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tutorialcup · 4 years ago

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Polymorphism in Java

Polymorphism in Java is another important feature of OOPs concept. We have seen an overview of polymorphism in the tutorial OOPs concepts in Java. In this tutorial, we will understand in detail about polymorphism and its different types. We will cover about Static Polymorphism, Dynamic Polymorphism, Runtime Polymorphism.

What is polymorphism in Java

Polymorphism in Java as the name suggests means the ability to take multiple forms. It is derived from the Greek words where Poly means many and morph means forms. In Java, polymorphism in java is that the same method can be implemented in different ways. To understand this, we need to have an idea of inheritance in java as well which we have learned in the previous tutorial.

Types of Polymorphism in Java

Below are the different types of polymorphism in java.

Static Polymorphism

When we overload a static method, we call it as static polymorphism. Since it resolves the polymorphism during compile time, we can call also name it as compile-time polymorphism. We can implement this type of polymorphism in java using either method overloading or operator overloading. In static polymorphism, during compile time it identifies which method to call based on the parameters we pass. Let's understand this in detail with the help of examples. Method overloading When there are many methods with the same name but different implementations, we call it a method overloading. We can implement method overloading in two different ways: - Different number of parameters - Different types of parameters. First, let's look at an example of method overloading with a different number of parameters. In this example, we have created a class Area that has 2 methods computeArea(int) and computeArea(int, int). The difference in both the method is one has 1 parameter and the other has 2 parameters even though the method name is the same. So when we call the method using 1 parameter, it calls computeArea(int) and when we call the method with 2 parameters, it calls computeArea(int, int).

class Area { public void computeArea(int a) { int area = a *a; System.out.println("Area of square: " + area); } public void computeArea(int a, int b) { int area = a*b; System.out.println("Area of rectangle: " + area); } } public class CalculateArea { public static void main(String args) { Area ar = new Area(); ar.computeArea(5); ar.computeArea(4, 2); } } Area of square: 25 Area of rectangle: 8 Below is an example of method overloading where the same method can have different types of parameters. In the below code, we have a method subtract with 2 different types of parameters. One has parameters of type int and the other of type double. Now, when we call the method using the class object, based on the type of data, it calls the corresponding method. Hence, the first method calls subtract(int, int) since we are passing integer numbers. The second method calls subtract(double, double) since we are passing double numbers.

class Difference { public void subtract(int a, int b) { int diff = a - b; System.out.println("Difference: " + diff); } public void subtract(double a, double b) { double diff = a - b; System.out.println("Difference: " + diff); } } public class MethodOverloadingDemo { public static void main(String args) { Difference d = new Difference(); d.subtract(10, 5); d.subtract(5.51, 2.21); } } Difference: 5 Difference: 3.3 Operator overloading Similar to method overloading, we can also override an operator which is also an example of static polymorphism in java. But we can overload only the "+" operator in Java and do not support the overloading of other operators. When we use String as operands between + it results in String concatenation. When we use numbers as operands between + it results in the addition of two numbers. Below is an example of + operator overloading. class Operatordemo { public void operator(String s1, String s2) { String s = s1+s2; System.out.println("String concatenation: " + s); } public void operator(int x, int y) { int sum = x + y; System.out.println("Sum: "+ sum); } } public class OperatorOverloading { public static void main(String args) { Operatordemo o = new Operatordemo(); o.operator("Good", "Evening"); o.operator(4, 9); } } String concatenation: GoodEvening Sum: 13

Dynamic Polymorphism

When polymorphism resolves during runtime, we call it dynamic or runtime polymorphism. In this type of polymorphism in java, it uses the reference variable of the superclass to call the overridden method. This means, based on the object that is referred by the reference variable, it calls the overridden method of that corresponding class. We can use the method overriding concept to implement Dynamic polymorphism. Method overriding When the subclass has the same method of the base class, we call it a method overriding which means that the subclass has overridden the base class method. Based on the type of object we create, it calls the method of that corresponding class. This means, if we create an object of the superclass and refer it using the subclass, then it calls the method of the subclass. Since it computes this during runtime, we call it as runtime polymorphism in java. You will be able to understand this concept clearly with the below example. We have created a parent class Vehicle and 2 subclasses Bike and Car. The parent class has a method speed and both the subclasses have overridden the base class method speed. If we create an object of instance Vehicle class and call the speed method (v.speed) it calls the parent class method. When we call the speed method using object created of instance Bike class(b.speed), it calls the Bike class method. Similarly, when we call the method using the object created of instance Car class(c.speed), it calls the Car class method.

class Vehicle { public void speed() { System.out.println("Default speed"); } } class Bike extends Vehicle { public void speed() { System.out.println("Bike speed"); } } class Car extends Vehicle { public void speed() { System.out.println("Car speed"); } } public class VehicleType { public static void main(String args) { //Create an instance of Vehicle Vehicle v = new Vehicle(); v.speed(); //Create an instance of Bike Vehicle b = new Bike(); b.speed(); //Create an instance of Car Vehicle c = new Car(); c.speed(); } } Default speed Bike speed Car speed Below is another example of runtime polymorphism where we can use the same object name to create multiple class instances. We first declare an object f of class Fruits. Then using the new keyword we instantiate the object variable for the class Fruits. When we use this object to call the taste method, it calls the taste method of Fruits class. Next, when we instantiate the object variable for the class Apple and call the taste method using this object, it calls the method of Apple class. Similarly, it happens for the Pineapple class. class Fruits { public void taste() { System.out.println("Fruits taste"); } } class Apple extends Fruits { public void taste() { System.out.println("Sweet taste"); } } class Pineapple extends Fruits { public void taste() { System.out.println("Sour taste"); } } public class FruitsDemo { public static void main(String args) { Fruits f; f = new Fruits(); f.taste(); f = new Apple(); f.taste(); f = new Pineapple(); f.taste(); } } Fruits taste Sweet taste Sour taste Let's consider another example below where we call the method of a subclass which does not have an overridden method. In such a case, it calls the parents class method. AnimalDemo class does not have the method animaltype. Hence it calls method of Cow class since AnimalDemo class extends the Cow class. class Animal { public void animaltype() { System.out.println("Animal"); } } class Cow extends Animal { public void animaltype() { System.out.println("Herbivorous"); } } public class AnimalDemo extends Cow{ public static void main(String args) { Animal a; a = new AnimalDemo(); a.animaltype(); } } Herbivorous Runtime polymorphism with data member In runtime polymorphism, only the method is overridden and not the data member or variable. From the below example, we can clearly understand this difference. When we create an object b of instance SBI class from the Bank class, it calls the superclass variable since data members are not overridden. In order to access the subclass variable, we need to create the object from the subclass i.e SBI s = new SBI(). class Bank { public double interestRate = 7.5; } class SBI extends Bank { public double interestRate = 6.4; } public class BankDemo { public static void main(String args) { Bank b = new SBI(); System.out.println(b.interestRate); SBI s = new SBI(); System.out.println(s.interestRate); } } 7.5 6.4

