Transcript
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1.B.NAGARAJU2.SRIDHAR REDDY3.P.SHAILAJA4.PADMAVATHI5.QURRATHUL AIN HAROON
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Week 1: E-R Model
E-R Model describes data as entities, relationships and attributes .The ER-Model is
important preliminary for its role in database design. ER Model is usually shown pictorially
using entity relationship diagrams.
Some basic concepts of E-R model are described below.
Entity: The object in the ERModel represents as an entity. Entity is the thing in the real world
with an independent existence. In ER modeling an entity is shown as a rectangle and name in it.
e.g.
Entity_Name
E-R Diagram: Shows the entities and relations pictorially. That is how the entities are relatedamong themselves.
e.g.Entity_1 Entity_2
Relates
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Attributes:
The properties that characterize an entity set are called its attributes. An attribute is
referred to by the terms data items, data element, data field item.
Ex: attributes for bus entity and ticket entity.
Types of attributes: There are different types of attributes
Domain Attribute: The set of possible values for an attribute is called the domain of the
attribute.
Key Attribute: The attribute (or combination of attributes) that is unique for every entity
instance.
Eg: The account number of account
The employee id of an employee
Simple Attribute:The attribute which cantbe broken-down into small components.
Divorced
Marital status
Single
Married Widowed
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Composite Attribute: The attribute which can be split into components.
Single valued Attribute: The attribute can take on only a single value for each entity instance.
Multi-valued Attribute: An attribute that may take more than one value for a given entity. It is
represented with double ellipse.
Stored Attribute: Attribute that need to be stored permanently.
Eg: Name of employee
Addr
Hno
Street City
Pincode
Day
Month
Year
Date
AGE
Phno
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Derived Attribute: Attribute that can be calculated based on other attributes. It is represented
with dotted lines.
Keys:
Candidate Key: It can be defined as an attribute or combination of attributes that identifiesthe record uniquely but none of its proper subsets can identify the record uniquely.
Primary Key: A candidate key that is used by the database designer for unique identificationof each row in a table is known as primary key. A primary key consists of one or more
attributes of the table.
EMP
Current
Date
Joining
Date
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Partial key: A weak entity type normally has a partial key which is the set of attributes thatcan uniquely identify weak entity that are related to the same owner entity.
Super Key: This can be defined as the combination of all keys.
Foreign key: A foreign key (FK) is an attribute or combination of attributes that is used toestablish and enforce a link between the data in two tables.
In a foreign key reference, a link is created between two tables when the column or columns
that hold the primary key value for one table are referenced by the column or columns in
another table. This column becomes a foreign key in the second table.
Composite key: Sometimes it requires more than one attribute to uniquely identify an entity. Aprimary key that made up of more than one attribute is known as a composite key.
For our database we identify entities as below.
o Bus: Primary key: BusNo and ServiceNoo Passenger: Primary key: PassengerId, Foreign key: TicketIdo Ticket: Primary key: TicketId, Foreign key: BusNoo Reservationo Cancellation
Week 2: Concept Design with E-R Model
Relationship: A relationship type between two entity types defines the set of all associations
between these entity types
Each instance of the relationship between members of these entity types is called a relationship
instance.
e.g.Customer creates an acount in bank, here creating account is a relation between Customer
and Bank entity.
Entity diagrams with Attributes:
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Customer Entity:
Account Entity:
Bank Entity:
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Administration Entity:
Deposit Entity:
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Withdraw Entity:
Transfer Entity:
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Cardinalities: Cardinality relationships can have different connectivity
A ratio, such as 1:1, 1: N, M: 1, and M: N, gives a cardinality constraint or numeric restriction
on the possible relationships.
One-to-one (1:1): relationship from entity type S to entity type T is one in which an entity from
S is related to at most one entity from T and vice versa.
One-to-many (1: N): relationship from entity type S to entity type T is one in which two or more
entities from T can be related to an entity from S.
Many-to- One (M: 1): relationship from entity type S to entity type T is one in which an entity
from S can be related to two or more entities from T.
