lecture10: data manipulation in sql , advanced sql … · outer joins in oracle sql lecture10 21...

LECTURE10:

DATA MANIPULATION IN SQL , ADVANCED

SQL QUERIES

1

Ref. Chapter5

From

“Database Systems: A Practical Approach to Design, Implementation and Management.”Thomas Connolly, Carolyn Begg.

I S 2 2 0 : D a t a b a s e F u n d a m e n t a l s

The Process of Database Design2

Conceptual Design (ERD)

Logical Design

(Relational Model)

Physical Design

Create schema

(DDL)

Load Data

(DML)

Sample Data in Customer Table

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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

Sample Data in Product Table

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prodNo prodName prodDes 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 300

Sample Data in Orders Table

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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

Dept Number Dept Name Location Mail Number

D1 Computer Science Bundoora 39

D2 Information Science Bendigo 30

D3 Physics Bundoora 37

D4 Chemistry Bendigo 35

DEPARTMENT

EMPLOYEE

Employee No. First Name Last Name Dept Number Salary

E1 Mandy Smith D1 50000

E2 Daniel Hodges D2 45000

E3 Shaskia Ramanthan D2 58000

E4 Graham Burke D1 44000

E5 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

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Sample Data in Customer Table

Table names and Column names

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Table name can be prefixed with the owner name.

eg, if table product is owned by user John, you can use

SELECT * FROM John.product;

Column names can be prefixed with table name,

SELECT product.prodNo

FROM product;

Alias

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SQL aliases are used to temporarily rename a table or a column heading.

Syntax for Columns

Syntax for Tables

SELECT column_name AS alias_name

FROM table_name;

SELECT column_name AS alias_name

FROM table_name;

SELECT column_name(s)

FROM table_name [AS] alias_name;

SELECT column_name(s)

FROM table_name [AS] alias_name;

Alias ( important note )

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Columns Alias:

For example, you might wish to know how much is the combined total salary of all employees whose salary is above $25,000 / year.

SELECT SUM(salary) AS "Total Salary"FROM employees

WHERE salary > 25000;

In this example, we've aliased the sum(salary) field as "Total Salary". As a result, "Total Salary" will display as the field name when the result set is returned.

Table Alias:

SELECT o.OrderID, o.OrderDateFROM Orders AS o;

Exercise

create table count_null ( a number, b

number );

insert into count_null values ( 1, 5);

insert into count_null values ( null, 7);

insert into count_null values ( null, null);

insert into count_null values ( 8, 2);

select count(a) as "count_a_not_null",

count(b) as "count_b_not_null", count(*)

as "count_all”

from count_null;

Output :

11

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JOIN

12

Lecture10

JOIN

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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,

computes a new table from those to be joined

the new table contains data in the matching rows of

the individual tables.

Types of JOIN

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types of JOIN:.

INNER JOIN: Return rows when there is at least one match in both tables

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 are unmatched in both the tables.

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

Types of JOIN

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SQL Examples of Joins ( 1)

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Simple Join

SELECT E.firstName, E.lastName, D.deptName

FROM EMPLOYEE E, DEPARTMENT D

WHERE E.deptNumber = D.deptNumber;












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Employee No.First Name Last Name Salary Dept Number Dept Number Dept Name LocationMail Number

E1 Mandy Smith 50000 D1 D1 Computer Science Bundoora 39

E2 Daniel Hodges 45000 D2 D2 Information Science Bendigo 30

E3 Shaskia Ramanthan 58000 D2 D2 Information Science Bundoora 37

E4 Graham Burke 44000 D1 D1 Computer Science Bundoora 39

E5 Annie Nguyen 60000 D1 D1 Computer Science Bundoora 39

This is the result from the matching

This is the final result:

FIRSTNAME LASTNAME DEPTNAME

----------------- ------------------- -------------------

Mandy Smith Computer Science

Annie Nguyen Computer Science

Graham Burke Computer Science

Shaskia Ramanthan Information Science

Daniel Hodges Information Science

SQL Examples of Joins ( 2 )

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ure1

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Joining more than two tables