Difference between Method overloading and Method overriding

Reference Read the full article

#Java

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blogkdmi-blog · 6 years ago

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Global Transmission Repair Market Size, Share, Growth, Trends and Forecast 2018-2023

KD Market Insights provides a forecast for Global Transmission Repair market for a span of 6 years i.e. between 2018 and 2023. This report presents an overview on Transmission Repair market and technologies used in it such as Passenger Car, Light Commercial Vehicle, Heavy Commercial Vehicle used for various Repair Type segments such as Transmission General Repair, Transmission Overhaul. Transmission Repair analyzed in this report include Components such as Gaskets and Seals, Fluids, Transmission Filters, O-Rings, Axles, Flywheels, Gears, Clutch Plates, Pressure Plates, Oil Pumps.

Transmission Repair research provides a detailed analysis of its global market and provides useful insights to understand the reason behind the popularity of this emerging technology along with its advantages and challenges. The report covers detailed analysis of key industry drivers, challenges, market trends as well as market structure. The report has been segregated on the basis of VehicleType, Repair Type, Components and global regions. This research also provides an assessment of key industry giants and their strategies that helps them to succeed in business.

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Transmission Repair market has been segmented by VehicleType, Repair Type, Components and by region. On the basis of Vehicle Type market has been divided as Passenger Car, Light Commercial Vehicle, Heavy Commercial Vehicle. By Repair Type, it is further divided as Transmission General Repair, Transmission Overhaul. On the basis of Components market has been divided as Gaskets and Seals, Fluids, Transmission Filters, O-Rings, Axles, Flywheels, Gears, Clutch Plates, Pressure Plates, Oil Pumps.

Coming to next segment, report provides an analysis of Transmission Repair market for global countries in the region. It covers a market overview for 2018-2023 and gives probable forecast with the context of Transmission Repair. This also covers new technological development and their role in the market. The research covers the pivotal trends within countries contributing to growth of the market, as well as analyses the factor due to which drivers impact the market in each region. Key regions and countries included in this report includes North America (U.S& Canada), Europe (Germany, U.K, France, Italy) Asia Pacific (China, India, Japan, South Korea, Indonesia, Taiwan and Rest of Asia), Middle East &Africa (GCC, North America, North America, South America).

The report also reflects the current scenario and the target of the Transmission Repair market. For this evaluation, 2017 considered as base year, 2018 as an estimated year, 2019-2023 as forecasted year. As already mentioned, the global Transmission Repair market is divided into a number of segments. All segments in terms of VehicleType, Repair Type, Components and different regions are examined in terms of base points to understand the relative contributions of each segments to market growth. This detailed level of facts & information is essential for the identification of various key factors in the global Transmission Repair market.

In the final section of the report, we have included a competitive landscape to provide clients a dashboard view based on categories of providers in the value chain, their presence in the Transmission Repair market, and key differentiators. This section is mainly designed to provide clients an objective and detailed comparative assessment of key providers specific to a market segment in the current scenario and the main competitors for the same. Report audiences can gain segment-specific vendor insights to identify and evaluate key competitors based on the in-depth assessment of capabilities and success in the marketplace. Detailed profiles of providers are also included in the scope of the report to evaluate their long-term and short-term strategies, key offerings, and recent developments in the Transmission Repair market. Some of the key competitors covered in the report are Allison Transmission, Schaeffler, Continental, ZF, Lee Myles Autocare & Transmission, Firestone Complete Auto Care, Cottman Transmission and Total Auto Care, Borgwarner, Mister Transmission, Aamco Transmissions etc, Others Major & Niche Players.

By Components

- Gaskets and Seals

- Fluids

- Transmission Filters

- O-Rings

- Axles

- Flywheels

- Gears

- Clutch Plates

- Pressure Plates

- Oil Pumps

By Repair Type

- Transmission General Repair

- Transmission Overhaul

By Vehicle Type

- Passenger Car

- Light Commercial Vehicle

- Heavy Commercial Vehicle

Competitive Landscape

- Allison Transmission

- Schaeffler

- Continental

- ZF

- Lee Myles Autocare & Transmission

- Firestone Complete Auto Care

- Cottman Transmission and Total Auto Care

- Borgwarner

- Mister Transmission

- Aamco Transmissions etc

- Others Major & Niche Players

Browse Full Report With TOC@ https://www.kdmarketinsights.com/product/global-transmission-repair-market