Many-to-many (M:N):relationship from entity type S to entity type T is one in which an entity
from S can be related to two or more entities from T, and an entity from T can be related to two
or more entities from
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E R Diagrams :
Create account
Close account
Check Balance
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Deposit
Withdraw
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Transfer
Management
Bank Management System
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Types of Entities:
There are 2 types of entities.
Strong Entity:A Strong Entity is one that exists on its own, independent of other entities. An
entity set that has primary key is termed as strong entity set.
Weak Entity: A weak entity can be identified uniquely only by considering the primary key of
another (owner) entity. Owner entity set and weak entity set must participate in a one-to-many
relationship set (one owner, many weak entities).Weak entity set must have total participation in
this identifying relationship set.
Types of Relationships:
Aggregation: Aggregation refers to an abstraction in which a relationship between objects is
regarded as a higher-level object.
Generalization: It is a relationship that exists between a high level entity set and one or more
lower level entity set.
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Specialization: Is the process of defining a set of subclasses of a super class .The set of
subclasses is based upon some distinguishing characteristics of the entities in the super class.
Super class/subclass relationships and specialization can be diagrammatically represented in ER
diagrams.
Week 3: Relational Model
Relation can be defined as a set of tuples. Here all tuples in a relation must be distinct, thatis no two tuple can have the same combination of values for all their attributes.
Schema is the basic design of a relation i.e. nothing but a table design.
In our database the schema of tables would be as follows:
Customer table schema:
cust_id cust_name Age DOB Gender Ph_no Address
Customer
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cust_id: number(11) PRIMARY KEY
cust_name: varcharchar2(20)
Age : number(2)
Gender : varchar2(6)
DOB : date
Ph_no : number(10)
Address : varchar2(50)
Account table schema:
Account_no Account_type Date_of_creating_account Date_of_closing_account
AccountAccoun_ no: varcha2(10) PRIMARY KEY
Account_type: varchar2(10)
Date_o_creating_account : dateDate_of_closing_account : date
Bank table schema:
IFSC Bank_name Branch Ph_no Address
Bank
IFSC : number(10) PRIMARY KEY
Bank_ name : varchar2(20)
Branch : varchar2(20)
Ph_no: number(10)
Address: varchar2(30)
Administration Schema:
Emp_id Emp_name Designation Department
Administration
Emp_id : number(10) PRIMARY KEY
Emp_name: varchar2(25)
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Designation: varchar2(15)
Department: varchar2(15)
Deposit Schema:
Account_no Cust_id Cust_name Date_of_deposit Amount
Deposit
Account_no : varchar(11) foreign key(Account_no) reference to Account(Account_no)
Cust_id number(11) foreign key(cust_id) reference to Customer(Cust_id)
Date_of_deposit : date,
Cust_name : varchar2(30)Amount: number(11)
Withdraw schema:
Account_no Cust_id Cust_name Date_of_withdraw Amount
Withdraw
Account_no : varchar(11) foreign key(Account_no) reference to Account(Account_no)
Cust_id number(11) foreign key(cust_id) reference to Customer(Cust_id)
Cust_name : varchar2(20)
Date_of_withdraw :date
Accoun_ no : number(11) references customer(cust_id)
Amount: number(11)
Tr ansfer schema:
Cust_name Date_of_transfer Amount
TO_ACCOUNT
_NO
FROM_ACCOUT_NO
Transfer
Cust_name : varchar2(20)
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Date_of_transfer :date
To account:number(10)
From account:number(11)
Week4: NormalizationDatabase normalization is a technique for designing relational database tables to
minimize duplication of information and, in doing so , to safeguard the database against certain
types of logical or structural problems namely data anomalies.
Normal Forms:
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1st Normal Form: 1NF requires that the values in each column of a table are atomic. Byatomic we mean that there are no sets of values within a column.
So the relation gets normalized under the 1ST normal form as below by splitting therecord into two records.