SELECT 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 Salary






Employee No. ProjNo

E1 1

E4 1

E5 2

E2 3

E3 1

ProjNo Project Title

1 Project A

2 Project B

3 Project C

EMPLOYEE

WORKS_ON

PROJECT

FIRSTNAME LASTNAME PROJTITLE

------------------ ------------------ -------------------

Mandy Smith Project A

Graham Burke Project A

Annie Nguyen Project B

Daniel Hodges Project C

Shaskia Ramanthan Project A

SQL Examples of Joins ( 3 )

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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;

custN

o

custN

ame

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 CUSTNAME CUSTST CUSTCITY

---------------- ----------------- ------------- ----------------

1 C1 Olaya St Jeddah

2 C2 Mains St Riyadh


SQL Examples of Joins ( 4 )20

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;

prod

No

prodN

ame

prodDes price

100 P0 Food 100


102 P2 200

103 P3 self_raising

flour,80%wheat

300


ordNo ordDate custNo prodN

o

quant

ity

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

Outer Joins in Oracle SQL

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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

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SELECT Student_Name, Advisor_Name

FROM Students, Advisors

WHERE Students.Advisor_ID= Advisors.Advisor_ID;

2) Left Outer Join SQL Example

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

3) Right Outer Join SQL Example

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

Student_1 advisor 1

Student_5 advisor 3

Student_7 advisor 3

Student_9 advisor 1

Student_10 advisor 3

null Advisor 5

4) Full Outer Join SQL Example

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FROM Students , Advisors

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

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Nested Queries (1)

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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 SUM

100 14

101 2

102 1

PRODNO SUM

-------------- --------

100 14


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

Nested Queries (2)

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

Subquery with equality ( < , >) :

Example: Find products with price more than average

SELECT prodNo, price

FROM product

WHERE price > (SELECT AVG(price)

FROM product);

AVG(PRICE)

200

PRODNO PRICE

----------------- ----------------

103 300

104 300


100 P0 Food 100


102 P2 200

103 P3 self_raising flour,80%wheat 300


Subquery

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Subquery with equality (=) :

Give a list with first and last names of employees who work in the department

with mail number = 39

SELECT firstName, lastName

FROM EMPLOYEE

WHERE deptNumber =(SELECT deptNumber

FROM DEPARTMENT

WHERE mailNumber = 39);

DEPTNUMBER

D1

FIRSTNAME LASTNAME

----------------- ---------

mandy smith

graham burke

Annie nguyen












Subquery

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Subquery with aggregate function:

Give a list with first, last names and salary of employees whose salary is greater than average salary for all employees.

SELECT firstName, lastName, salary

FROM EMPLOYEE

WHERE salary > (SELECT avg(salary)

FROM EMPLOYEE);

AVG

(SALARY)

51400

FIRSTNAME LASTNAME SALARY

----------------- -------------------- ----------

Shaskia Ramanthan 58000

Annie Nguyen 60000












Subquery

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32

Nested Subquery (use of IN):

Give a list with first, last names of employees whose departments are located in

“Bundoora”.


FROM EMPLOYEE

WHERE deptNumber IN (SELECT deptNumber

FROM DEPARTMENT

WHERE location = ‘Bundoora’);

deptNumber

D1

D3












FIRSTNAME LASTNAME

---------------- - -----------------

Annie nguyen

graham burke

mandy smith

Subquery

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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

----------

102

100

100

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Queries using EXISTS or NOT EXISTS

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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 );

SELECT column_name

FROM table_name

WHERE EXISTS|NOT EXISTS ( subquery );

Queries using EXISTS or NOT EXISTS

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Example


FROM EMPLOYEE E

WHERE EXISTS (SELECT * FROM DEPARTMENT D

WHERE E.deptNumber = D.deptNumber

AND D.location = ‘Bendigo’); Employee No. First Name Last Name Dept Number Salary











FIRSTNAME LASTNAME

---------------- ----------------

Shaskia Ramanthan

Daniel Hodges

Example . EXISTS

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Find all customers who have ordered some products.SELECT *

FROM customer c

WHERE exists (SELECT *

FROM orders o

WHERE o.custNo =c.custNo);

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







ordNo ordDate custNo prodNo quanti

ty

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 CUNAME CUSTST CUSTCITY AGE

------------ ------------ ---------- --------------- ------------




Example . NOT EXISTS

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

SELECT *

FROM customer c

WHERE NOT EXISTS (SELECT *

FROM orders o

WHERE o.custNo = c.custNo

AND quantity <2);

custNo custName custSt custCity ag

e






ordNo ordDate custNo prodNo quanti

ty

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 CUTNAME CUSTST CUSTCITY AGE

-------------- -------------- ------------- -------------- ---------




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UNION

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The UNION operator is used to combine the result-set of two or more

SELECT statements.

