CLASS XII COMPUTER SCIENCE CHAPTER 11 & 12 DATABASE AND SQL

CHAPTER 11& 12
DATABASE CONCEPTS AND SQL

Data :- Raw facts and figures which are useful to an organization. We cannot take decisions on the basis of data.
Information:- Well processed data is called information. We can take decisions on the basis of information.
Field:   Set of characters that represents specific data element.
Record:Collection of fields is called a record. A record can have fields of different data types.
File:     Collection of similar types of records is called a file.
Table: Collection of rows and columns that contains useful data/information is called a table.
A table generally refers to the passive entity which is kept in secondary storage device.
Relation:Relation (collection of rows and columns) generally refers to an active entity on which   we can perform various operations.
Database:Collection of logically related data along with its description is termed as database.
Tuple: A row in a relation is called a tuple.
Attribute:A column in a relation is called an attribute. It is also termed as field or data item.
Degree:Number of attributes in a relation is called degree of a relation.
Cardinality:Number of tuples in a relation is called cardinality of a relation.
Primary Key:Primary key is a key that can uniquely identifies the records/tuples in a relation. This          key can never be duplicated and NULL.
Foreign Key:  Foreign Key is a key that is defined as a primary key in some other relation. This key is used to enforce referential integrity in RDBMS.
Candidate Key: Set of all attributes which can serve as a primary key in a relation.
Alternate Key: All the candidate keys other than the primary keys of a relation are alternate keys for a relation.
DBA:  Data Base Administrator is a person (manager) that is responsible for defining the data base schema, setting security features in database, ensuring proper functioning of the data bases etc.
           
Structured Query Language
SQL is a non procedural language that is used to create, manipulate and process the databases(relations).

Characteristics of SQL
  1. It is very easy to learn and use.
  2. Large volume of databases can be handled quite easily.
  3. It is non procedural language. It means that we do not need to specify the procedures to accomplish a task but just to give a command to perform the activity.
  4. SQL can be linked to most of other high level languages that makes it first choice for the database programmers.

Processing Capabilities of SQL
The following are the processing capabilities of SQL
  1. Data Definition Language (DDL)
DDL contains commands that are used to create the tables, databases, indexes, views, sequences and synonyms etc.
e.g: Create table, create view, create index, alter table etc.
  1. Data Manipulation Language (DML)
DML contains command that can be used to manipulate the data base objects and to query    the       databases for information retrieval.
e.g Select, Insert, Delete, Update etc.
  1. View Definition:
DDL contains set of command to create a view of a relation.
e.g :create view
  1. Data Control Language:
         This language is used for controlling the access to the data. Various commands like  GRANT, REVOKE etc are available in DCL.
      5.   Transaction Control Language (TCL)
      TCL include commands to control the transactions in a data base system. The commonly used commands in TCL are COMMIT, ROLLBACK etc.

Data types of SQL
Just like any other programming language, the facility of defining data of various types is available in SQL also. Following are the most common data types of SQL.
1)      NUMBER
2)      CHAR
3)      VARCHAR / VARCHAR2
4)      DATE
5)      LONG
6)      RAW/LONG RAW


1. NUMBER
Used to store a numeric value in a field/column. It may be decimal, integer or a real value. General syntax is
                        Number(n,d)
Where n specifies the number of digits and
d specifies the number of digits to the right of the decimal point.
e.g       marks  number(3)    declares marks to be of type number with maximum value 999.
pct number(5,2)      declares pct to be of type number of 5 digits with two digits to the right of decimal point.
2. CHAR
Used to store character type data in a column. General syntax is
                        Char (size)
where size represents the maximum number of characters in a column. The CHAR type data can hold at most 255 characters.
e.g       name char(25)     declares a data item name of type character of upto 25 size long.

3. VARCHAR/VARCHAR2
This data type is used to store variable length alphanumeric data. General syntax is
                        varchar(size) / varchar2(size)
where size represents the maximum number of characters in a column. The maximum allowed size in this data type is 2000 characters.
e.g   address varchar(50);        address is of type varchar of upto 50 characters long.