Cust_id cust_name Age gender DOB Phno ADDRESS
1010054211
Preetam 30 M 03-04-1981 9949887766
9873838272
#1-345, Narayanaguda,
Hyderabad
1010054212
Sreeram 27 M 02-12-1984 9440565656 #3-11/A, Bowenpally,
Hyderabad
1010054213
Shankar 32 M 23-07-1979 9440440440 #4-12/1/2, Malakpet,
Hyderabad
1010054214
Ashutosh 31 M 05-08-1980 9866324324 #1-234, Chikkadpally,
Hyderabad
1010054215
Aruna 20 F 04-05-1991 8011478531 #1-234, Chikkadpally,
Hyderabad
1010054216
Anvesh 20 M 23-01-1991 8011478532 #2-89/A/1, Santoshnar,
Hyderabad
1010054217
Saidatta 31 M 03-12-1982 8066952326 #2-45/6/2, Lungarhouse,
Hyderabad
Normalized Relation:
Cust_id cust_name Age gender DOB PHN_N0 ADDRESS
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1010054211 Preetam 30 M
03-04-1981 9949887766 #1-345, Narayanaguda,
Hyderabad
1010054211 Preetam 30 M
03-04-1981 9873838272 #1-345, Narayanaguda,
Hyderabad
1010054212 Sreeram 27 M 02-12-19849440565656
#3-11/A, Bowenpally,
Hyderabad
1010054213 Shankar 32 M
23-07-1979 9440440440 #4-12/1/2, Malakpet,
Hyderabad
1010054214
Ashutosh
31 M
05-08-1980 9866324324 #1-234, Chikkadpally,
Hyderabad
1010054215Aruna
20 F23-01-1991 8011478531 #1-234, Chikkadpally,
Hyderabad
1010054216
Anvesh
20 M
23-01-1991 8011478532 #2-89/A/1, Santoshnar,
Hyderabad
1010054217
Saidatta
29 M
03-12-1982 8066952326 #2-89/A/1, Santoshnar,
Hyderabad
2nd Normal Form: where the 1NF deals with atomicity of data, the 2NF deals withrelationships between composite key columns and non-key columns. To achieve 2NF thetables should be in 1NF. The 2NF any non-key columns must depend on the entire
primary key. In case of a composite primary key, this means that non-key column cant
depend on only part of the composite key.
That is noPartial Functional Dependencies shoud exist. In our entities Deposit entity is having the keys with the combination of
account_no and cust_id . But there exists the following partial functional
dependency:
ACCOUNT_NO CUST_ID,CUST_NAME
So in order to avoid this we normalize the Deposit table as below.
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ACCOUNT_NO Cust_id Cust_name Date of deposit Amount
1234A21012 1010054211 Preetam 12-09-2010 120000
123J90187K 1010054212 Sreeram 09-12-2010 130000
123FGT5678 1010054213 Shankar 22-08-2010 3000
1234M4567D 1010054214 Ashutosh 23-04-2010 40000
Normalized Relations:
ACCOUNT_NO CUST_ID,CUST_NAME
Account_no Cust_id Cust_name
1234A21012 1010054211 Preetam
123J90187K 1010054212 Sreeram
123FGT5678 1010054213 Shankar
1234M4567D 1010054214 Ashutosh
ACCOUNT_NO DATE OF DEPOSIT,AMOUNT
ACCOUNT_NO DATE OF
DEPOSIT
AMOUNT
1234A21012 12-09-2010 120000
123J90187K 09-12-2010 130000
123FGT5678 22-08-2010 3000
1234M4567D 23-04-2010 40000
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3rd Normal Form: 3NF requires that all columns depend directly on the primary key.Tables violate the third normal form when one column depends an another column,
which in turn depends on the primary key(transitive dependency). One way to identify
transitive dependency is to look at your tables and see if any columns would require
updating if another column in the table was updated. If such a column exists, it probably
violates 3NF.
In our entities CUSTOMER entity faces the Transitive Dependency problem.Thefollowing transitive dependencies exist:
CUST_ID DOB
DOB AGE
To eliminate this problem the relation is normalized as follows.
