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1. Consider the Insurance database given below. The primary keys are underlined and the

data types are specified:

PERSON (driver id #: String, name: string, address: string)

CAR (regno: string, model: string, year: int)

ACCIDENT (report-number: int, accd-date: date, location: string)

OWNS (driver-id #:string, Regno:string)

PARTICIPATED (driver-id: string, Regno:string, report-number:int, damage amount:int)(i) Create the above tables by properly specifying the primary keys and the foreign

keys.

create table person

(

driver_id varchar(10) primary key,

name varchar(10),

address varchar(20)

);

create table car

(

regno varchar(12) primary key,

model varchar(10),

year number(4)

);

create table accident

(

repno number(3) primary key,

accd_date date,

location varchar(12)

);

create table owns

(

driver_id varchar(10) references person(driver_id),

regno varchar(12) references car(regno),

primary key(driver_id,regno)

);

create table participated

(

driver_id varchar(10) references person(driver_id),

regno varchar(12) references car(regno),

repno number(3) references accident(repno),damage_amt number(10,2),

primary key(driver_id,regno,repno)

);

(ii) Enter at least five tuples for each relation.insert into person values

('dr1','salman','bapunagar, bombay');

insert into person values

('dr2','diganth','rrnagar, bangalore');

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insert into person values

('dr3','vijay','goripalya, mysore');

insert into person values

('dr4','darshan','gubbi, mandya');

insert into person values

('dr5','sudeep','jaynagar, bangalore');

insert into car values

('dl01-h1234','innova','2008');

insert into car values

('ka01-m1234','santro','2007');

insert into car values

('ka09-s1234','maruthi','2005');

insert into car values

('ka11-f1234','innova','2008');

insert into car values

('ka01-h1234','innova','2006');

insert into accident values

(3,'01-jan-2008','delhi');

insert into accident values

(6,'01-jun-2007','bangalore');

insert into accident values

(9,'01-nov-2007','tumkur');

insert into accident values

(12,'01-sep-2008','bangalore');

insert into accident values

(15,'01-apr-2006','bangalore');

insert into owns values('dr1','dl01-h1234');

insert into owns values

('dr2','ka01-m1234');

insert into owns values

('dr3','ka09-s1234');

insert into owns values

('dr4','ka11-f1234');

insert into owns values

('dr5','ka01-h1234');

insert into participated values

('dr1','dl01-h1234',3,704245);insert into participated values

('dr2','ka01-m1234',6,35653);

insert into participated values

('dr3','ka09-s1234',9,54323);

insert into participated values

('dr4','ka11-f1234',12,17439);

insert into participated values

('dr5','ka01-h1234',15,42174);

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(iii) Demonstrate how you

a. Update the damage amount to 25000 for the car with a specific Regno in theACCIDENT table with report number 12.

update participated

set damage_amt=25000

where repno=12;

b. Add a new accident to the database.insert into accident values

(18,'01-mar-2008','bangalore');

(iii) Find the total number of people who owned cars that were involved in accidentsin 2008.

select count(*) as cnt_acc

from accident,participated,personwhere

(

(participated.repno=accident.repno) and

(participated.driver_id=person.driver_id) and

(accident.accd_date like '%08')

);

(iv) Find the number of accidents in which cars belonging to a specific model wereinvolved.

select count(*) as cnt

from car

where model='innova';

(vi) Generate suitable reports.

(vii)Create suitable front end for querying and displaying the results.