Table of Contents@

Research Methodology

Market Definition and List of Abbreviations

1. Executive Summary

2. Growth Drivers & Issues in Global Transmission Repair Market

3. Global Transmission Repair Market Trends

4. Opportunities in Global Transmission Repair Market

5. Recent Industry Activities, 2017

6. Porter's Five Forces Analysis

7. Market Value Chain and Supply Chain Analysis

8. Global Transmission Repair Market Size (USD Million), Growth Analysis and Forecast, (2017-2023)

9. Global Transmission Repair Market Segmentation Analysis, By Components

9.1. Introduction

9.2. Market Attractiveness, By Components

9.3. BPS Analysis, By Components

9.4. Gaskets and Seals

9.5. Fluids

9.6. Transmission Filters

9.7. O-Rings

9.8. Axles

9.9. Flywheels

9.10. Gears

9.11. Clutch Plates

9.12. Pressure Plates

9.13. Oil Pumps

10. Global Transmission Repair Market Segmentation Analysis, By Repair Type

10.1. Introduction

10.2. Market Attractiveness, By Repair Type

10.3. BPS Analysis, By Repair Type

10.4. Transmission General Repair

10.5. Transmission Overhaul

11. Global Transmission Repair Market Segmentation Analysis, By Vehicle Type

11.1. Introduction

11.2. Market Attractiveness, By Vehicle Type

11.3. BPS Analysis, By Vehicle Type

11.4. Passenger Car

11.5. Light Commercial Vehicle

11.6. Heavy Commercial Vehicle

12. Geographical Analysis

12.1. Introduction

12.2. North America Transmission Repair Market Size (USD Million), 2017-2023

12.2.1. By Components

12.2.2. By Repair Type

12.2.3. By Vehicle Type

12.2.4. By Country

12.2.4.1. Market Attractiveness, By End-user

12.2.4.2. BPS Analysis, By End-User

12.2.4.3. U.S. Market Size (USD Million), 2017-2023

12.2.4.4. Canada Market Size (USD Million), 2017-2023

12.3. Europe Transmission Repair Market Size (USD Million), 2017-2023

12.3.1. By Components

12.3.2. By Repair Type

12.3.3. By Vehicle Type

12.3.4. By Country

12.3.4.1. Market Attractiveness, By Country

12.3.4.2. BPS Analysis, By Country

12.3.4.3. Germany Market Size (USD Million), 2017-2023

12.3.4.4. United Kingdom Market Size (USD Million), 2017-2023

12.3.4.5. France Market Size (USD Million), 2017-2023

12.3.4.6. Italy Market Size (USD Million), 2017-2023

12.3.4.7. Spain Market Size (USD Million), 2017-2023

12.3.4.8. Russia Market Size (USD Million), 2017-2023

12.3.4.9. Rest of Europe Market Size (USD Million), 2017-2023

12.4. Asia Pacific Transmission Repair Market Size (USD Million), 2017-2023

12.4.1. By Components

12.4.2. By Repair Type

12.4.3. By Vehicle Type

12.4.4. By Country

12.4.4.1. Market Attractiveness, By Country

12.4.4.2. BPS Analysis, By Country

12.4.4.3. China Market Size (USD Million), 2017-2023

12.4.4.4. India Market Size (USD Million), 2017-2023

12.4.4.5. Japan Market Size (USD Million), 2017-2023

12.4.4.6. South Korea Market Size (USD Million), 2017-2023

12.4.4.7. Indonesia Market Size (USD Million), 2017-2023

12.4.4.8. Taiwan Market Size (USD Million), 2017-2023

12.4.4.9. Australia Market Size (USD Million), 2017-2023

12.4.4.10. New Zealand Market Size (USD Million), 2017-2023

12.4.4.11. Rest of Asia Pacific Market Size (USD Million), 2017-2023

Continue…

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About KD Market Insights

KD Market Insights has come with the idea of helping business by intelligent decision making and thorough understanding of the industry. We offer a comprehensive database of syndicated research, customized reports as well as consulting services to help a business grow in their respective domain. At KD Market Insights, we offer our client a deep Market research reports accompanied by business consulting services that can help them to reach on top of the corporate world. Our customized reports are built by keeping all factors of the industry in mind.

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#Transmission Repair Market #Transmission Repair Market Size #Transmission Repair Market Share #Transmission Repair Market Trends #Transmission Repair Market Research

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margdarsanme · 5 years ago

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NCERT Class 12 Computer Science Chapter 5 Structured Query Language

NCERT Class 12 Computer Science Python Solutions for Chapter 5 :: Structured Query Language

Short Answer Type Questions

Question 1:Write queries for (i) to (iv) and find ouputs for SQL queries (v) to (viii), which are based on the tables.

Table : VEHICLE

Note:

PERKS is Freight Charges per kilometer.

Km is kilometers Travelled

NOP is number of passangers travelled in vechicle.

To display CNO, CNAME, TRAVELDATE from the table TRAVEL in descending order of CNO.

To display the CNAME of all customers from the table TRAVEL who are travelling by vechicle with code Vo1 or Vo2

To display the CNO and CNAME of those customers from the table TRAVEL who travelled between ‘2015-1231’ and ‘2015-05-01’.

To display all the details from table TRAVEL for the customers, who have travel distacne more than 120 KM in ascending order of NOE

SELECT COUNT (*), VCODE FROM TRAVEL GROUP BY VCODE HAVING COUNT (*) > 1;

SELECT DISTINCT VCODE FROM TRAVEL :

SELECT A.VCODE, CNAME, VEHICLETYPE FROM TRAVEL A, VEHICLE B WHERE A. VCODE = B. VCODE and KM < 90;

SELECT CNAME, KM*PERKM FROM TRAVEL A, VEHICLE B WHERE A.VCODE = B.VCODE AND A. VCODE ‘V05’;

Answer:

Question 2:Consider the following tables SCHOOL and ADMIN and answer this question :Give the output the following SQL queries :

Select Designation Count (*) From Admin Group By Designation Having Count (*) <2;

SELECT max (EXPERIENCE) FROM SCHOOL;

SELECT TEACHER FROM SCHOOL WHERE EXPERIENCE >12 ORDER BY TEACHER;

SELECT COUNT (*), GENDER FROM ADMIN GROUP BY GENDER;

Answer:

Question 3:Write SQL qureries for (i) to (iv) and find outputs for SQL queries (v) to (viii), which are based on the tables TRANSPORT and TRIE

Note:

PERKS is Freight Charages per kilometer

TTYPE is Transport Vehicle Type

Note:

NO is Driver Number

KM is Kilometer travelled

NOP is number of travellers travelled in vehicle

TDATE is Trip Date

To display NO, NAME, TDATE from the table TRIP in descending order of NO.

