Everything about OOPs (Object-oriented programming) in this slide we cover the all details about object-oriented programming using C++. we also discussed why C++ is called a subset of C.
Everything about OOPs (Object-oriented programming) in this slide we cover the all details about object-oriented programming using C++. we also discussed why C++ is called a subset of C.
Introduction, The & and * operator, Declaration of pointer, Pointer to pointer, Pointer arithmetic, Pointer and array, Pointer with multidimensional array, Pointer and strings, Array of pointer with string, Dynamic memory allocation.
Introduction to control structure in C Programming Language include decision making (if statement, if..else statement, if...else if...else statement, nested if...else statement, switch...case statement), Loop(for loop, while loop, do while loop, nested loop) and using keyword(break, continue and goto)
Breadth First Search & Depth First SearchKevin Jadiya
The slides attached here describes how Breadth first search and Depth First Search technique is used in Traversing a graph/tree with Algorithm and simple code snippet.
Introduction, The & and * operator, Declaration of pointer, Pointer to pointer, Pointer arithmetic, Pointer and array, Pointer with multidimensional array, Pointer and strings, Array of pointer with string, Dynamic memory allocation.
Introduction to control structure in C Programming Language include decision making (if statement, if..else statement, if...else if...else statement, nested if...else statement, switch...case statement), Loop(for loop, while loop, do while loop, nested loop) and using keyword(break, continue and goto)
Breadth First Search & Depth First SearchKevin Jadiya
The slides attached here describes how Breadth first search and Depth First Search technique is used in Traversing a graph/tree with Algorithm and simple code snippet.
This presentation from 2008 is a good summary of Design by Contract and its application to PL/SQL as I have adopted and recommend others to try as well.
Builder.ai Founder Sachin Dev Duggal's Strategic Approach to Create an Innova...Ramesh Iyer
In today's fast-changing business world, Companies that adapt and embrace new ideas often need help to keep up with the competition. However, fostering a culture of innovation takes much work. It takes vision, leadership and willingness to take risks in the right proportion. Sachin Dev Duggal, co-founder of Builder.ai, has perfected the art of this balance, creating a company culture where creativity and growth are nurtured at each stage.
Connector Corner: Automate dynamic content and events by pushing a buttonDianaGray10
Here is something new! In our next Connector Corner webinar, we will demonstrate how you can use a single workflow to:
Create a campaign using Mailchimp with merge tags/fields
Send an interactive Slack channel message (using buttons)
Have the message received by managers and peers along with a test email for review
But there’s more:
In a second workflow supporting the same use case, you’ll see:
Your campaign sent to target colleagues for approval
If the “Approve” button is clicked, a Jira/Zendesk ticket is created for the marketing design team
But—if the “Reject” button is pushed, colleagues will be alerted via Slack message
Join us to learn more about this new, human-in-the-loop capability, brought to you by Integration Service connectors.
And...
Speakers:
Akshay Agnihotri, Product Manager
Charlie Greenberg, Host
Dev Dives: Train smarter, not harder – active learning and UiPath LLMs for do...UiPathCommunity
💥 Speed, accuracy, and scaling – discover the superpowers of GenAI in action with UiPath Document Understanding and Communications Mining™:
See how to accelerate model training and optimize model performance with active learning
Learn about the latest enhancements to out-of-the-box document processing – with little to no training required
Get an exclusive demo of the new family of UiPath LLMs – GenAI models specialized for processing different types of documents and messages
This is a hands-on session specifically designed for automation developers and AI enthusiasts seeking to enhance their knowledge in leveraging the latest intelligent document processing capabilities offered by UiPath.
Speakers:
👨🏫 Andras Palfi, Senior Product Manager, UiPath
👩🏫 Lenka Dulovicova, Product Program Manager, UiPath
Generating a custom Ruby SDK for your web service or Rails API using Smithyg2nightmarescribd
Have you ever wanted a Ruby client API to communicate with your web service? Smithy is a protocol-agnostic language for defining services and SDKs. Smithy Ruby is an implementation of Smithy that generates a Ruby SDK using a Smithy model. In this talk, we will explore Smithy and Smithy Ruby to learn how to generate custom feature-rich SDKs that can communicate with any web service, such as a Rails JSON API.
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Clients don’t know what they don’t know. What web solutions are right for them? How does WordPress come into the picture? How do you make sure you understand scope and timeline? What do you do if sometime changes?
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Epistemic Interaction - tuning interfaces to provide information for AI supportAlan Dix
Paper presented at SYNERGY workshop at AVI 2024, Genoa, Italy. 3rd June 2024
https://alandix.com/academic/papers/synergy2024-epistemic/
As machine learning integrates deeper into human-computer interactions, the concept of epistemic interaction emerges, aiming to refine these interactions to enhance system adaptability. This approach encourages minor, intentional adjustments in user behaviour to enrich the data available for system learning. This paper introduces epistemic interaction within the context of human-system communication, illustrating how deliberate interaction design can improve system understanding and adaptation. Through concrete examples, we demonstrate the potential of epistemic interaction to significantly advance human-computer interaction by leveraging intuitive human communication strategies to inform system design and functionality, offering a novel pathway for enriching user-system engagements.
Kubernetes & AI - Beauty and the Beast !?! @KCD Istanbul 2024Tobias Schneck
As AI technology is pushing into IT I was wondering myself, as an “infrastructure container kubernetes guy”, how get this fancy AI technology get managed from an infrastructure operational view? Is it possible to apply our lovely cloud native principals as well? What benefit’s both technologies could bring to each other?