Notice that each SELECT statement within the UNION must

1. 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_name1

UNION

SELECT column_name(s) FROM table_name2;


UNION

SELECT column_name(s) FROM table_name2;

UNION

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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


UNION ALL



UNION ALL


UNION Example 1

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list all the different employees in Norway and USA

SELECT E_Name FROM Employees_Norway

UNION

SELECT E_Name FROM Employees_USA;

E_Name

Hansen, Ola

Svendson, Tove

Svendson, Stephen

Pettersen, Kari

Turner, Sally

Kent, Clark

Scott, Stephen

E_ID E_Name

01 Hansen, Ola

02 Svendson, Tove

03 Svendson, Stephen

04 Pettersen, Kari

E_ID E_Name

01 Turner, Sally

02 Kent, Clark

03 Svendson, Stephen

04 Scott, Stephen

"Employees_Norway" “Employees_USA”

UNION Example 2

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SELECT custNo FROM customer

WHERE custCity=‘Riyadh’

UNION

SELECT 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

MINUS

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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


MINUS

SELECT column_name(s) FROM table_name2 ;


MINUS


MINUS Example 1

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Example: List the products that had never been ordered by customers

SELECT prodNo FROM product

MINUS

SELECT 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


100 P0 Food 100


102 P2 200

103 P3 self_raising

flour,80%wheat

300


PRODNO

---------------

103

104

INTERSECT

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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


INTERSECT



INTERSECT


INTERSECT

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Example: List the customers from Riyadh city who ordered product 102

SELECT custNo FROM customer

WHERE custCity=‘Riyadh'

INTERSECT

SELECT custNo FROM orders

WHERE prodNo=102; // intersect of the two queries

custN

o

custName custSt custCity age






ordNo ordDate custNo prodNo quantit

y

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

DEPENDENT

EMPLOYEE

Employee No. First Name Last Name Date of Birth

E1 Joshua Smith 12-Jun-1998

E3 Jay Ramanthan 04-Jan-1996

E1 Jemima Smith 08-Sep-2000





Examples

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Examples

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SELECT employeeNo, firstName, lastName

FROM EMPLOYEE

UNION

SELECT employeeNo, firstName, lastName

FROM DEPENDENT;

SELECT employeeNo

FROM EMPLOYEE

INTERSECT

SELECT employeeNo

FROM DEPENDENT

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EMPLOYEE Table Example1

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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 20

7499 ALLEN KEVIN 670 7698 1600 300 30

7505 DOYLE JEAN 671 7839 2850 NULL 30

7506 DENNIS LYNN 671 7839 2750 NULL 30

7507 BAKER LESLIE 671 7839 2200 NULL 40

7521 WARK CYNTHIA 670 7698 1250 500 30

DEPARTMENT_ID COUNT(*) MAX(SALARY) MIN(SALARY)

------------------------- ------------ -------------------- -------------------

20 1 800 800

40 1 2200 2200

30 4 2850 1250


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SELECT Employee_ID, FIRST_NAME,DEPARTMENT_ID

FROM employee

WHERE salary=(SELECT max(salary) FROM employee);


7369 SMITH JOHN 667 7902 800 NULL 20

7499 ALLEN KEVIN 670 7698 1600 300 30

7505 DOYLE JEAN 671 7839 2850 NULL 30

7506 DENNIS LYNN 671 7839 2750 NULL 30


7521 WARK CYNTHIA 670 7698 1250 500 30

Employee_ID FIRST_NAME DEPARTMENT_ID

---------------- ------------------- -------------------------

7505 JEAN 30


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SELECT Employee_ID

FROM employee

WHERE department_id IN (SELECT department_id

FROM department WHERE name=’SALES’);