To display the NAME of the drivers from the table TRIP who are traveling by transport vehicle with code 101 or 103.

To display the NO and NAME of those drivers from the table TRIP who travelled between ‘2015-02-10’ and ‘2015-04-01’.

To display all the details from table TRIP in which the distance travelled is more than 100 KM in ascending order of NOP

SELECT COUNT (*), TCODE From TRIPGROUP BY TCODE HAVNING COUnT (*) > 1;

SELECT DISTINCT TCODE from TRIP;

SELECT A.TCODE, NAME, TTYPEFROM TRIP A, TRANSPORT BWHERE A. TCODE = B. TCODE AND KM < 90;

SELECT NAME, KM *PERKMFROM TRIP A, TRANSPORT BWHERE A. TCODE = B. TCODE AND A. TCODE = 105′;

Answer:

SELECT NO, NAME, TDATE FROM TRIP ORDER BY NO;

SELECT NAME FROM TRIPWHERE TCODE = 101 OR TCODE = 103;

SELECT NO AND NAME FROM TRIPWHERE ‘2015-02-10’ < TDATE < ‘2015-04-01’;

SELECT NO, NAME, TDATE, KM, TCODE FROM TRIPWHERE KM >100 ORDER BY NOP;

TO DISPLAY THE MORE THAN ONE COUNT OF TCODE FROM THE TABLE TRIP

TO DISPALY SEPERATE TCODE OF TABLE TRIP

TO DISPAY THE NAME AND CODE OF THOSE TRANS PORTERS, WHO HAVE TRAVELLED MORE THAN 90 KMS.

TO DISPLAY THE NAME AND EXPENDITARE OF A TRANSPORTER WHO HAVE TCODE AS 105.

Question 4:Write SQL query to add a column total price with datatype numeric and size 10, 2 in a table product.Answer:ALTER TABLE PRODUCT ADD TOTAL PRICE NUMBER (10,2).

Question 5:Sonal needs to display name of teachers, who have “0” as the third character in their name. She wrote the following query.SELECT NAME FROM TEACHER WHERE NAME = “$$0?”;But the query is’nt producing the result. Identify the problem.Answer:The wildcards are incorrect. The corrected query is SELECT NAME FROM TEACHER WHERE NAME LIKE ‘_ _0%’.

Question 6:Deepika wants to remove all rows from the table BANK. But he needs to maintain the structure of the table. Which command is used to implement the same?Answer:DELETE FROM BANK.

Question 7:While creating table ‘customer’, Rahul forgot to add column ‘price’. Which command is used to add new column in the table. Write the command to implement the same.Answer:ALTER TABLE CUSTOMER ADD PRICE NUMBER (10, 2).

Question 8:What is the use of wildcardAnswer:The wildcard operators are used with the LIKE operator to search a value similar to a specific pattern in a column. There are 2 wildcard operators.% – represents 0,1 or many characters – – represents a single number or character

Question 9:Differentiate between DELETE and DROP table commands ?Answer:DELETE command is used to remove infor¬mation from a particular row or rows. If used without condition, it will delete all row information but not the structure of the table. It is a DML command.DROP table command is used to remove the entire structure of the table and information. It is a DDL command

Long Answer Type Questions

Question 1:Write SQL commands for the queries (i) to (iv) and output for (v) & (viii) based on a table COMPANY and CUSTOMER.

To display those company name which are having prize less than 30000.

To display the name of the companies in reverse alphabetical order.

To increase the prize by 1000 for those customer whose name starts with „S?

To add one more column totalprice with decimal] 10,2) to the table customer

SELECT COUNT(*) , CITY FROM COMPANY GROUP BY CITY;

SELECT MIN(PRICE), MAX(PRICE) FROM CUSTOMER WHERE QTY>10;

SELECT AVG(QTY) FROM CUSTOMER WHERE NAME LIKE “%r%;

SELECT PRODUCTNAME,CITY, PRICEFROM COMPANY, CUSTOMER WHERECOMPANY. CID=CUSTOMER.CID ANDPRODUCTNAME=”MOBILE”;

Answer:

SELECT NAME FROM COMPANY WHERE COMPANY.CID=CUSTOMER. CID ANDPRICE < 30000;

SELECT NAME FROM COMPANY ORDER BY NAME DESC;

UPDATE CUSTOMERSET PRICE = PRICE + 1000WHERE NAME LIKE ‘S%’;

ALTER TABLE CUSTOMERADD TOTALPRICE DECIMAL(10,2);

50000,70000

Question 2:Consider the following tables SCHOOL and ADMIN and answer this question :

Write SQL statements for the following:

To display TEACHERNAME, PERIODS of all teachers whose periods are more than 25.

To display all the information from the table SCHOOL in descending order of experience.

To display DESIGNATION without dupli¬cate entries from the table ADMIN.

To display TEACHERNAME, CODE and corresponding DESIGNATION from tables SCHOOL and ADMIN of Male teachers.

Answer:

SELECT TEACHERNAME, PERIODSFROM SCHOOL WHERE PERIODS>25:

SELECT * FROM SCHOOL;

SELECT DISTINCT DESIGNATION FROM ADMIN;

SELECT TEACHERNAME.CODEDESIGNATION FROMSCHOOL.CODE = ADMIN.CODEWHERE GENDER = MALE;

Question 3:Write SQL commands for the queries (i) to (iv) and output for (v) to (viii) based on the tables Watches’ and Sale given below.