Let me take this questions and provide you a short journey through existing deployment models and use cases for AI software. On practical examples, we discuss what cloud/on-premise strategy we may need for applying it to our own infrastructure to get it to work from an enterprise perspective. I want to give an overview about infrastructure requirements and technologies, what could be beneficial or limiting your AI use cases in an enterprise environment. An interactive Demo will give you some insides, what approaches I got already working for real.
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Neuro-symbolic (NeSy) AI is on the rise. However, simply machine learning on just any symbolic structure is not sufficient to really harvest the gains of NeSy. These will only be gained when the symbolic structures have an actual semantics. I give an operational definition of semantics as “predictable inference”.
All of this illustrated with link prediction over knowledge graphs, but the argument is general.
Encryption in Microsoft 365 - ExpertsLive Netherlands 2024Albert Hoitingh
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Welcome to UiPath Test Automation using UiPath Test Suite series part 4. In this session, we will cover Test Manager overview along with SAP heatmap.
The UiPath Test Manager overview with SAP heatmap webinar offers a concise yet comprehensive exploration of the role of a Test Manager within SAP environments, coupled with the utilization of heatmaps for effective testing strategies.
Participants will gain insights into the responsibilities, challenges, and best practices associated with test management in SAP projects. Additionally, the webinar delves into the significance of heatmaps as a visual aid for identifying testing priorities, areas of risk, and resource allocation within SAP landscapes. Through this session, attendees can expect to enhance their understanding of test management principles while learning practical approaches to optimize testing processes in SAP environments using heatmap visualization techniques
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1. Insights into SAP testing best practices
2. Heatmap utilization for testing
3. Optimization of testing processes
4. Demo
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Execution from the test manager
Orchestrator execution result
Defect reporting
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Deepak Rai, Automation Practice Lead, Boundaryless Group and UiPath MVP
GDG Cloud Southlake #33: Boule & Rebala: Effective AppSec in SDLC using Deplo...James Anderson
Effective Application Security in Software Delivery lifecycle using Deployment Firewall and DBOM
The modern software delivery process (or the CI/CD process) includes many tools, distributed teams, open-source code, and cloud platforms. Constant focus on speed to release software to market, along with the traditional slow and manual security checks has caused gaps in continuous security as an important piece in the software supply chain. Today organizations feel more susceptible to external and internal cyber threats due to the vast attack surface in their applications supply chain and the lack of end-to-end governance and risk management.
The software team must secure its software delivery process to avoid vulnerability and security breaches. This needs to be achieved with existing tool chains and without extensive rework of the delivery processes. This talk will present strategies and techniques for providing visibility into the true risk of the existing vulnerabilities, preventing the introduction of security issues in the software, resolving vulnerabilities in production environments quickly, and capturing the deployment bill of materials (DBOM).
Speakers:
Bob Boule
Robert Boule is a technology enthusiast with PASSION for technology and making things work along with a knack for helping others understand how things work. He comes with around 20 years of solution engineering experience in application security, software continuous delivery, and SaaS platforms. He is known for his dynamic presentations in CI/CD and application security integrated in software delivery lifecycle.
Gopinath Rebala
Gopinath Rebala is the CTO of OpsMx, where he has overall responsibility for the machine learning and data processing architectures for Secure Software Delivery. Gopi also has a strong connection with our customers, leading design and architecture for strategic implementations. Gopi is a frequent speaker and well-known leader in continuous delivery and integrating security into software delivery.
GDG Cloud Southlake #33: Boule & Rebala: Effective AppSec in SDLC using Deplo...
Plsql programs(encrypted)
1. http://placementkit.blogspot.com/
PL / SQL PROGRAMS
P1. WRITE A NESTED PROGRAM TO ADD AND TO MULTIPLY TWO NUMBERS.
DECLARE
N1 number;
N2 number;
Sum number ;
BEGIN
Sum := N1+N2 ;
<< inner_block >>
DECLARE
Prod number;
BEGIN
Prod := N1 * N2 ;
Dbms_output.put_line( Product Value = prod );
END inner_block ;
Dbms_output.put_line( Sum Value = sum);
END;
P2. WRITE A PROGRAM TO CALCULATE THE SIMPLE INTEREST AND COMPUND
INTEREST ,
IF P, N, R ARE GIVEN.
declare
p number(9,2) ;
n number(9,2) ;
r number(9,2) ;
si number(9,2) := 0;
ci number(9,2) := 0;
begin
p := &principal_amount;
n := &no_of_years;
r := &rate_of_interest;
si := p*n*r/100;
ci := p*(1+r/100)**n;
dbms_output.put_line('simple interset =' si);
dbms_output.put_line('compound interset =' ci);
end;
SQL> /
Enter value for principal_amount: 10000
old 8: p:=&principal_amount;
new 8: p:=10000;
Enter value for no_of_years: 5
old 9: n:=&no_of_years;
new 9: n:=5;
Enter value for rate_of_interest: 10.5
old 10: r:=&rate_of_interest;
new 10: r:=10.5;
simple interset =5250
compound interset =16474.47
PL/SQL procedure successfully completed.
PROGRAM BASED ON IF LOOP
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P3 . Write a program to check greatest of two numbers :
declare
a number(3) :=20;
b number(3) :=10;
begin
if a>b then
dbms_output.put_line('a is the greatest : ' a);
else
dbms_output.put_line('B is the greatest : ' b);
end if;
end;
SQL> /
a is the greatest : 20
PL/SQL procedure successfully completed.