4. DATE
Date data type is used to store dates in columns. SQL supports the various date formats other that the standard DD-MON-YY.
e.g       dob     date;    declares dob to be of type date.


5. LONG
This data type is used to store variable length strings of upto 2 GB size.
e.g description  long;  

6. RAW/LONG RAW
To store binary data (images/pictures/animation/clips etc.) RAW or LONG RAW data type is used. A column LONG RAW type can hold upto 2 GB of binary data.
e.g       image raw(2000);

SQL Commands
a. CREATE TABLE Command:
Create table command is used to create a table in SQL. It is a DDL type of command. The general syntax of creating a table is

Creating Tables
The syntax for creating a table is
create table <table> (
<column 1> <data type> [not null] [unique] [<column constraint>],
. . . . . . . . .
<column n> <data type> [not null] [unique] [<column constraint>],
[<table constraint(s)>]
);
For each column, a name and a data type must be specified and the column name must be unique within the table definition. Column definitions are separated by comma. Uppercase and lowercase letters makes no difference in column names, the only place where upper and lower case letters matter are strings comparisons. A not null Constraint means that the column cannot have null value, that is a value needs to be supplied for that column. The keyword unique specifies that no two tuples can have the same attribute value for this column.

Operators in SQL:
The following are the commonly used operators in SQL
1.      Arithmetic Operators                    +,   -,    *,   / 
2.      Relational Operators                     =,  <,   >,   <=,   >=,   <>
3.      Logical Operators                         OR,  AND,  NOT
Arithmetic operators are used to perform simple arithmetic operations.
Relational Operators are used when two values are to be compared and Logical operators are used to connect search conditions in the WHERE Clause in SQL.

Constraints:
Constraints are the conditions that can be enforced on the attributes of a relation. The constraints come in play when ever we try to insert, delete or update a record in a relation.
  1. NOT NULL
  2. UNIQUE
  3. PRIMARY KEY
  4. FOREIGN KEY
  5. CHECK
  6. DEFAULT
Not null ensures that we cannot leave a column as null. That is a value has to be supplied for that column.
e.g       name    varchar(25)      not null;
Unique constraint means that the values under that column are always unique.
e.g       Roll_no number(3)      unique;
Primary key constraint means that a column can not have duplicate values and not even a null value.
e.g.      Roll_no number(3)      primary key;
The main difference between unique and primary key constraint is that a column specified as unique may have null value but primary key constraint does not allow null values in the column.
Foreign key is used to enforce referential integrity and is declared as a primary key in some other table.
e.g       cust_id varchar(5) references master(cust_id);
it declares cust_id column as a foreign key that refers to cust_id field of table master. That means we cannot insert that value in cust_id filed whose corresponding value is not present in cust_id field of master table.
Check constraint limits the values that can be inserted into a column of a table.
e.g       marks   number(3)       check(marks>=0);
The above statement declares marks to be of type number and while inserting or updating the value in marks it is ensured that its value is always greater than or equal to zero.
Default constraint is used to specify a default value to a column of a table automatically. This default value will be used when user does not enter any value for that column.
e.g  balance  number(5)           default = 0;
CREATE TABLE student (
Roll_no           number(3) primary key,
Name               varchar(25) not null,
Class                varchar(10),
Marks              number(3) check(marks>0),
City                 varchar(25)  );

Data Modifications in SQL
After a table has been created using the create table command, tuples can be inserted into the table, or tuples can be deleted or modified.
INSERT Statement
The simplest way to insert a tuple into a table is to use the insert statement
insert into <table> [(<column i, . . . , column j>)] values (<value i, . . . , value j>);

INSERT INTO student VALUES(101,'Rohan','XI',400,'Jammu');
While inserting the record it should be checked that the values passed are of same data types as the one which is specified for that particular column.