TO DISPLAY ALL THE DETAILS OF THOSE WATCHES WHOSE NAME ENDS WITH ‘TIME’

TO DISPLAY WATCH’S NAME AND PRICE OF THOSE WATCHES WHICH HAVE PRICE RANGE IN BE-TWEEN 5000-15000.

TO DISPLAY TOTAL QUANTITY IN STORE OF UNISEX TYPE WATCHES.

TO DISPLAY WATCH NAME AND THEIR QUANTITY SOLD IN FIRST QUARTER;

SELECT MAX (PRICE), MIN(QTY_STORE) FROM WATCHES;

SELECT QUARTER, SUM(QTY SOLD) FROM SALE GROUP BY QUARTER;

SELECT WATCH_NAME, PRICE, TYPE FROM WATCHES W, SALE S WHERE W. WAT£H1D!=S.WATCHID; (viii) SELECT WATCH_NAME, QTYSTORE, SUM (QTY_SOLD), QTY_STORESUM (QTYSOLD) “STOCK” FROM WATCHES W, SALE S WHERE W. WATCHID = S.WATCHID GROUP BY S.WATCHID;

Answer:

SELECT * FROM WATCHES WHERE WATCH_NAME LIKE ‘%TIME’(Vi mark for SELECT query) (Vi mark for where clause)

SELECT WATCH_NAME, PRICE WATCH WHERE PRICE BETWEEN 5000 AND 15000;(Vi mark for SELECT query) (Vz mark for where clause)

SELECT SUM (QTY STORE) FROM WATCHES WHERE TYPE LIKE ‘UNISEX’;(Vz mark for SELECT query) (Vi mark for where clause)

SELECT WATCHNAME, QTY SOLD FROM WATCHES W,SALE S WHERE W. WATCHID = S. WATCHIDAND QUARTER = 1;

Question 4:Answer the questions (a) and (b) on the basis of the following tables SHOP and ACCESSORIES.

(a) Write the SQL queries:

To display Name and Price of all the Accessories in ascending order of their Price.

To display Id and SName of all Shop located in Nehru Place.

To display Minimum and Maximum Price of each Name of Accessories.

To display Name, Price of all Accessories and their respective SName where they are available.

(b) Write the output of the following SQL

SELECT DISTINCT NAME FROM ACCESSORIES WHERE PRICE> =5000;

SELECT AREA, COUNT(*) FROM SHOPPE GROUP BY AREA;

SELECT COUNT (DISTINCT AREA) FROM SHOPPE;

SELECT NAME, PRICE*0.05 DISCOUNT FROM ACCESSORIES WHERE SNO IN (‘S02‘,S03‘);

Answer:(a)

SELECT Name, Price FROM ACCESSORIES ORDER BY Price Asc;

SELECT ID SName FROM SHOP WHERE Area=”Nehru Place”;

SELECT Name, max (Price); min(Price) FROM ACCESSORIES, Group By Name;

SELECT Name,price, Sname FROMACCESSORIES, SHOP WHERE SHOEID=ACCESSORIES.ID;

(b)

Question 5:Write SQL queries for:

To display name, fee, gender, joinyear about the applicants, who have joined before 2010.

To display names of applicants, who are playing fee more than 30000.

To display names of all applicants in ascending order of their joinyear.

To display the year and the total number of applicants joined in each YEAR from the table APPLICANTS.

To display the C_ID (i.e., CourselD) and the number of applicants registered in the course from the APPLICANTS and table.

To display the applicant’s name with their respective course’s name from the tables APPLICANTS and COURSES.

Give the output of following SQL statements:

SELECT Name, Joinyear FROM APPLICANTSWHERE GENDER=’F’ and C_ID=’A02′;

SELECT MIN (Joinyear) FROMAPPLICANTSWHERE Gender=’m’;

SELECT AVG (Fee) FROM APPLICANTSWHERE C_ID=’A0T OR C_ID=’A05′;

SELECT SUM- (Fee), C_ID FROM C_ IDGROUP BY C_IDHAVING COUNT(*)=2;

Answer:

SELECT NAME,FEE,GENDER,JOINYEARFROM APPLICANTSWHERE J OINYE AR <2010

SELECT NAME FROM APPLICANTS WHERE FEE >30000

SELECT NAME FROM APPLICANTS ORDERBY JOINYEAR ASC

SELECT YEAR, COUNT]*) FROMAPPLICANTS GROUP BY YEAR;

SELECT C_ID, COUNT]*) FROMAPPLICANTS, COURSES GROUP BY IDWHERE APPLICANTS.C_ID=COURSES. C_ID

SELECT NAME,COURSE FROMAPPLICANTS, COURSESWHERE APPLICANTS. C_ID=COURSES. C_ID

Avisha 2009

2009

55000 A01

Question 6:Write SQL queries for (a) to (g) and write the output for the SQL queries mentioned shown in (hi) to (h4) parts on the basis of table ITEMS and TRADERS :

To display the details of all the items in ascending order of item names (i.e., INAME).

To display item name and price of all those items, whose price is in the range of 10000 and 22000 (both values inclusive).

To display the number of items, which are traded by each trader. The expected output of this query should be:

To display the price, item name and quantity (i.e., qty) of those items which have quantity more than 150.

To display the names of those traders, who are either from DELHI or from MUMBAI.

To display the names of the companies and the names of the items in descending order of company names.

Obtain the outputs of the following SQL queries based on the data given in tables ITEMS and TRADERS above.