P4. Given 2 sides of a rectangle .Write a program to find out its area is gr
eater than its perimeter or not .
declare
l number;
b number;
ar number;
pr number;
begin
l := &l;
b := &b;
ar := l*b;
pr := 2*(l+b);
if ar > pr then
dbms_output.put_line('the area iS > its perimeter' 'area = ' ar 'perimet
er = ' pr);
else
dbms_output.put_line('the area iS < its perimeter' ' area = ' ar ' perim
eter = ' pr);
end if;
end;
Enter value for l: 10
old 7: l:=&l;
new 7: l:=10;
Enter value for b: 6
old 8: b:=&b;
new 8: b:=6;
the area is > its perimeter area = 60 perimeter = 32
PL/SQL procedure successfully completed.
P5. WRITE A PROGRAM TO INPUT A SINGLE DIGIT NO: CONVERT IT INTO WORDS.
Declare
a number;
t varchar2(10);
begin
a :=&a;
if a=1 then
3. http://placementkit.blogspot.com/
t := 'one';
elsif a=2 then
t := 'two';
elsif a= 3 then
t := 'three';
elsif a= then
t := 'four';
elsif a=5 then
t := 'five';
elsif a=6 then
t := 'six';
elsif a=7 then
t := 'seven';
elsif a=8 then
t := 'eight';
elsif a=9 then
t := 'nine';
else
t := 'zero';
end if;
dbms_output.put_line(a '=' t);
end;
Enter value for a: 2
old 5: a:=&a;
new 5: a:=2;
2 = two
PL/SQL procedure successfully completed.
P6 . Write a program to check the given number is +ve or ve :
SQL>
declare
n number(2):=12;
begin
if n>0 then
dbms_output.put_line('the given number is positive' n);
else
dbms_output.put_line('the given number is negative' n);
end if;
end;
/
SQL> save posneg.sql
Created file posneg.sql
SQL >The given number is positive 12
PL/SQL procedure successfully completed.
PROGRAM BASED ON NESTED IF LOOP
P7. WRITE A PROGRAM TO INPUT 2 NUMBERS IF THE 1st No >2nd No THEN SWAP
IT, ELSE IF 1st No < 2nd No RAISE IT TO ITS POWER , ELSE DOUBLES IT.
declare
a number(5);
b number(5);
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t number(5);
begin
a := &a;
b := &b;
dbms_output.put_line( 'initial value of a = ' a 'b=' b)
if a>b
then
t:= a;
a:= b;
b:=t;
elsIF A<B
then
a:=a**a;
b:=b**b;
ELSE
a:=2*a;
b:=2*b;
end if;
dbms_output.put_line('final value of a = ' a ' b = ' b);
end;
SQL> /
Enter value for a: 4
old 6: a:=&a;
new 6: a:=4;
Enter value for b: 5
old 7: b:=&b;
new 7: b:=5;
initial value of a = 4 b = 5
final value of a = 256 b = 3125
PL/SQL procedure successfully completed.
SQL> /
Enter value for a: 5
old 6: a:=&a;
new 6: a:=5;
Enter value for b: 4
old 7: b:=&b;
new 7: b:=4;
initial value of a = 5 b = 4
final value of a = 4 b = 5
PL/SQL procedure successfully completed.
SQL> Enter value for a: 5
old 6: a:=&a;
new 6: a:=5;
Enter value for b: 5
old 7: b:=&b;
new 7: b:=5;
initial value of a = 5 b = 5
final value of a = 10 b = 10
PL/SQL procedure successfully completed.
P8. A bank accepts fixed deposits for one or more years and the policy it ad
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opts on interest is as follows:
If a deposit is < Rs 2000 and for 2 or more years , the interest rat
e is 5% compounded annually.
If a deposit is Rs.2000 or more but less than 6000 and for 2 or more
years , the interest is 7 % compounded annually.
If a deposit is Rs.6000 and for 1 or more years , the interest is 8
% compounded annually.
On all deposits for 5 years or more , interest is 10% compounded ann
ually.
On all other deposits not covered by above conditions , the interest
is 3% compounded annually.
Given the amount deposited and the number of years , Write a program to calc
ulate the amount on maturity.
declare
p number(9,2);
r number(9,2);
t number(9,2);
ci number(9,2);
begin
p := &p;
t := &t;
if p<2000 and t>=2 then
r := 5;
elsif p>=2000 and p<6000 and t>=2 then
r := 7;
elsif p>6000 and t>=1 then
r := 8;
elsif t>=5 then
r := 10;
else
r := 3;
end if;
ci := p*(1+r/100)**t - p;
dbms_output.put_line('The Principal amount is =' p);
dbms_output.put_line('The rate of interest is =' r);
dbms_output.put_line('The time period is =' t);
dbms_output.put_line('The compund interst is =' ci);
end;
Enter value for p: 10000
old 7: p:=&p;
new 7: p:=10000;
Enter value for t: 5
old 8: t:=&t;
new 8: t:=5;
The Principal amount is =10000
The rate of interest is =8
The time period is =5
The compund interst is =4693.28
PL/SQL procedure successfully completed.
Enter value for p: 1500
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old 7: p:=&p;
new 7: p:=1500;
Enter value for t: 3
old 8: t:=&t;
new 8: t:=3;
The Principal amount is =1500
The rate of interest is =5
The time period is =3
The compound interest is =236.44
PL/SQL procedure successfully completed.