Cust_id cust_name Age gender DOB PHN_N0 ADDRESS
1010054211 Preetam 30 M
03-04-1981 9873838272 #1-345, Narayanaguda,
Hyderabad
1010054212 Sreeram 27 M 02-12-19849440565656
#3-11/A, Bowenpally,
Hyderabad
1010054213 Shankar 32 M
23-07-1979 9440440440 #4-12/1/2,Malakpet,
Hyderabad
1010054214
Ashutosh
31 M
05-08-1980 9866324324 #1-234, Chikkadpally,
Hyderabad
1010054215
Aruna
20 F
23-01-1991 8011478531 #1-234, Chikkadpally,
Hyderabad
1010054216
Anvesh
20 M
23-01-1991 8011478532 #2-89/A/1, Santoshnar,
Hyderabad
1010054217
Saidatta
29 M
03-12-1982 8066952326 #2-89/A/1, Santoshnar,
Hyderabad
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CUST_ID DOB
DOB AGE
Cust_id cust_name gender DOB PHN_N0 ADDRESS
1010054211 Preetam M
03-04-1981 9873838272 #1-345,
Narayanaguda,
Hyderabad
1010054212 Sreeram M02-12-1984
9440565656
#3-11/A,
Bowenpally,
Hyderabad
1010054213 Shankar M
23-07-1979 9440440440 #4-
12/1/2,Malakpet,
Hyderabad
1010054214
Ashutosh
M
05-08-1980 9866324324 #1-234,
Chikkadpally,
Hyderabad
Cust_id cust_name gender Age PHN_N0 ADDRESS
1010054211 Preetam M 30
9873838272 #1-345,
Narayanaguda,
Hyderabad
1010054212 Sreeram M 27
9440565656
#3-11/A,
Bowenpally,
Hyderabad
1010054213 Shankar M 32
9440440440 #4-
12/1/2,Malakpet,
Hyderabad
1010054214Ashutosh
M 319866324324 #1-234,
Chikkadpally,
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Boyce Codd Normal Form:
Takes care of candidate keys.A relation in a Boyce Codd normal form iff every determinant in
the relation is a candidate key. That is this normal form eliminates the overlapping candidate
keys.
Week 5: Installation of MySQL
Download MySQL software from the web sitehttp://www.soft82.com/get/download/mysqlworkbench
Save the file and then run the set up after downloading. Before running Mysqlworkbench we have to install .net frameworkof version 2.0 (or) higher.Now we have
to download the .net frameworkfromthe web site:
http://www.brothersoft.com/net.framework-download-112623.html
Before installing the .net frame workwe have to download the Windows Installer 3.1to support the .net frame workfrom the web site:
http://www.microsoft.com/downloads/
Windows I nstaller I nstallation:
After downloading the Windows Installer 3.1 software run the set up.
Hyderabad
http://www.soft82.com/get/download/mysqlworkbenchhttp://www.soft82.com/get/download/mysqlworkbenchhttp://www.brothersoft.com/net.framework-download-112623.htmlhttp://www.brothersoft.com/net.framework-download-112623.htmlhttp://www.microsoft.com/downloads/http://www.microsoft.com/downloads/http://www.microsoft.com/downloads/http://www.brothersoft.com/net.framework-download-112623.htmlhttp://www.soft82.com/get/download/mysqlworkbench7/30/2019 DDB Lab Record
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Start the set up wizard and clickNEXT. Next click on the LICENSE agreement to ACCEPT and again clickNEXT. Now updating the system process starts. After installing the Windows Installer 3.1 software click on FINISH.
.NET Framework I nstallation:
Now start the installation process of.net frame workto install Mysql Work Bench. Start the set up wizard. Next ACCEPT the LICENCE AGREEMENT and clickINSTALL. Now the DOWNLOAD and INSTALLATIONprocess starts.
Before INSTALLATION, it first DOWNLOADS the components of.net frame workfrom internet and then starts the INSTALLATIONprocess.
After completing the INSTALLATIONprocess click on EXIT and the setup iscompleted.
MySQL Work Bench Installation:
Run the setup. It will show setup wizard like the following window.
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Now click on NEXT to start the INSTALLATION process. Now select the Setup Type.
Complete
Custom Click on COMPLETE to install all program feature and clickNEXT. Now click on INSTALL to start the installation process. Now the INSTALLATIONprocess starts. Now click on FINISH to complete the setup wizard. The overall INSTALLATIONprocess has been completed. And we can perform
QUERIESby using MySQL Work Bench.
We are going to execute the queries by using MySQL server.