SELECT MAX (PRICE), MIN (PRICE) FROM ITEMS;

SELECT PRICE*QTYFROM ITEMS WHERE CODE-1004;

SELECT DISTINCT TCODE FROM ITEMS;

SELECT INAME, TNAME FROM ITEMS I, TRADERS T WHERE I.TCODE=T.TCODE AND QTY< 100;

Answer:

SELECT INAME FROM ITEMS ORDER BYINAME ASC;

SELECT INAME, PRICE FROM ITEMS WHERE PRICE => 10000 AND PRICE =< 22000; (c) SELECT TCODE, COUNT (CODE) FROM ITEMS GROUP BY TCODE;

SELECT PRICE, INAME, QTY FROM ITEMS WHERE (QTY> 150);

SELECT TNAME FROM TRADERS WHERE (CITY = “DELHI”) OR (CITY = “MUMBAI”)

SELECT COMPANY, INAME FROM ITEMSORDER BY COMPANY DESC;

(hi) 380001200(h2)1075000(h3)T01T02TO3(h4) LED SCREEN 40 DISP HOUSE INC CAR GPS SYSTEM ELECTRONICS SALES

Question 7:Write SQL queries for (a) to (f) and write the outputs for the SQL queries mentioned shown in (gl) to (g4) parts on the basis of tables PRODUCTS and SUPPLIERS

To display the details of all the products in ascending order of product names (i.e., PNAME).

To display product name and price of all those products, whose price is in the range of 10000 and 15000 (both values inclusive).

To display the number of products, which are supplied by each suplier. i.e., the expected output should be;

S01 2

S02 2

S03 1

To display the price, product name and quantity (i.e., qty) of those products which have quantity more thhn 100.

To display the names of those suppliers, who are either from DELHI or from CHENNAI.

To display the name of the companies and the name of the products in descending order of company names.

Obtain the outputs of the following SQL queries based on the data given in tables PRODUCTS and SUPPLIERS above.

SELECT DISTINCT SUPCODE FROM PRODUCTS;

SELEC MAX (PRICE), MIN (PRICE) FROM PRODUCTS;

SELECT PRICE*QTYFROM PRODUCTS WHERE PID = 104; (g4)

SELECT PNAME, SNAMEFROM PRODUCTS P, SUPPLIERS S WHERE E SUPCODE = S. SUPCODEAND QTY>100;

Answer:

SELECT * FROM PRODUCTS ORDER BY PNAME ASC;

SELECT PNAME, PRICE FROM PRODUCTS WHERE ((PRICE => 10000) AND (PRICE = < 15000));

SELECT SUPCODE, COUNT (PID) FROM PRODUCTS GROUP BY SUPCODE;

SELECT PRICE, PNAME, QTY FROM PRODUCTS WHERE (QTY > 100);

SELECT SNAME FROM SUPPLIERS WHERE ((CITY = “DELHI”) OR (CITY = “CHENNAI”));

SELECT COMPANY, PNAME FROM PRO-DUCTS ORDER BY COMPANY DESC; 4

SOI1(gl)s02 s03(g2) 280001100(g3) 550000(g4) PNAME SNAME ViDIGITAL CAMERA 14 X GETALL INCPENDRIVE16 GB GETALL INC

Question 8:Consider the following tables CARDEN and CUSTOMER and answer (b) and (c) parts of this question:

Give a suitable example of a table with sample data and illustrate Primary and Alternate Keys in it.

Write SQL commands for the following statements:

To display the names of all the silver coloured cars.

To display names of car, make and capacity of cars in descending order of their sitting capacity.

To display the highest charges at which a vehicle can be hired from CARDEN.

To display the customer name and the corresponding name of the cars hired by them.

Give the output of the following SQL queries:

SELECT COUNT(DISTINCT Make) FROM CARDEN;

SELECT MAX(Charges), MIN (Charges) FROM CARDEN;

SELECT COUNTS), Make FROM CARDEN;

Answer:

Primary Key of CARDEN = Ccode CARDENAlternate Key = CarName:Primary key of Customer = CodeAlternate Key of Customer = Cname 2

SELECT CarName From CARDENWHERE Color = “SILVER”;

SELECT CarName, Make, Capacity FromCARDEN ORDER BY Capacity DESC;

SELECT MAX(Charges) Frm CARDEN;

ELECT Cname, CarName FromCARDEN, CUSTOMER WHERECARDEN. Ccode = CUSTOMER. Ccode;

(i) 4(ii) MAX(Charges) MIN (Charges)35 112(iii) 5(iv) SX4C Class

Question 9:Consider the following tables CABHUB and CUSTOMER and answer (b) and (c) parts of this question :

Give a suitable example of a table with sample data and illustrate Primary and Candidate Keys in it.

Write SQL commands for the following statements:

To display the names of all the white coloured vehicles.

To display name of vehicle name and capacity of vehicles in ascending order of their sitting capacity.

To display the highest charges at which a vehicle can be hired from CABHUB.

To display the customer name and the corresponding name of the vehicle hired by them.

Give the output of the following SQL queries :

SELECT COUNT(DISTINCT Make) FROM CABHUB;

SELECT MAX(Charges), MIN(Charges)

FROM CABHUB;

SELECT COUNT (*) Make FROM CABHUB;

SELECT Vehicle FROM CABHUB WHERE Capacity=4;

Answer:

Primary key of CABHUB = Vcode alternate key of CABHUB = Vehicle Name. Primary key of Customer = Ccode Alternate Key of CUSTOMER = Cname.

(i) SELECT VehicleName FROM CABHUBWHERE Colour = “WHITE”;

SELECT VehicleName, capacity From CABHUB ORDER BY Capacity ASC;

SELECT MAX(Charges) FROM CABHUB;

SELECT Cname,VehicleName FROM CABHUB, CUSTOMER WHERE CUSTOMER. Vcode=CABHUB. Vcode;

(i) 4(ii) MAX(Charges) MIN (Charges)35 12(iii) 5(iv) SX4C Class

Question 10:Consider the following tables EMPLOYEE and DEPARTMENT and answer (a) and (b) parts of this question.

Write SQL commands for the following statements:

To display all DepName along with the DepCde in descending order of DepCde.

To display the average age of Employees in DepCde as 103.

To display the name of DepHead of the Employee named “Sanjeev P”

To display the details of all employees who has joined before 2007 from EMPLOYEE table.