P9. WRITE A PROGRM TO INPUT 3 NUMBERS FIND THE 1st Greatest, 2nd Greatest,
3rd Greatest.
declare
a number(3);
b number(3);
c number(3);
f number(3);
s number(3);
t number(3);
begin
a :=&a;
b :=&b;
c :=&c;
if a>b and a>c then
f:= a;
if b>c then
s:=b;
t:=c;
else
s:=c;
t:=b;
end if;
elsif b>a and b>c then
f:= b;
if a>c then
s:=a;
t:=c;
else
s:=c;
t:=a;
end if;
else
f:= c;
if a>b then
s:=a;
t:=b;
else
s:=b;
t:=a;
end if;
end if;
dbms_output.put_line('first largest = ' f);
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dbms_output.put_line('second largest = ' s);
dbms_output.put_line('third largest = ' t);
end;
/
Enter value for a: 5
old 9: a:=&a;
new 9: a:=5;
Enter value for b: 8
old 10: b:=&b;
new 10: b:=8;
Enter value for c: 9
old 11: c:=&c;
new 11: c:=9;
first largest = 9
second largest = 8
third largest = 5
PL/SQL procedure successfully completed.
P10 . WRITE A PROGRAM TO INPUT 2 NUMBERS AND AN OPERATOR , AND
DISPLAY THE RESULT.
declare
a number(3) ;
b number(3) ;
c number(3) ;
op char(1) ;
begin
a := &a ;
b := &b ;
op := &op ;
if op='+'
then
c:=a+b;
elsif op='-'
then
c:=a-b;
elsif op='*'
then
c:=a*b;
else
c:=a/b;
end if;
dbms_output.put_line('result=' c);
end;
Enter value for a: 5
old 7: a:=&a;
new 7: a:=5;
Enter value for b: 6
old 8: b:=&b;
new 8: b:=6;
Enter value for op: '*'
old 9: op:=&op;
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new 9: op:='*';
result=30
PL/SQL procedure successfully completed.
P11. Write a program to calculate the commission of the sales man.
If salesmade Comm
>10000 500
10000 20000 1000
>20000 1500
declare
sman varchar(10);
sm number(9,2);
com number(9,2);
begin
sman := &sman;
sm := &sm;
if sm > 10000 then
com := 500;
elsif sm > 20000 then
com := 1000;
else
com := 1500;
end if;
dbms_output.put_line(' the sales man name is :' sman);
dbms_output.put_line(' the sales made is :' sm);
dbms_output.put_line(' the sales commission is :' com);
end;
Enter value for sman: 'Mathew'
old 6: sman:=&sman;
new 6: sman:='Mathew';
Enter value for sm: 15000
old 7: sm:=&sm;
new 7: sm:=15000;
The sales man name is :Mathew
The sales made is :15000
The sales commission is :500
PL/SQL procedure successfully completed.
PROGRAM BASED ON WHILE LOOP
P12. TO GENERATE NUMBERS FROM 0 TO 25 IN STEP OF 5
declare
i number :=10 ;
begin
dbms_output.put_line(' THE WHILE LOOP begins');
WHILE I<=25 LOOP
dbms_output.put_line(to_char(i));
i:=i+5;
end loop;
End;
THE WHILE LOOP BEGINS
10
15
9. http://placementkit.blogspot.com/
20
25
PL/SQL procedure successfully completed.
P13. Write a program to find the sum of the digits of the number:
DECLARE
N number ;
S NUMBER :=0;
R NUMBER;
begin
n:=&N;
WHILE N<>0 LOOP
R := MOD(N,10);
S := S + R;
N := TRUNC(N/10);
end loop;
dbms_output.put_line('THE SUM OF THE DIGITS = ' S);
end;
SQL> Enter value for n: 375
old 7: n:=&N;
new 7: n:=375;
THE SUM OF THE DIGITS = 15
PL/SQL procedure successfully completed.
PROGRAM BASED ON NESTED WHILE LOOP
PROGRAM BASED ON FOR LOOP
P14. WRITE A PROGRAM CODE TO PRINT THE MULTIPLICATION TABLE OF A GIVEN
NO:
declare
t number(3) := 3;
begin
T := &T;
FOR I IN 1..3 LOOP
dbms_output.put_line(t 'X' i '=' i*t );
end loop;
end;
p15. write aprogram to generate even numbers from 2 to 50, and find its sum.
declare
i number(5);
n number(5);
v number(5);
s number(5):=0;
begin
n := &terminal_number;
for i in 1 .. n/2 loop
v := i*2;
s := s+v;
dbms_output.put_line(v);
end loop;
dbms_output.put_line('the sum of numbers from 2 to ' n ' = ' s);
end;
SQL> /
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Enter value for terminal_number: 10
old 7: n:=&terminal_number;
new 7: n:=10;
2
4
6
8
10
The sum of numbers from 2 to 10 = 30
PL/SQL procedure successfully completed.
P16 . WRITE A PROGRAM TO GENERATE FIRST 25 TERMS OF THE FIBONACCIS
SERIES.
declare
a number:= 0 ;
b number:= 1;
c number;
begin
dbms_output.put(a ' ' b ' ');
for i in 3..10 loop
c := a + b;
dbms_output.put(c ' ');
a := b;
b := c;
end loop;
dbms_output.put_line(' ');
end;
0 1 1 2 3 5 8 13 21 34
PL/SQL procedure successfully completed.