DEPARTMENT

Department_ID Name Location_ID

10 ACCOUNTING 122

20 RESEARCH 124

30 SALES 123

40 OPERATIONS 167


7369 SMITH JOHN 667 7902 800 NULL 20

7499 ALLEN KEVIN 670 7698 1600 300 30

7505 DOYLE JEAN 671 7839 2850 NULL 30

7506 DENNIS LYNN 671 7839 2750 NULL 30


7521 WARK CYNTHIA 670 7698 1250 500 30

EMPLOYEE_ID

---------------------

7521

7506

7505

7499


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SELECT name

FROM department d

WHERE NOT EXISTS (SELECT last_name

FROM employee e

WHERE d.department_id=e.department_id);

DEPARTMENT


10 ACCOUNTING 122

20 RESEARCH 124

30 SALES 123

40 OPERATIONS 167


7369 SMITH JOHN 667 7902 800 NULL 20

7499 ALLEN KEVIN 670 7698 1600 300 30

7505 DOYLE JEAN 671 7839 2850 NULL 30

7506 DENNIS LYNN 671 7839 2750 NULL 30


7521 WARK CYNTHIA 670 7698 1250 500 30

NAME

----------------

ACCOUNTING

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LAST_NAME DEPARTMENT_ID Department_ID Name

SMITH 20 20 RESEARCH

ALLEN 30 30 SALES

DOYLE 30 30 SALES

DENNIS 30 30 SALES

BAKER 40 40 OPERATIONS

WARK 30 30 SALES


ACCOUNTING Department Not exist


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SELECT last_name, d.department_id, d.name

FROM employee e, department d

WHERE e.department_id (+)= d.department_id AND d.department_id in (SELECT

department_id FROM department WHERE name IN (‘ACCOUNTING’ , ‘OPERATIONS’));

DEPARTMENT


10 ACCOUNTING 122

20 RESEARCH 124

30 SALES 123

40 OPERATIONS 167


7369 SMITH JOHN 667 7902 800 NULL 20

7499 ALLEN KEVIN 670 7698 1600 300 30

7505 DOYLE JEAN 671 7839 2850 NULL 30

7506 DENNIS LYNN 671 7839 2750 NULL 30


7521 WARK CYNTHIA 670 7698 1250 500 30

LAST_NAME DEPARTMENT_ID NAME

------------------ ------------------------ ----------

BAKER 40 OPERATIONS

10 ACCOUNTING


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SELECT last_name, d.department_id, d.name

FROM employee e, department d

WHERE e.department_id (+)= d.department_id AND d.department_id in (SELECT

department_id FROM department WHERE name IN (‘ACCOUNTING’ , ‘OPERATIONS’));



ALLEN 30 30 SALES

DOYLE 30 30 SALES

DENNIS 30 30 SALES


WARK 30 30 SALES

10 ACCOUNTINGLAST_NAME DEPARTMENT_ID NAME

------------------ ------------------------ ----------

BAKER 40 OPERATIONS

10 ACCOUNTING



ALLEN 30 30 SALES

DOYLE 30 30 SALES

DENNIS 30 30 SALES


WARK 30 30 SALES

10 ACCOUNTING

LAST_NAME Department_ID Name

BAKER 40 OPERATIONS

10 ACCOUNTING


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SELECT employee_id, First_name, Last_name, Salary

FROM employee

WHERE last_name like ‘D%’;


7369 SMITH JOHN 667 7902 800 NULL 20

7499 ALLEN KEVIN 670 7698 1600 300 30

7505 DOYLE JEAN 671 7839 2850 NULL 30

7506 DENNIS LYNN 671 7839 2750 NULL 30


7521 WARK CYNTHIA 670 7698 1250 500 30

EMPL OYEE_ID FIRST_NAME LAST_NAME SALARY

-------------------- ------------------ ------------------- --------------

7505 JEAN DOYLE 2850

7506 LYNN DENNIS 2750

Lecture1059

lecture10: data manipulation in sql , advanced sql … · outer joins in oracle sql lecture10 21...

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