Give the output of the following SQL queries:

SELECT COUNT (DISTINCT DepCde) FROM EMPLOYEE;

SELECT MAX(JoinDate), MIN (JointDate) FROM EMPLOYEE;

SELECT TName, DepHead FROM EMPLOYEE E, DEPARTMENT DWHERE E.DepCde = D.DepCde;

SELECT COUNT (*) FROM EMPLOYEE WHERE Salary > 60000 AND Age > 30;

Answer:

(a)

SELECT DEPNAME, DEPARTME-NT.DepCde FROM EMPLOYEE, DEPARTMENT WHERE EMPLOYEE. DepCDE=DEPARTMENT. DepCde Order by DepCde DESC;

Select AVG (Age) from EMPLOYEE WHERE DepCde=”103″;

SELECT DeptHead FROM DEPARTMENT WHERE Employee. TName=“Sanjeev P” AND EMPLOYEE. DepCde= DEPARTMENT. DepCde;

SELECT * from EMPLOYEE WHEREjoinDate<’01-JAN-2007′;

Question 11:Consider the following tables WORKER and PAYLEVEL and answer (a) and (b) parts of this question:

(a) Write SQL commands for the following statements:

To display the name of all Workers in descending order of DOB.

To display NAME and DESIGN of those Workers, whose PLEVEL is either P001 or

To display the content of all the workers table, whose DOB is in between ’19-JAN- 1984′ and ’18-JAN-1987′.

To add a new row with the following:19, ‘DayaKishore’, ‘Operator’, ‘P003′, ’19- Sep-2008’, ‘ll-Jul-1984’

(b) Give the output of the following SQL queries :

SELECT COUNT (PLEVEL), PLEVEL FROM WORKER GROUP BY PLEVEL;

SELECT MAX(DOB), MIN(DOJ) FROM WORKER;

SELECT Name,PAY FROM WORKERW,PAYLEVEL P WHERE W.LEVEL=P.PLEVEL AND W.ECODE<13;

SELECT PLEVEL, PAYLEVELWHERE PLEVEL=”POO3″;

Answer:

(a)

SELECT NAME FROM WORKERORDER BY DOBDESC;

SELECT NAME, DESIGN FROM WORKER WHERE PLEVEL=”POOO1″ ORPLEVEL=”POO2″;

SELECT * FROM WORKER WHEREDOB BETWEEN ’19-JAN-1984 AND ’18-JAN-1987′;

INSERT INTO WORKER VALUES (19,”DayaKISHORE”, “oPERATOR”, “P0003”,’19-Sep-2008′,’11-Jul-1984′)’

(b)

Question 12:Consider the following tables EMPLOYEE and SALGRADE and answer (b) and (c) parts of this question:

(a) What do you understand by Selection and Projection operations in relational algebra ?(b) Write SQL commands for the following statements :

To display the details of all EMPLOYEES in descending order of DOJ.

To display NAME and DESIGN of those EMPLOYEES, whose SAL-GRADE is either S02 or S03.

TO display the content Of all the EMPLOYEES table, whose DOJ is in between ’09-Feb-2006′ and ’08-Aug-2009′.

To add a new row with the following:109, ‘HarishRoy’, ‘HEAD-IT’, ‘SOX, ’09-Sep-2007′, ’21-Apr-1983’

SELECT COUNT(SGRADE), SGRADEFROM EMPLOYEE GROUP BYSGRADE;

SELECT MIN(DOB), MAX(DOJ) FROM EMPLOYEE;

SELECT NAME, SALARY FROMEMPLOYEE E, SAL-GRADE S WHEREE.SGRADE= S.SGRADE AND E.ECODE<103′;

SELECT SGRADE, SALARY +HRA FROM SALGRADE WHERE SGRADE =SGRADE=’S02;’

Answer:Projection(ff): In relational algebra, a projection is a unary operafion. The result of such projecion is defined as the set obtained when the components Of the tuple R are restriceted to the set {a1……an}. It discards (or excludes) the other attributes.Selection(): In relational algebra, a selectionis a unary operation written as (R) or (R) where:

a and b are attribute names.

p is a binary operation in the set.

v is a value constant.

R is a relation.

The selection (R) selects all those tuples in R for which 9 holds between the a and the b

(b)

SELECT FROM EMPLOYEE ORDER BY DOJ DESC;

SELECT NAME, DESIGN FROM EMPLOYEE WHERE SGRADE – “S02” OR SGRADE = “SO3;

SELECT * FROM EMPLOYEE WHERE DOJ BETWEEN ’09-FEB-2006′ AND ’08- AUG -200%

INSERT INTO EMPLOYEE VALUES(109, “HARSH RAY”, “HEAD-IT.S02”, ’09-SEP-2007′, ’21-APR-1983′);

Question 13:Consider the following tables GAMES and PLAYER and answer (b) and (c) parts of this question :

(a)What do you understand by primary key and candidate keys ?(b)Write SQL commands for the following statements:

To display the name of all GAMES with their GCodes.

To display details of those GAMES which are having PrizeMoney more than 7000.

To display the content of the GAMES table in ascending order of Schedule Date.

To display sum of PrizeMoney for each type of GAMES.

SELECT COUNT(DISTINCT Number) FROM GAMES;

SELECT MAX(ScheduleDate), MIN(Sche- duleDate) FROM GAMES

SELECT Name, GameName FROM GAMES G, PLAYER PWHERE (G.Gcode=PGcode AND G.Pri- zeMoney>10000);

SELECT DISTINCT Geode FROM PLAYER;

Answer:

(a) An attribute or set of attributes which are used to identify a tuple uniquely is known as a primary key. If a table has more than one such attributes which identify a tuple uniquely than all such attributes are known as candidate keys.

(b)

SELECT GameName, GCode FROM GAMES;

SELECT * FROM Games WHERE PrizeMoney >7000;

SELECT * FROM Games ORDER BY ScheduleDate;

SELECT SUM(Pnzemoney) FROM Games GROUPBY Type;

(c)

19-Mar-2004 12-Dec-2003

Ravi Sahai Lawn Tennis

101 108 103

Question 14:Consider the following tables ACTIVITY and COACH and answer (a) and (b) parts of this question :

(a) Write SQL commands for the following statements:

To display the names of all activities with their Acodes in descending order.