P17. Write a program to find the factorial of a number :
declare
n number(2);
i number(2);
f number(5):=1;
begin
n :=&n;
for i in 1..n loop
f := f * i;
end loop;
dbms_output.put_line(' the factorial value = ' f);
end;
Enter value for n: 5
old 6: n:=&n;
new 6: n:=5;
the factorial value = 120
PROGRAM BASED ON NESTED FOR LOOP
P18. Write a program to print the following design:
11. http://placementkit.blogspot.com/
1
12
123
1234
12345
declare
i number ;
j number;
n number;
begin
n :=&n;
for i in 1..n loop
for j in 1..i loop
dbms_output.put(j);
end loop;
dbms_output.put_line(' ');
end loop;
end;
Enter value for n: 5
old 6: n:=&n;
new 6: n:=5;
P19. WRITE A PROGRAM TO DISPLAY NUMBERS OF THE FORM
00000
12345
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
DECLARE
I NUMBER;
J NUMBER ;
K NUMBER;
BEGIN
FOR I IN 0 .. 5 LOOP
FOR J IN 1..5 LOOP
K := I*J;
DBMS_OUTPUT.PUT(K);
END LOOP;
DBMS_OUTPUT. PUT_LINE (' ');
END LOOP;
END;
P20. WRITE A PL/SQL CODE TO ACCEPT THE TEXT AND REVERSE THE GIVEN TEXT.
CHECK THE TEXT IS PALINDROME OR NOT
DECLARE
G VARchar2(20);
r VARchar2(20);
BEGIN
G:='&g';
dbms_output.put_line('THE GIVEN TEXT :' G);
for i in REVERSE 1.. length(G) loop
R:= R substr(G,i,1);
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end loop;
dbms_output.put_line('THE REVERSED TEXT :' R);
IF R=G THEN
dbms_output.put_line('THE GIVEN TEXT IS PALINDROME ');
ELSE
dbms_output.put_line('THE GIVEN TEXT IS NOT PALINDROME ');
END IF;
end;
SQL> /
Enter value for g: MALAYALAM
old 5: G:='&g';
new 5: G:='MALAYALAM';
THE GIVEN TEXT :MALAYALAM
THE REVERSED TEXT :MALAYALAM
THE GIVEN TEXT IS PALINDROME
PL/SQL procedure successfully completed.
/
Enter value for g: HELLO
old 5: G:='&g';
new 5: G:='HELLO';
THE GIVEN TEXT :HELLO
THE REVERSED TEXT :OLLEH
THE GIVEN TEXT IS NOT PALINDROME
PL/SQL procedure successfully completed.
P21. Write a program to print the following design:
declare
i number ;
j number;
n number;
k number;
m number;
begin
n := &n;
for i in 1..n loop
for j in 1..n-i loop
dbms_output.put('~');
end loop;
for k in 1..i loop
dbms_output.put(k);
end loop;
for k in reverse 1..i-1 loop
dbms_output.put(k);
end loop;
dbms_output.put_line(' ');
end loop;
end;
Enter value for n: 5
old 8: n:=&n;
new 8: n:=5;
~~~~1
~~~121
~~12321
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~1234321
123454321
PL/SQL procedure successfully completed.
P22. Write a program to print the following design :
declare
i number ;
j number;
n number;
k number;
m number;
begin
n := &n;
for i in 1..n loop
for j in 1..n-i loop
dbms_output.put('~');
end loop;
for k in 1..i loop
dbms_output.put(k);
end loop;
for k in reverse 1..i-1 loop
dbms_output.put(k);
end loop;
dbms_output.put_line(' ');
end loop;
for i in reverse 1..n-1 loop
for j in 1..n-i loop
dbms_output.put('~');
end loop;
for k in 1..i loop
dbms_output.put(k);
end loop;
for k in reverse 1. . i-1 loop
dbms_output.put(k);
end loop;
dbms_output.put_line(' ');
end loop;
end;
Enter value for n: 5
old 8: n:=&n;
new 8: n:=5;
~~~~1
~~~121
~~12321
~1234321
123454321
~1234321
~~12321
~~~121
~~~~1
PL/SQL procedure successfully completed.
PROGRAM BASED ON FOR & IF
P23. GENERATE ODD NOS: FROM 1 TO 10 AND FIND ITS SUM.
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DECLARE
I NUMBER(4);
S NUMBER(4):=0;
BEGIN
FOR I IN 1..10 LOOP
IF MOD(I,2) <> 0 THEN
S := S+I;
DBMS_OUTPUT.PUT_LINE(I);
END IF;
END LOOP;
DBMS_OUTPUT.PUT_LINE(' THE SUM OF ODD NOS FROM 1 TO 10 = ' S);
END;
SQL>/
1
3
5
7
9
THE SUM OF ODD NOS FROM 1 TO 10 = 25
PL/SQL procedure successfully completed.
P24 . WRITE A PROGRAM TO ALL ODD NUMBERS FORM 10 TO 1 IN REVERSE
ORDER.
declare
i number(5);
n number(5);
v number(5);
s number(5) :=0;
begin
S := &STARTING_NUMBER;
n := &terminal_number;
for i in REVERSE S .. N loop
IF MOD (I,2) <>0 THEN
s := s + I;
dbms_output.put_line(I);
end loop;
dbms_output.put_line('the sum of numbers from 2 to ' n ' = ' s);
end;
Enter value for starting number : 1
Old 6: s=&starting_number:1
New 6: s:=1;
Enter value for Terminal_number: 10
old 7: n := &terminal_number;
new 7: n:=10;
9
7
5
3
1
The sum of numbers from 1 to 9 = 25
PL / SQL procedure successfully completed.