For inserting a row interactively (from keyboard) & operator can be used.
e.g       INSERT INTO student VALUES(&Roll_no’,’&Name’,’&Class’,’&Marks’,’&City’);
In the above command the values for all the columns are read from keyboard and inserted into the table student.

NOTE:- In SQL we can repeat or re-execute the last command typed at SQL prompt by typing “/” key and pressing enter.
Roll_no
Name
Class
Marks
City
101
Rohan
XI
400
Jammu
102
Aneeta Chopra
XII
390
Udhampur
103
Pawan Kumar
IX
298
Amritsar
104
Rohan
IX
376
Jammu
105
Sanjay
VII
240
Gurdaspur
113
Anju Mahajan
VIII
432
Pathankot
Queries:
To retrieve information from a database we can query the databases. SQL SELECT statement is used to select rows and columns from a database/relation.

SELECT Command
This command can perform selection as well as projection.
Selection:This capability of SQL can return you the tuples form a relation with all the attributes.
Projection: This is the capability of SQL to return only specific attributes in the relation.
*SELECT * FROM student; command will display all the tuples in the relation student
*SELECT * FROM student WHERE Roll_no <=102;
The above command display only those records whose Roll_no less than or equal to 102.
Select command can also display specific attributes from a relation.
*SELECT name, class FROM student;         
The above command displays only name and class attributes from student table.
*SELECT count(*) AS “Total Number of Records” FROM student;
Display the total number of records with title as “Total Number of Records” i.e an alias
We can also use arithmetic operators in select statement, like
*SELECT Roll_no, name, marks+20 FROM student;
*SELECT name, (marks/500)*100 FROM student WHERE Roll_no > 103;

Eliminating Duplicate/Redundant data

DISTINCT keyword is used to restrict the duplicate rows from the results of a SELECT statement.
e.g. SELECT DISTINCT name FROM student;
The above command returns
Name
Rohan
Aneeta Chopra
Pawan Kumar

Conditions based on a range
SQL provides a BETWEEN operator that defines a range of values that the column value must fall for the condition to become true.
e.g. SELECT Roll_no, name FROM student WHERE Roll_no BETWENN 100 AND 103;
The above command displays Roll_no and name of those students whose Roll_no lies in the range 100 to 103 (both 100 and 103 are included in the range).

Conditions based on a list
To specify a list of values, IN operator is used. This operator select values that match any value in the given list.
e.g.  SELECT * FROM student WHERE city IN (‘Jammu’,’Amritsar’,’Gurdaspur’);
The above command displays all those records whose city is either Jammu or Amritsar or Gurdaspur.

Conditions based on Pattern
SQL provides two wild card characters that are used while comparing the strings with LIKE operator.
a. percent(%)               Matches any string
b.Underscore(_)          Matches any one character
e.g SELECT Roll_no, name, city FROM student WHERE Roll_no LIKE “%3”;
displays those records where last digit of Roll_no is 3 and may have any number of characters in front.
e.g SELECT Roll_no, name, city FROM student WHERE Roll_no LIKE “1_3”;
displays those records whose Roll_no starts with 1 and second letter may be any letter but ends with digit 3.

ORDER BY Clause
ORDER BY clause is used to display the result of a query in a specific order(sorted order).
The sorting can be done in ascending or in descending order. It should be kept in mind that the actual data in the database is not sorted but only the results of the query are displayed in sorted order.
e.g.      SELECT name, city FROM student ORDER BY name;
The above query returns name and city columns of table student sorted by name in increasing/ascending order.
e.g.      SELECT  * FROM student ORDER BY city DESC;
It displays all the records of table student ordered by city in descending order.
Note:-  If order is not specifies that by default the sorting will be performed in ascending order.