To display sum of PrizeMoney for the Activities played in each of the Stadium separately.

To display the coach’s name and acodes in ascending order of Acode from the table Coach.

To display the content of the Activity table whose schedule date earlier than 01-01-2004 in ascending order of Participants Num.

(b) Give the output of the following SQL queries:

SELECT COUNT (DISTINCT Participants Num) FROM ACTIVITY;

SELECT MAX (Schedule Date), Min (Schedule Date) FROM ACTIVITY;

SELECT Name, Activity Name FROM ACTIVITY A, COACH CWHERE A.Acode=C.Acode AND A.Parti- cipants Num=10;

SELECT DISTINCT Acode FROM COACH;

Answer:

(a)

SELECT Acodes, ActivityName FROM ACTIVITY ORDER BY ACode DESC;

SELECT SUM(PrizeMoney) FROM ACTIVITY GROUP BY Stadium;

SELECT Name, Acode FROM COACH ORDER BY Acode;

SELECT * FROM ACTIVITY WHERE SchduleDate < ’01-Jan-2004′ ORDER BY ParticipantsNum;

(b)

12-Dec-2003 19-Mar-2004

Ravinder Discuss Throw

100110081003

Question 15:Consider the following tables RESORT and OWNEDBY and answer (a) and (b) parts of this question:

(a)Write SQL commands for the following statements:

To display the RCODE and PLACE of all ‘5 STAR’ resorts in the alphabetical order of the place from table RESORT.

To display the maximum and minimum rent for each type of resort from table RESORT.

To display the details of all resorts which are started after 31-DEC-05 from table RESORT.

Display the OWNER of all ‘5 STAR’ resorts from tables RESORT and OWNEDBY.

(b)Give output for the following SQL queries:

SELECT MIN(RENT) FROM RESORT Where PLACE = ‘KERALA’;

SELECT TYPE, START DATE FROM RESORT Where TYPE ‘2 STAR’ORDERBY STARTDATE,

SELECT PLACE, OWNER FROMOWNEDBY Where PLACE LIKE “%A”;

SELECT RCODE, RENT FROM RESORT, OWNEDBY WHERE (RESORT PLACE= OWNEDBY. PLACE AND TYPE = ‘3 STAR’);

Answer: (a)

SELECT RCODE, PLACE FROM RESORT mere TYPE = “5 STAR” ORDERBY PLACE;

SELECr MAX (RENT), MIN (RENT) FROM RESORT GROUP BY TYPE;

SELECT FROM RESORT WHERE OSWAAL (BSE Question Bank. COMPUTER SCIENCE – PYTHON, STARTDATE > ’31-DEC-05′;

SELECT OWNER FROM RESOR OWNEDBY B WHERE (A.TYPE START’ AND A.PLACE B.PLACE);

(b)

Question 16:Consider the following tables STORE and SUPPLIERS and answer (a) and (b) parts of this question:

(a) Write SQL commands for the following statements:

To display details of all the items in the STORE table in ascending order of LastBuy.

To display ItemNo and Item name of those items from STORE table whose Rate is more than 15 Rupees.

To display the details of those items whose supplier code (Scode) is 22 or Quantity in Store (Qty) is more than 110 from the table Store.

To display minimum Rate of items for each supplier individually as per Scode from the table STORE.

(b) Give the output of the following SQL queries:

SELECT COUNT(DISTINCT Scode) FROM STORE;

SELECT Rate* Qty FROM STORE WHERE ItemNo=2004;

SELECT Item, Sname FROM STORE S, Suppliers P

WHERE S.Scode=PScode AND ItemNo=2006;

SELECT MAX(LastBuy) FROM STORE;

Answer:

(a)

SELECT * FROM STORE ORDER BY LastBuy ASC;

SELECT ItemNo, Item FROM STORE WHERE Rate > 15;

SELECT * FROM STORE WHERE (Scode = ’22’ OR Qty >’110′);

SELECT Sname, MIN(Rate) FROM STORE, SUPPLIERS WHERE STORE. Scode = SUPPLIERS.Scode GROUP BY Sname;

(b)

880

Item SnameGel Pen Classic Premium Stationers

24-Feb-10

Question 17:Consider the following tables STOCK and DEALERS and answer (a) and (b) parts of this question:

(a)Write SQL commands for the following statements:

To display the details of all Items in the STOCK table in ascending order of StockDate.

To display ItemNo and Item name of those items from STOCK table whose UnitPrice is more than Rupees 10.

To display the details of those items whose dealer code (Dcode) is 102 or quantity in STOCK (Qty) is more than 100 from the table Stock.

To display maximum UnitPrice of items for each dealer individually as per Dcode from the table STOCK.

(b)Give the output of the following SQL queries:

SELECT COUNT(DISTINCT Dcode)FROM STOCK;

SELECT Qty* UnitPrice FROM STOCK WHERE ItemNo=5006;

SELECT Item, Dname FROM STOCK S, Dealers D WHERE S.Dcode=D.Dcode AND ItemNo = 5004;

SELECT MIN (StockDate) FROM STOCK;

Answer:

(a)

SELECT*FROM STOCK ORDER BY StockDate;

SELECT Item No, Item FROM STOCK WHERE UnitPrice >10;

SELECT *FROM DEALERS, STOCKWHERE (DEALERS.Dcode=”102″OR STOCK.Qty >100 and DEALERS. DCODE = STOCK.DCODE);

SELECT MAX (Unitprice) FROM DEALERS, STOCK ORDER BY STOCK. Dcode WHERE DEALERS.Dcode = STOCK.Dcode;

(b)

4400

Item DnameEraser Big Clear Deals

01-Jan-09

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