P25. Write a program to check the given no: is prime or not:
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declare
n number;
i number;
pr number(2):=1;
begin
n:= &n;
for i in 2..n/2 loop
if mod(n,i) = 0 then
pr := 0;
end if;
end loop;
if pr = 1 then
dbms_output.put_line(' the given no: is prime: ' n);
else
dbms_output.put_line(' the given no: is not prime: ' n);
end if;
end;
Enter value for n: 7
old 6: n:=&n;
new 6: n:=7;
the given no: is prime: 7
PL/SQL procedure successfully completed.
SQL> /
Enter value for n: 25
old 6: n:=&n;
new 6: n:=25;
the given no: is not prime: 25
PL/SQL procedure successfully completed.
P26. WRITE A PROGRAM TO PRINT ASCII TABLE :
DECLARE
I NUMBER;
BEGIN
FOR I IN 33..256 LOOP
DBMS_OUTPUT.PUT (( TO_CHAR (I,'000')) ':' CHR(I) ' ' );
IF MOD (I, 8) = 0 THEN
DBMS_OUTPUT.PUT_LINE(' ');
END IF;
END LOOP;
END;
033:! 034:" 035:# 036:$ 037:% 038:& 039:' 040:(
041:) 042:* 043:+ 044:, 045:- 046:. 047:/ 048:0
049:1 050:2 051:3 052:4 053:5 054:6 055:7 056:8
249:ù 250:ú 251:û 252:ü 253:ý 254:þ 255:ÿ 256:
:: :: :: :: :: ::
:: :: :: :: :: ::
PL/SQL procedure successfully completed.
P27. Write a program to generate prime nos from 1 to 10:
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declare
n number;
i number;
pr number;
begin
for n in 1..10 loop
pr:=1;
for i in 2 .. n/2 loop
if mod(n,i) = 0 then
pr := 0;
end if;
end loop;
if pr = 1 then
dbms_output.put_line(n);
end if;
end loop;
end;
1
2
3
5
7
P28 . Write a program to check the square root of a number is prime or not.
declare
n number;
i number;
pr number;
begin
pr := 1;
n := &n;
n := sqrt(n);
for i in 2 .. n/2 loop
if mod(n,i) = 0 then
pr := 0;
end if;
end loop;
if pr = 1 then
dbms_output.put_line('the square root of the given number is prime' n*n);
else
dbms_output.put_line('the square root of the given number is not prime' n*n
);
end if;
end;
SQL> Enter value for n: 81
old 7: n:=&n;
new 7: n:=81;
The square root of the given number is not prime81
PL/SQL procedure successfully completed.
PROGRAM BASED ON LOOP... END LOOP
P29. Write a program to reverse the digits of the number:
17. http://placementkit.blogspot.com/
DECLARE
N number ;
S NUMBER : = 0;
R NUMBER;
K number;
begin
N := &N;
K := N;
LOOP
EXIT WHEN N = 0 ;
S := S * 10;
R := MOD(N,10);
S := S + R;
N := TRUNC(N/10);
end loop;
dbms_output.put_line( ' THE REVERSED DIGITS ' ' OF ' K '=' S);
end;
Enter value for n: 4567
old 7: N:=&N;
new 7: N:=4567;
THE REVERSED DIGITS OF 4567 = 7654
PL/SQL procedure successfully completed.
PROGRAM BASED ON RECORDS
P30. WRITE PL/SQL SCRIPT TO CREATE A RECORD TYPE THAT CAN HOLD SALES IN
4 DIFFERENT QUARTERS.
Structure of the table :
SALES ( SNO, NAME, QUAD1, QUAD2, QUAD3, QUAD4 )
SALES2004 ( Q1, Q2, Q3, Q4 )
DECLARE
TYPE SALEREC IS RECORD
( Q1 NUMBER,
Q2 NUMBER,
Q3 NUMBER,
Q4 NUMBER,
);
BEGIN
SREC SALEREC ;
SELECT SUM(QUAD1) , SUM(QUAD2) , SUM(QUAD3), SUM(QUAD4) INTO SREC FROM
SALES
;
INSERT INTO SALES2004 VALUES (SREC.Q1 ,SREC.Q2, SREC.Q3, SREC.Q4);
END;
PROGRAM BASED ON DATABASE INTERACTION
P31 . Write PL/SQL SCRIPT to determine the salary of employee in the EMP tab
le. If salary >7500 give an increment of 15% , otherwise display the message NO
INCREMENT . Get the empno from the user.
DECLARE
ENO EMP.EMPNO%TYPE;
SALARY EMP.SAL%TYPE;
BEGIN
ENO := &ENO;
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SELECT SAL INTO SALARY FROM EMP WHERE EMPNO=ENO;
IF SALARY >7500 THEN
UPDATE EMP SET SAL =SAL+SAL* 15/100 WHERE EMPNO=ENO;
ELSE
DBMS_OUTPUT.PUT_LINE(NO INCREMENT)
END IF;
END;
P32 . GIVEN A TABLE STUDENT ( SID INTEGER PRIMARY KEY, NAME CHAR(30),
AGE INTEGER , GPA FLOAT ). WRITE A PL/SQL SCRIPT TO GO THROUGH SID 142-
857 AND SET ALL GPA UNDER 4.0 TO 4.0 .
DECLARE
TSID STUDENT.SID%TYPE;
TGPA STUDENT.GPA%TYPE;
BEGIN
TSID:=142;
LOOP
EXIT WHEN TSID > 857 ;
SELECT GPA INTO TGPA FROM STUDENT WHERE SID=TSID;
IF TGPA< 4.0 THEN
UPDATE STUDENT SET GPA=4.0 WHERE SID=TSID;
END IF;
TSID:=TSID+1;
END LOOP;
END;
P33 . DISPLAY AVERAGE SALARY , TOTOL NO: OF EMP, AND TOTAL SALARY FOR
FIRST 5 DEPARTMENTS IN EMP TABLE. DEPT NO ARE 10,20,30,40,5090.