MySQL Server I nstallation:
Step 1: Download mysql server essential from the websitewww.mysql.com/downloads and savethe .exe file.
http://www.mysql.com/downloads%20and%20save%20thehttp://www.mysql.com/downloads%20and%20save%20thehttp://www.mysql.com/downloads%20and%20save%20thehttp://www.mysql.com/downloads%20and%20save%20thehttp://www.mysql.com/downloads%20and%20save%20thehttp://www.mysql.com/downloads%20and%20save%20the7/30/2019 DDB Lab Record
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Steps 2&3: Double click on the mysql.exe file to start installation.
Steps 4&5:
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:
Steps: 6&7
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Steps 8&9:
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Steps: 10&11
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Step: 12&13
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Steps: 14&15
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Steps: 16&17
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Steps: 18&19
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DDL Commands
Data Definition Language commands are used for the basic operations like creating,
deleting etc..of database tables. There are many DDL Commands some of them are described
below.
Create: It is used to create the tables in the database.On a specific database a new table is
created using create command.
Syntax: CREATE TABLE TABLE_NAME(ATTRIBUTE_LIST);
e.g. CREATE TABLE CUSTOMER (CUST_ID NUMBER(11) PRIMARY KEY,
CUST_NAME VARCHAR2(20), AGE NUMBER(2) ,DOB DATE, GENDER VARCHAR2(1),
PH_NO NUMBER(10),ADDRESS VARCHAR2(40));
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Above example creates a table called CUSTOMER. With DESC command we can view
the basic table scheme.
DESC CUSTOMER;
Above command shows the table design as follows.
* Lets consider our DB is having a table called DEMO.
Rename: It is used to change the table name in the database.
Syntax: RENAME TABLE OLD_TABLE_NAME TO NEW_TABLE_NAME;
e.g. RENAME TABLE DEMO TO DEMO1;
Alter: It is used to add or remove the columns/fields from existing table.
Syntax: ALTER TABLE TABLE_NAME ADD/REMOVE FIELD_NAME;
e.g. ALTER TABLE DEMO1 ADD EXTRA_FIELD CHAR(10);
Above example adds one extra field EXTRA_FIELD to the DEMO1 table.
Truncate: It is used to delete only the records of the table in the database
Syntax: TRUNCATE TABLE TABLE_NAME;
e.g. TRUNCATE TABLE DEMO1;
Above command deletes all records of DEMO1 table.
Drop: It is used to delete the entire table from the database.
Syntax: DROP TABLE TABLE_NAME;
e.g. DROP TABLE DEMO1;
Field Type Null Key Default Extra
CUST_ID
CUST_NAME
AGE
DOB
GENDER
PH_NO
ADDRESS
NUMBER(11)
VARCHAR2(20)
NUMBER(2)
DATE
VARCHAR2(2)
NUMBER(10)
VARCHAR2(40)
NO
YES
YES
YES
YES
YES
YES
PRI
Not Null
Not Null
Not Null
Not Null
Not Null
Not Null
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Above table deletes DEMO1 table completely from database.
Week 6: DML Commands
Data Manipulation Language commands are used for access/ manage the data items in
database table. Some of the DML commamds are described below.
Insert: It is used to insert the data items in to the created tables in the database.
Syntax: INSERT INTO TABLE_NAME VALUES(values);
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e.g. INSERT INTO CUSTOMER VALUES (1010054211,Preetham,30,m,03-01-1981,
9949887766, #1-345, Narayanaguda, Hyderabad);
INSERT INTO CUSTOMER VALUES (1010054212,Sreeram,27,m,02-12-1984,
9440565656, #3-11/A, Bowenpally, Hyderabad);
Above 2 queries insert the values in CUSTOMER table.
Select: It is used to select/retrieve the data from the tables in the database.
Syntax: SELECT FIELD(S) FROM TABLE_NAME;
e.g. SELECT * FROM CUSTOMER;
Above command retrieves all the records of BUS table and shows like the follow.
Here * indicates all fields values. into two records.