GROUP BY Clause
The GROUP BY clause can be used in a SELECT statement to collect data across multiple records and group the results by one or more columns.
The syntax for the GROUP BY clause is:
SELECT column1, column2, ... column_n, aggregate_function (expression)
FROM tables
WHERE conditions
GROUP BY column1, column2, ... column_n;
aggregate_function can be a function such as SUM, COUNT, MAX, MIN, AVG etc.

e.g       SELECT name, COUNT(*) as "Number of employees"
FROM student
WHERE marks>350
GROUP BY city;

HAVING Clause
The HAVING clause is used in combination with the GROUP BY clause. It can be used in a SELECT statement to filter the records that a GROUP BY returns.
The syntax for the HAVING clause is:
SELECT column1, column2, ... column_n, aggregate_function (expression)
FROM tables
WHERE predicates
GROUP BY column1, column2, ... column_n
HAVING condition1 ... condition_n;
e.g       SELECT SUM(marks) as "Total marks"
FROM student
GROUP BY department
HAVING SUM(sales) > 1000;
Note: select statement can contain only those attribute which are already present in the group by clause.
Functions available in SQL

SQL provide large collection of inbuilt functions also called library functions that can be used directly in SQL statements.
1.      Mathematical functions
2.      String functions
3.      Date & Time functions

1.Mathematical functions
Some of the commonly used mathematical functions are sum() avg(), count(), min(), max() etc.
e.g. SELECT sum(marks) FROM student;
displays the sum of all the marks in the table student.
e.g. SELECT min(Roll_no), max(marks) FROM student;
displays smallest Roll_no and highest marks in the table student.


2.String functions
These functions are used to deal with the string type values like
ASCII, LOWEWR, UPPER, LEN, LEFT, RIGHT, TRIM, LTRIM, RTRIM etc.
ASCII : Returns the ASCII code value of a character(leftmost character of string).
Syntax: ASCII(character)
SELECT ASCII('a') returns       97
SELECT ASCII('A') returns      65
SELECT ASCII('1') returns       49
SELECT ASCII('ABC') returns  65
For Upper character 'A' to 'Z' ASCII value 65 to 90
For Lower character 'A' to 'Z' ASCII value 97 to 122
For digit '0' to '9' ASCII value 48 to 57
NOTE:           If no table name is specified then SQL uses Dual table which is a dummy table used for performing operations.
LOWER : Convert character strings data into lowercase.
Syntax: LOWER(string)
SELECT LOWER('STRING FUNCTION') returns       string function
UPPER : Convert character strings data into Uppercase.
Syntax: UPPER(string)
SELECT UPPER('string function') returns       STRING FUNCTION
LEN : Returns the length of the character string.
Syntax: LEN(string)
SELECT LEN('STRING FUNCTION') returns       15
REPLACE : Replaces all occurrences of the second string(string2) in the first string(string1) with a third string(string3).
Syntax: REPLACE('string1','string2','string3')
SELECT REPLACE('STRING FUNCTION','STRING','SQL') returns       SQL Function
Returns NULL if any one of the arguments is NULL.
LEFT : Returns left part of a string with the specified number of characters counting from left.LEFT function is used to retrieve portions of the string.
Syntax: LEFT(string,integer)
SELECT LEFT('STRING FUNCTION', 6) returns  STRING
RIGHT : Returns right part of a string with the specified number of characters counting from right.RIGHT function is used to retrieve portions of the string.
Syntax: RIGHT(string,integer)
SELECT RIGHT('STRING FUNCTION', 8) returns FUNCTION
LTRIM : Returns a string after removing leading blanks on Left side.(Remove left side space or blanks)
Syntax: LTRIM(string)
SELECT LTRIM('   STRING FUNCTION') returns       STRING FUNCTION
RTRIM : Returns a string after removing leading blanks on Right side.(Remove right side space or blanks)
Syntax: RTRIM( string )
SELECT RTRIM('STRING FUNCTION   ') returns       STRING FUNCTION
REVERSE : Returns reverse of a input string.
Syntax: REVERSE(string)
SELECT REVERSE('STRING FUNCTION') returns       NOITCNUF GNIRTS
REPLICATE : Repeats a input string for a specified number of times.
Syntax: REPLICATE (string, integer)
SELECT REPLICATE('FUNCTION', 3) returns       FUNCTIONFUNCTIONFUNCTION
SPACE : Returns a string of repeated spaces. The SPACE function is an equivalent of using REPLICATE function to repeat spaces.
Syntax: SPACE ( integer) (If integer is negative, a null string is returned.)
SELECT ('STRING') + SPACE(1) + ('FUNCTION') returns       STRING FUNCTION
SUBSTRING : Returns part of a given string.
SUBSTRING function retrieves a portion of the given string starting at the specified character(startindex) to the number of characters specified(length).
Syntax: SUBSTRING (string,startindex,length)
SELECT SUBSTRING('STRING FUNCTION', 1, 6) returns       STRING
SELECT SUBSTRING('STRING FUNCTION', 8, 8) returns       FUNCTION