DECLARE
DNO NUMBER:=10;
DPNAME DEPT. DNAME%TYPE;
ASAL EMP. SAL%TYPE;
ACOUNT NUMBER;
SSAL EMP. SAL%TYPE;
BEGIN
LOOP
EXIT WHEN DNO > 50;
SELECT DNAME INTO DPNAME FROM DEPT WHERE DEPTNO=DNO;
SELECT AVG(SAL), COUNT(*), SUM(SAL) INTO ASAL, ACOUNT,SSAL
FROM EMP WHERE DEPTNO = DNO;
DBMS_OUTPUT.PUT_LINE( DEPARTMENT NAME DPNAME);
DBMS_OUTPUT.PUT_LINE( DEPARTMENT COUNT ACOUNT);
DBMS_OUTPUT.PUT_LINE( AVERAGE SALARY ASAL);
DBMS_OUTPUT.PUT_LINE( SUM SALARY SSAL);
DNO := DNO+10;
END LOOP;
END;
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P34. WRITE A PL/SQL SCRIPT TO COUNT THE NO: OF EMPLOYEES. IF TOTAL <10
DISPLAY SMALL COMPANY, ELSE IF TOTAL BETWEEN 10-50 DISPLAY MEDIUM
ELSE DISPLAY LARGE
.
DECLARE
TOT NUMBER (3);
BEGIN
SELECT COUNT(*) INTO TOT FROM EMP;
IF TOT <10 THEN
TEXT := SMALL ;
ELSIF TOT<50 THEN
TEXT := MEDIUM;
ELSE
TEXT := LARGE;
END IF;
DBMS_OUTPUT.PUT_LINE( TEXT COMPANY);
END;
P35. Find the employee drawing minimum salary. Add his details like empno, n
ame, sal into the table newsal, after incrementing Rs.700.
DECLARE
ENO EMP.EMPNO%TYPE;
NAME EMP.ENAME%TYPE;
SALARY EMP.SAL%TYPE;
BEGIN
SELECT EMPNO, ENAME, SAL INTO ENO, NAME, SALARY
FROM EMP WHERE SAL = (SELECT MIN(SAL) FROM EMP);
SALARY := SALARY+700;
INSERT INTO NEWSAL VALUES ( ENO,NAME,SALARY);
END;
P36. Write a program to update the salary of the employee by 25% if he earns
salary >4000, otherwise if salary <4000 update by 10%, else update by 15%.
DECLARE
S NUMBER(8,2) ;
NAME VARCHAR2(10);
BEGIN
NAME:=&NAME;
SELECT SAL INTO S FROM EMP WHERE ENAME = NAME;
IF S >4000 THEN
UPDATE EMP SET SAL := SAL+SAL*20/100 WHERE ENAME=NAME;
ELSIF S<4000
UPDATE EMP SET SAL := SAL+SAL*10/100 WHERE ENAME=NAME;
ELSE
UPDATE EMP SET SAL := SAL+SAL*15/100 WHERE ENAME=NAME;
END IF;
END;
P37 .Write PL/SQL code to increase the sal of an employee by 5% whose salary
is more than 4000 . Get the empno from the user.
DECLARE
ENO NUMBER(4);
BEGIN
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ENO := &ENO;
UPDATE EMP SET SAL =SAL+SAL*5/100 WHERE SAL>4000 AND EMPNO=ENO;
END;
P38. GIVEN A TABLE TEMP ( SNO NUMBER(3) , DNO NUMBER,TEXT CHAR (4) .
WRITE A
PL/SQL SCRIPT TO INSERT 10 RECORDS IN TABLE TEMP AS PER THE FOLLOWING
SPECIFICA
TIONS:
SNO DNO TEXT
1 50 ODD
2 100 EVEN
3 150 ODD
DECLARE
SNO NUMBER := 1;
DNO NUMBER ;
TEXT CHAR(5) ;
BEGIN
LOOP
EXIT WHEN SNO > 10;
DNO := SNO * 50;
IF MOD ( SNO, 2 ) = 0
TEXT := EVEN;
ELSE
TEXT := ODD;
INSERT INTO TEMP VALUES (SNO,DNO,TEXT);
SNO := SNO + 1;
END LOOP;
COMMIT;
END;
P39 . Write PL/SQL code to DELETE the record of an employee whose salary is
more than 4000 . Get the empno from the user.
