each SELECT statement must be in the same

order.• Combines the results of two SELECT statements into one result set,

and then eliminates any duplicate rows from that result set.• SQL UNION Syntax

SELECT column_name(s) FROM table_name1UNIONSELECT column_name(s) FROM table_name2

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UNION

• Note: The UNION operator selects only distinct values by default. To allow duplicate values, use UNION ALL.

• UNION ALL Combines the results of two SELECT statements into one result set.

• SQL UNION ALL Syntax

SELECT column_name(s) FROM table_name1UNION ALLSELECT column_name(s) FROM table_name2

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UNION Example 1

• list all the different employees in Norway and USASELECT E_Name FROM Employees_Norway

UNIONSELECT E_Name FROM Employees_USA; E_Name

Hansen, OlaSvendson, Tove

Svendson, StephenPettersen, Kari

Turner, SallyKent, Clark

Scott, Stephen

E_ID E_Name

01 Hansen, Ola02 Svendson, Tove03 Svendson, Stephen04 Pettersen, Kari

E_ID E_Name01 Turner, Sally02 Kent, Clark03 Svendson, Stephen04 Scott, Stephen

"Employees_Norway" “Employees_USA”

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UNION Example 2

SELECT custNo FROM customer WHERE custCity=‘Riyadh' UNIONSELECT custNo FROM orders WHERE prodNo=102; // union of the two queries








1 01-jan-2003 1 100 2

2 02-jan-2003 1 101 1

3 01-jan-2003 2 102 1

4 01-jan-2003 3 100 2

5 03-jan-2003 1 101 1

6 06-mar-2003 2 100 10

custNo

2

3

5

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MINUS

• the MINUS operator returns only unique rows returned by the first query but not by the second.

• Takes the result set of one SELECT statement, and removes those rows that are also returned by a second SELECT statement.

• SQL MINUS Syntax

SELECT column_name(s) FROM table_name1MINUSSELECT column_name(s) FROM table_name2

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MINUS Example 1

SELECT prodNo FROM productMINUSSELECT prodNo FROM orders; //difference from the two queries


1 01-jan-2003 1 100 2

2 02-jan-2003 1 101 1

3 01-jan-2003 2 102 1

4 01-jan-2003 3 100 2

5 03-jan-2003 1 101 1

6 06-mar-2003 2 100 10

prodNo prodName prodDes price

100 P0 Food 100


102 P2 200


300

104 P4 network 80x 300

ProdNo

103

104

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INTERSECT

• the INTERSECT operator returns only those rows returned by both queries.

• Returns only those rows that are returned by each of two SELECT statements.

• SQL INTERSECT Syntax

SELECT column_name(s) FROM table_name1INTERSECTSELECT column_name(s) FROM table_name2

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INTERSECT

SELECT custNo FROM customer WHERE custCity=‘Riyadh' INTERSECTSELECT custNo FROM orders WHERE prodNo=102; // intersect of the two queries

custNo

custName custSt custCity age







1 01-jan-2003 1 100 2

2 02-jan-2003 1 101 1

3 01-jan-2003 2 102 1

4 01-jan-2003 3 100 2

5 03-jan-2003 1 101 1

6 06-mar-2003 2 100 10

CustNo

2

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DEPENDENT

EMPLOYEE

Employee No. First Name Last Name Date of BirthE1 Joshua Smith 12-Jun-1998E3 Jay Ramanthan 04-Jan-1996E1 Jemima Smith 08-Sep-2000

Employee No. First Name Last Name Dept Number SalaryE1 Mandy Smith D1 50000E2 Daniel Hodges D2 45000E3 Shaskia Ramanthan D2 58000

Examples

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Examples

SELECT employeeNo, firstName, lastNameFROM EMPLOYEEUNIONSELECT employeeNo, firstName, lastNameFROM DEPENDENT;

SELECT employeeNoFROM EMPLOYEEINTERSECTSELECT employeeNoFROM DEPENDENT

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EMPLOYEE Table Example1

SELECT department_id, count(*), max(salary), min(salary)FROM employee GROUP BY department_id;

EMPLOYEE_ID LAST_NAME FIRST_NAME JOB_ID MANAGER_ID SALARY COMM DEPARTMENT_ID

7369 SMITH JOHN 667 7902 800 NULL 207499 ALLEN KEVIN 670 7698 1600 300 307505 DOYLE JEAN 671 7839 2850 NULL 307506 DENNIS LYNN 671 7839 2750 NULL 307507 BAKER LESLIE 671 7839 2200 NULL 407521 WARK CYNTHIA 670 7698 1250 500 30

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SELECT Employee_ID, FIRST_NAME,DEPARTMENT_ID FROM employeeWHERE salary=(SELECT max(salary) FROM employee);



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SELECT Employee_IDFROM employeeWHERE department_id IN (SELECT department_idFROM department WHERE name=’SALES’);

DEPARTMENT

Department_ID Name Location_ID

10 ACCOUNTING 12220 RESEARCH 12430 SALES 12340 OPERATIONS 167



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SELECT nameFROM department dWHERE NOT EXISTS (SELECT last_nameFROM employee eWHERE d.department_id=e.department_id);

DEPARTMENT





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SELECT last_name, d.department_id, d.nameFROM employee e, department dWHERE e.department_id (+)= d.department_id AND d.department_id in (SELECT department_id FROM department WHERE name IN (‘RESEARCH’ , ’OPERATIONS’));

DEPARTMENT





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SELECT employee_id, First_name, Last_name, SalaryFROM employeeWHERE last_name like ‘D%’;



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References

• “Database Systems: A Practical Approach to Design, Implementation and Management.” Thomas Connolly, Carolyn Begg. 5th Edition, Addison-Wesley, 2009.

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Lecture8:Data Manipulation in SQL Advanced SQL queries

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