DELETE Command
To delete the record fro a table SQL provides a delete statement. General syntax is:-

DELETE FROM <table_name> [WHERE <condition>];
e.g. DELETE FROM student WHERE city = ‘Jammu’;
This command deletes all those records whose city is Jammu.
NOTE:              It should be kept in mind that while comparing with the string type values lowercase and uppercase letters are treated as different. That is ‘Jammu’ and ‘jammu’ is different while comparing.

UPDATE Command
To update the data stored in the data base, UOPDATE command is used.

e. g.     UPDATE student SET marks = marks + 100;
Increase marks of all the students by 100.
e. g.     UPDATE student SET City = ‘Udhampur’ WHERE city = ‘Jammu’;
changes the city of those students to Udhampur whose city is Jammu.

We can also update multiple columns with update command, like
e. g.     UPDATE student set marks = marks + 20, city = ‘Jalandhar’
WHERE city NOT IN (‘Jammu’,’Udhampur’);

CREATE VIEW Command
In SQL we can create a view of the already existing table that contains specific attributes of the table.
e. g. the table student that we created contains following fields:
Student (Roll_no, Name, Marks, Class, City)
Suppose we need to create a view v_student that contains Roll_no,name and class of student table, then Create View command can be used:
CREATE VIEW v_student AS SELECT Roll_no, Name, Class FROM student;
The above command create a virtual table (view) named v_student that has three attributes as mentioned and all the rows under those attributes as in student table.

We can also create a view from an existing table based on some specific conditions, like
CREATE VIEW v_student AS SELECT Roll_no, Name, Class FROM student WHERE City <>’Jammu’;
The main difference between a Table and view is that
A Table is a repository of data. The table resides physically in the database.
A View is not a part of the database's physical representation. It is created on a table or another view. It is precompiled, so that data retrieval behaves faster, and also provides a secure accessibility mechanism.

ALTER TABLE Command
In SQL if we ever need to change the structure of the database then ALTER TABLE command is used. By using this command we can add a column in the existing table, delete a column from a table or modify columns in a table.

Adding a column
The syntax to add a column is:-

ALTER TABLE table_name
ADD column_name datatype;

e.g ALTER TABLE student ADD(Address varchar(30));
The above command add a column Address to the table atudent.
If we give command
SELECT * FROM student;
The following data gets displayed on screen:

Roll_no
Name
Class
Marks
City
Address
101
Rohan
XI
400
Jammu

102
Aneeta Chopra
XII
390
Udhampur

103
Pawan Kumar
IX
298
Amritsar

104
Rohan
IX
376
Jammu

105
Sanjay
VII
240
Gurdaspur

113
Anju Mahajan
VIII
432
Pathankot


Note that we have just added a column and there will be no data under this attribute. UPDATE command can be used to supply values / data to this column.

Removing a column

ALTER TABLE table_name 
DROP COLUMN column_name;

e.g       ALTER TABLE Student
DROP COLUMN Address;
The column Address will be removed from the table student.