DECLARE
ENO NUMBER(4);
BEGIN
ENO := &ENO;
DELETE FROM EMP WHERE SAL>4000 AND EMPNO=ENO;
END;
P40 .Write PL/SQL code to insert a new record in the table emp after obtaining
values ( empno, ename, hiredate, sal ) from user.
declare
tempno number(4);
tename varchar(10);
thiredate varchar2(12);
tsal number(7,2);
begin
tempno :=&tempno;
tename :='&tename';
thiredate :='&thiredate';
tsal := &tsal;
insert into emp (empno,ename,hiredate,sal) values(tempno,tename,to_date(thir
21. http://placementkit.blogspot.com/
edate,'DD-mon-yyyy'),tsal);
end;
Enter value for tempno: 8000
old 7: tempno:=&tempno;
new 7: tempno:=8000;
Enter value for tename: mathew
old 8: tename :='&tename';
new 8: tename :='mathew';
Enter value for thiredate: 10-jan-2004
old 9: thiredate :='&thiredate';
new 9: thiredate :='10-jan-2004';
Enter value for tsal: 8000
old 10: tsal := &tsal;
new 10: tsal := 8000;
PL/SQL procedure successfully completed.
P41. WRITE A PROGRAM TO CREATE A EMP %ROWTYPE RECORD .ACCEPT THE
EMPNO FROM
THE USER, AND DISPLAY ALL THE INFORMATION ABOUT THE EMPLOYEE.
declare
erec emp%rowtype;
eno emp.empno%type;
begin
eno := &eno;
select * into erec from emp where empno=eno;
dbms_output.put_line( 'Emp no : ' erec.empno) ;
dbms_output.put_line( 'Name : ' erec.ename) ;
dbms_output.put_line( 'Salary : ' erec.sal);
dbms_output.put_line( 'Deptno : ' erec.deptno);
dbms_output.put_line( 'hiredate: ' erec.hiredate);
dbms_output.put_line( 'job : ' erec.job);
end;
SQL> /
Enter value for eno: 7788
old 5: eno := &eno;
new 5: eno := 7788;
Emp no : 7788
Name : SCOTT
Salary : 1452
Deptno : 20
hiredate: 19-APR-87
job : ANALYST
PL/SQL procedure successfully completed.
PROGRAM BASED ON EXCEPTION
P42. Write PL/SQL script that traps an invalid data type value given and displays a
custom error message.
DECLARE
ENO EMP. EMPNO%TYPE;
SNO VARCHAR2(5);
NAME EMP.ENMAE%TYPE;
BEGIN
SNO := &SNO ;
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SELECT EMPNO,ENAME INTO ENO,NAME FROM EMP WHERE EMPNO=SNO;
DBMS_OUTPUT.PUT_LINE ( NAME ENAME EMPNO EMPNO);
EXCEPTION
WHEN INVALID_NUMBER THEN
DBMS_OUTPUT.PUT_LINE(SNO IS INVALID DATA FOR EMPLOYEE ID);
END;
PROGRAM BASED ON FUNCTIONS
P43 . write a function to create factorial of a number.
CREATE OR REPLACE FUNCTION FACT (N NUMBER)
RETURN NUMBER
IS
I NUMBER(10);
F NUMBER :=1;
BEGIN
FOR I IN 1.. N LOOP
F:= F*I;
END LOOP;
RETURN F;
END;
SQL> /
Function created.
SQL> select fact(8) from dual;
FACT(8)
---------
40320
PROGRAM BASED ON PROCEDURES
P44. Write a Procedure to increase the salary for all the employees in the EMP
table :
SQL> select * from emp;
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
--------- ---------- --------- --------- --------- --------- --------- -----
----
7369 SMITH CLERK 7902 17-DEC-80 800 20
7499 ALLEN SALESMAN 7698 20-FEB-81 1600 300 30
14 rows selected.
SQL> create or replace procedure inc(i number)
is
begin
update emp set sal =sal+i;
end;
/
Procedure created.
To execute the procedures in the PL/SQL block:
SQL> declare
begin
inc(100);
end;
/
PL/SQL procedure successfully completed.
SQL> select * from emp;
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
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7369 SMITH CLERK 7902 17-DEC-80 800 20
7499 ALLEN SALESMAN 7698 20-FEB-81 1700 300 30
14 rows selected.
P45. WRITE A PROCEDURE TO INCREASE THE SALARY FOR THE SPECIFIED
EMPLOLEE USING EMPNO IN THE EMP TABLE BASED ON THE FOLLOWING
CRITERIA: INCREASE THE SALARY BY 5% FOR CLERKS, 7% FOR SALESMAN , 10%
FOR ANALYST, 20 % FOR MANAGER and 25% FOR PRESIDENT. ACTIVATE USING
PL/SQL BLOCK.
SQL> SELECT * FROM EMP;
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPT
7369 SMITH CLERK 7902 17-DEC-80 800 20
CREATE OR REPLACE PROCEDURE DESIGNATION(ENO NUMBER)
IS
BEGIN
UPDATE EMP SET SAL=SAL+SAL*5/100 WHERE JOB ='CLERK' AND EMPNO=ENO;
UPDATE EMP SET SAL=SAL+SAL*7/100 WHERE JOB='SALESMAN' AND EMPNO=ENO;
UPDATE EMP SET SAL=SAL+SAL*10/100 WHERE JOB='ANALYST' AND EMPNO=ENO;
UPDATE EMP SET SAL=SAL+SAL*20/100 WHERE JOB='MANAGER' AND EMPNO=ENO;
UPDATE EMP SET SAL=SAL+SAL*25/100 WHERE JOB='PRESIDENT' AND EMPNO=ENO;
END;
SQL> /
Procedure created.
SQL> DECLARE
2 BEGIN
3 DESIGNATION(7369);
4 END;
5/
PL/SQL procedure successfully completed.
SQL> SELECT * FROM EMP;
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPT
7369 SMITH CLERK 7902 17-DEC-80 840 20__