CUST_ID
CUST_NA
ME
AG
E
GEND
ER DOB PHNO ADDRESS
1010054211Preetam 30 M 03-04-1981 9949887766 #1-345, Narayanaguda,
Hyderabad
1010054212
Sreeram 27 M 02-12-1984 9440565656 #3-11/A, Bowenpally,
Hyderabad
1010054213
Shankar 32 M 23-07-1979 9440440440 #4-12/1/2, Malakpet,
Hyderabad
1010054214
Ashutosh 31 M 05-08-1980 9866324324 #1-234, Chikkadpally,
Hyderabad
1010054215
Aruna 20 F 04-05-1991 8011478531 #1-234, Chikkadpally,
Hyderabad
1010054216
Anvesh 20 M 23-01-1991 8011478532 #2-89/A/1, Santoshnar,
Hyderabad
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1010054217
Saidatta 31 M 03-12-1982 8066952326 #2-45/6/2, Lungarhouse,
Hyderabad
Update: It is used to update the existing data within a table in the database.
Syntax: UPDATE TABLE_NAME SET FIELD=VALUE WHERE CONDITION;
e.g. UPDATE CUSTOMER SET CUST_NAME=nagaraju WHERE CUST_ID=1010054211;
Above query sets name nagaraju in CUST_NAME field in the 1010054211 CUST_ID
record in CUSTOMER table.
If we select the records from BUS table, it will show table as follows.
CUST_ID
CUST_NA
ME
AG
E
GEND
ER DOB PHNO ADDRESS
1010054211
Nagaraju 30 M 03-04-1981 9949887766 #1-345, Narayanaguda,
Hyderabad
1010054212
Sreeram 27 M 02-12-1984 9440565656 #3-11/A, Bowenpally,
Hyderabad
1010054213
Shankar 32 M 23-07-1979 9440440440 #4-12/1/2, Malakpet,
Hyderabad
1010054214
Ashutosh 31 M 05-08-1980 9866324324 #1-234, Chikkadpally,
Hyderabad
1010054215
Aruna 20 F 04-05-1991 8011478531 #1-234, Chikkadpally,
Hyderabad
1010054216
Anvesh 20 M 23-01-1991 8011478532 #2-89/A/1, Santoshnar,
Hyderabad
1010054217
Saidatta 31 M 03-12-1982 8066952326 #2-45/6/2, Lungarhouse,
Hyderabad
Delete: It is used to delete the records from the tables and the space will remain same in the
table with empty space.
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Syntax: DELETE TABLE_NAME WHERE CONDITION;
e.g. DELETE CUSTOMER WHERE CUST_ID=1010054217;
Above query will delete record related to the customer id is 1010054217 from CUSTOMER
table.
CUST_ID
CUST_NA
ME
AG
E
GEND
ER DOB PHNO ADDRESS
1010054211
Nagaraju 30 M 03-04-1981 9949887766 #1-345, Narayanaguda,
Hyderabad
1010054212
Sreeram 27 M 02-12-1984 9440565656 #3-11/A, Bowenpally,
Hyderabad
1010054213
Shankar 32 M 23-07-1979 9440440440 #4-12/1/2, Malakpet,
Hyderabad
1010054214
Ashutosh 31 M 05-08-1980 9866324324 #1-234, Chikkadpally,
Hyderabad
1010054215
Aruna 20 F 04-05-1991 8011478531 #1-234, Chikkadpally,
Hyderabad
1010054216Anvesh 20 M 23-01-1991 8011478532 #2-89/A/1, Santoshnar,
Hyderabad
DataBase Tables creation
Account Table
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MySQL> CREATE TABLE ACCOUNT (ACCOUNT_NO REFERENCES
CUSTOMER(CUST_ID), ACCOUNT_TYPE varchar2(10), IFSC REFERENCES
BANK(IFSC), DATE_OF_CREATING_ACCOUNT date,
DATE_OF_DELETING_ACCOUNT date, AMOUNT number(11));
MySQL> INSERT INTO ACCOUNT VALUES (1010054211, savings, 23239, 11-jan-
1998, NULL,15000);
MySQL> INSERT INTO ACCOUNT VALUES (1010054212, savings, 23239, 11-jun-