DROP TABLE Command
Sometimes you may need to drop a table which is not in use. DROP TABLE command is used to Delete / drop a table permanently. It should be kept in mind that we can not drop a table if it contains records. That is first all the rows of the table have to be deleted and only then the table can be dropped. The general syntax of this command is:-
DROP TABLE <table_name>;
e.g       DROP TABLE student;
This command will remove the table student


SECTION B: CONCEPT BASED QUESTIONS
(Very Short Answer questions 1 & 2 Marks)



Q2. Define the terms:
i. Database Abstraction
ii. Data inconsistency
iii. Conceptual level of database implementation/abstraction
iv. Primary Key
v. Candidate Key
vi. Relational Algebra
vii Domain
Ans
i. Database Abstraction
Ans: Database system provides the users only that much information that is required
by them, and hides certain details like, how the data is stored and maintained in
database at hardware level. This concept/process is Database abstraction.
ii. Data inconsistency
Ans: When two or more entries about the same data do not agree i.e. when one of
them stores the updated information and the other does not, it results in data
inconsistency in the database.
iii. Conceptual level of database implementation/abstraction
Ans: It describes what data are actually stored in the database. It also describes the
relationships existing among data. At this level the database is described logically in
terms of simple data-structures.
iv. Primary Key
Ans : It is a key/attribute or a set of attributes that can uniquely identify tuples within the relation.
v. Candidate Key
Ans : All attributes combinations inside a relation that can serve as primary key are candidate key as they are candidates for being as a primary key or a part of it.
vi. Relational Algebra
Ans : It is the collections of rules and operations on relations(tables). The various operations are selection, projection, Cartesian product, union, set difference and intersection, and joining of relations.
vii. Domain
Ans : it is the pool or collection of data from which the actual values appearing in a given column are drawn.

SECTION C: HOTS

Q1      Write SQL commands for (i) to (viii) on the basis of relations given below:


BOOKS

book_id       Book_name        author_name            Publishers      Price   Type     qty
k0001           Let us C              Sanjay mukharjee       EPB                 450      Comp  15
p0001              Genuine             J. Mukhi                    FIRST PUBL. 755      Fiction  24
m0001             Mastering c++   Kanetkar                   EPB                 165      Comp  60
n0002              Vc++ advance    P. Purohit                 TDH                250      Comp  45
k0002              Near to heart      Sanjeev                     FIRST PUBL. 350      Fiction  30

ISSUED
Book_ID
Qty_Issued
L02
13
L04
5
L05
21
i. To show the books of FIRST PUBL Publishers written by P.Purohit.
ii. To display cost of all the books written for FIRST PUBL.
iii. Depreciate the price of all books of EPB publishers by 5%.
iv. To display the BOOK_NAME,price of the books whose more than 3 copies have been issued.
v. To show total cost of books of each type.
vi. To show the detail of the most costly book.

Answers:
Q1.
Ans i: select * from books where publishers=’FIRST PUBL’
Ans ii: select sum(price*qty) from books where publishers=’FIRST PUBL’;
Ansiii: update books set price=price-0.5*price where publishers=’EPB’;
Ans iv: select BOOK_NAME,price from books, issued where
books.book_id=issued.book_id and quantity_issued>3;
Ans v: select sum(price*qty) from books group by type;
Ans vi: select * from books where price=(select max(price) from books));

Q2. Write SQL commands for (a) to (f) and write output for (g) on the basis of PRODUCTS relation given below:

PRODUCT  TABLE
PCODE PNAME       COMPANY  PRICE  STOCK  MANUFACTURE    WARRANTY
P001    TV                        BPL            10000           200     12-JAN-2008                 3
P002    TV                       SONY         12000           150        23-MAR-2007            4
P003    PC                      LENOVO    39000           100        09-APR-2008              2
P004    PC                      COMPAQ    38000           120        20-JUN-2009              2
P005    HANDYCAM                SONY         18000            250       23-MAR-2007            3
a) To show details of all PCs with stock more than 110.
b) To list the company which gives warranty for more than 2 years.
c) To find stock value of the BPL company where stock value is sum of the products of price and stock.
d) To show number of products from each company.
e) To count the number of PRODUCTS which shall be out of warranty on 20-NOV-2010.
f) To show the PRODUCT name which are within warranty as on date.
g). Give the output of following statement.
(i) Select COUNT(distinct company) from PRODUCT.
(ii) Select MAX(price)from PRODUCT where WARRANTY<=3

Answers:
Ans a: select * from products where pname=’TV’ and stock>110;
Ans b: select company from products where warranty>2;
Ans c: select sum(price*stock) from PRODUCTS where company=’BPL’;
Ans d: select company,COUNT(*) from products group by company;
Ans e: select count(*) from products where (‘20-NOV-2010’- manufacture)/365>warranty;
Ans f: select pname from products where (sysdate- manufacture)/365<warranty;

Ansg (i): 4
Ans(ii): 39000

Some practice questions from Database and SQL :

2 marks questions
   1.  What is relation? What is the difference between a tuple and an attribute?
2.  Define the following terminologies used in Relational Algebra:                       
(i) selection     (ii) projection     (iii) union      (iv) Cartesian product
3.  What are DDL and DML?
4.  Differentiate between primary key and candidate key in a relation?
5.  What do you understand by the terms Cardinality and Degree of a relation in relational  database?
6. Differentiate between DDL and DML.  Mention the 2 commands for each caterogy.

6 marks questions
   1.
Table : SchoolBus
Rtno
Area_overed
Capacity
Noofstudents
Distance
Transporter
Charges
1
Vasant kunj
100
120
10
Shivamtravels
100000
2
Hauz Khas
80
80
10
Anand travels
85000
3
Pitampura
60
55
30
Anand travels
60000
4
Rohini
100
90
35
Anand travels
100000
5
Yamuna Vihar
50
60
20
Bhalla Co.
55000
6
Krishna Nagar
70
80
30
Yadav Co.
80000
7
Vasundhara
100
110
20
Yadav Co.
100000
8
Paschim Vihar
40
40
20
Speed travels
55000
9
Saket
120
120
10
Speed travels
100000
10
Jank Puri
100
100
20
Kisan Tours
95000
(b)    To show all information of students where capacity is more than the no of student in order of rtno.
(c)     To show area_covered for buses covering more than 20 km., but charges less then 80000.
(d)    To show transporter wise total no. of students traveling.
(e)     To show rtno, area_covered and average cost per student for all routes where average cost per student is - charges/noofstudents.
(f)     Add a new record with following data:
            (11, “ Moti bagh”,35,32,10,” kisan tours “, 35000)   
(g)    Give the output considering the original relation as given:
       (i)  select sum(distance) from schoolbus where transporter= “ Yadav travels”;
       (ii) select min(noofstudents) from schoolbus;
       (iii) select avg(charges) from schoolbus where transporter= “ Anand  travels”;
(i)        select distinct transporter from schoolbus;

2.
TABLE : GRADUATE
S.NO
NAME
STIPEND
SUBJECT
AVERAGE
DIV.
1
KARAN
400
PHYSICS
68
I
2
DIWAKAR
450
COMP. Sc.
68
I
3
DIVYA
300
CHEMISTRY
62
I
4
REKHA
350
PHYSICS
63
I
5
ARJUN
500
MATHS
70
I
6
SABINA
400
CEHMISTRY
55
II
7
JOHN
250
PHYSICS
64
I
8
ROBERT
450
MATHS
68
I
9
RUBINA
500
COMP. Sc.
62
I
10
VIKAS
400
MATHS
57
II
(a)                                                                                                   List  the names of those students who have obtained DIV 1 sorted by NAME.
(b)         Display  a report, listing NAME, STIPEND, SUBJECT  and amount of stipend received in a year assuming that the STIPEND  is paid every month.
(c)          To count the number of students who are either PHYSICS or COMPUTER SC graduates.
(d)         To insert a new row in the GRADUATE table:    11,”KAJOL”, 300, “computer sc”, 75, 1
   (e)     Give the output of following sql statement based on table GRADUATE:
(i)     Select MIN(AVERAGE) from GRADUATE where SUBJECT=”PHYSICS”;
(ii)   Select SUM(STIPEND) from GRADUATE WHERE div=2;
(iii) Select AVG(STIPEND) from GRADUATE where AVERAGE>=65;
(iv) Select COUNT(distinct SUBDJECT)  from GRADUATE;
(f)     Assume that there is one more table GUIDE in the database as shown below:
Table: GUIDE
MAINAREA
ADVISOR
PHYSICS
VINOD
COMPUTER SC
ALOK
CHEMISTRY
RAJAN
MATHEMATICS
MAHESH






g) What will be the output of the following query:
SELECT NAME, ADVISOR FROM GRADUATE,GUIDE WHERE SUBJECT= MAINAREA;

3.Write SQL command for (i)  to (vii) on the basis of the table SPORTS

   Table: SPORTS
Student NO
Class
Name
Game1
Grade
Game2
Grade2
10
7
Sammer
Cricket
B
Swimming
A
11
8
Sujit
Tennis
A
Skating
C
12
7
Kamal
Swimming
B
Football
B
13
7
Venna
Tennis
C
Tennis
A
14
9
Archana
Basketball
A
Cricket
A
15
10
Arpit
Cricket
A
Atheletics
C

(a)    Display the names of the students who have grade ‘C’  in either Game1 or Game2 or both.
(b)   Display the number of students getting grade ‘A’ in Cricket.
(c)    Display the names of  the students who have same game for both Game1 and Game2.
(d)    Display the games taken up by the students, whose name starts with ‘A’.
(e)     Assign a value 200 for Marks for all those who are getting grade ‘B’ or  grade ‘A’ in both Game1 and  Game2.
(f)    Arrange the whole table in the alphabetical order of Name.
(g)    Add a new column named ‘Marks’. 
(h)    
4. Write SQL command for (i)  to (vii) on the basis of the table Employees & EmpSalary
Table: Employees
h
Empid
Firstname
Lastname
Address
City
010
Ravi
Kumar
Raj nagar
GZB
105
Harry
Waltor
Gandhi nagar
GZB
152
Sam
Tones
33 Elm St.
Paris
215
Sarah
Ackerman
440 U.S. 110
Upton
244
Manila
Sengupta
24 Friends street
New Delhi
300
Robert
Samuel
9 Fifth Cross
Washington
335
Ritu
Tondon
Shastri Nagar
GZB
400
Rachel
Lee
121 Harrison St.
New York
441
Peter
Thompson
11 Red Road
Paris
                                                            Table: EmpSalary
Empid
Salary
Benefits
Designation
010
75000
15000
Manager
105
65000
15000
Manager
152
80000
25000
Director
215
75000
12500
Manager
244
50000
12000
Clerk
300
45000
10000
Clerk
335
40000
10000
Clerk
400
32000
7500
Salesman
441
28000
7500
salesman

Write the SQL commands for the following :
(i)  To show firstname,lastname,address and city of all employees living in paris.          
(ii)To display the content of Employees table in descending order of Firstname.          
(iii) To display the firstname,lastname and total salary of all managers from the tables Employee and empsalary , where total salary is calculated as salary+benefits.         
(iv)             To display the maximum salary among managers and clerks from the table Empsalary.
  Give the Output of following SQL commands:                                                       
(i)      Select firstname,salary from employees ,empsalary where designation = ‘Salesman’  and Employees.empid=Empsalary.empid;
(ii)    Select count(distinct designation) from empsalary;
(iii)  Select designation, sum(salary) from empsalary group by designation having count(*)>2;

(iv)  Select sum(benefits) from empsalary where designation =’Clerk’;

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