2003, NULL,14000);
MySQL> INSERT INTO ACCOUNT VALUES (1010054213, savings, 11007 30-jan-
2008, NULL,23000);
MySQL> SELECT * FROM ACCOUNT;
ACCOUNT_NO ACCOUNT
_TYPE
IFSC DATE_OF_CREA
TING_ACCOUN
T
DATE_OF_DE
LETING_ACC
OUNT
AMOUNT
1234A21012 Savings 23239 11-jan-1998 NULL 15,000
123J90187K Joint 23239 11-jun-2003 NULL 14,000
123FGT5678 Savings 11007 30-jan-2008 NULL 23,000
1234M4567D Current 11557 28-apr-2003 NULL 11,000
1234GH56710 Savings 45379 17-aug-2009 NULL 29,000
1230978GFW Savings 11007 15-sep-2006 NULL 15,000
Bank Table
MySQL> CREATE TABLE BANK(BANK_ NAME varchar2(20),
IFSC number(5) PRIMARYKEY, BRANCH varchar2(20), ADDRESS varchar2(30),
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PH_NO number(10));
MySQL> INSERT INTO BANK
VALUES(SBI,11007,Dilshuknagar,DSNR,9441808080);
MySQL> INSERT INTO BANK VALUES(SBI,11667,Gaddiannaram,Gaddiannaram ,
9440566022);
MySQL> INSERT INTO BANK VALUES(SBI,23239,BNReddy Nagar,BNReddy
Nagar, 9490997654);
MySQL> INSERT INTO BANK VALUES(SBI,45379,Uppal,Uppal Xroads,9030547823);
MySQL> SELECT * FROM BANK;
BANK_NAME IFSC BRANCH ADDRESS PH_NO
SBI 11007 Dilshuknagar DSNR 9441808080
SBI 11557 Gaddiannaram Gaddiannaram 9440566022
SBI 23239 BNReddy Nagar BNReddyNagar 9490997654
SBI 45379 Uppal Uppal X roads 9030547823
ADMINISTRATION TABLE
EMP_ID EMP_NAME DESIGNATION DEPARTMENT
10023 Nagaraju MANAGER
10052 Raghava Asst MANAGER
10072 Preetham CASHIER
10084 Nagarjuna
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DEPOSITE TABLE
ACCOUNT_NO Cust_id Cust_name Date of deposit Amount
1234A21012 1010054211 Preetam 12-09-2010 120000
123J90187K 1010054212 Sreeram 09-12-2010 130000
123FGT5678 1010054213 Shankar 22-08-2010 3000
1234M4567D 1010054214 Ashutosh 23-04-2010 40000
WITHDRAW TABLE
ACCOUNT_NO Cust_id Cust_name Date of
withdraw
Amount
1234A21012 1010054211 Preetam 12-09-2010 120000
123J90187K 1010054212 Sreeram 09-12-2010 130000
123FGT5678 1010054213 Shankar 22-08-2010 3000
1234M4567D 1010054214 Ashutosh 23-04-2010 40000
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TRANSFER TABLE
CUST_NAME TO_ACCOUNT_
NO
FROM_ACCOUNT
_NO
DATE_OF_TR
ANSFER
AMOUNT
Nagaraju 1010054211 1010054215 18-AUG-2009 5,000
Aruna 1010054215 1010054213 31-DEC-2009 4,000
Shankar 1010054213 1010054211 27-NOV-2007 8,000
Anvesh 1010054216 1010054213 19-JUL-2007 6,000
Week 7: Querying
Practicing the some queries
1. Display CUSTOMER NAME, CUSTOMER ID of all customers.MySQL> SELECT CUST_ID,CUST_NAME FROM CUSTOMER;
CUSTOMER_ID CUST_NAME
1010054211 Nagaraju
1010054212 Sreeram
1010054213 Shankar
1010054214 Ashutosh
1010054215 Aruna
1010054216 Anvesh
2. Display all the names of male customers.
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MySQL> SELECT CUST_NAME CUSTOMER WHERE GENDER=M;
CUST_NAME
Nagaraju
Sreeram
Shankar
Ashutosh
Anvesh
3. Display names and address of all the customers.MySQL> SELECT CUST_NAME,ADDRESS FROM CUSTOMER;
CUST_NAME ADDRESS
Nagaraju #1-345, Narayanaguda, Hyderabad
Sreeram #3-11/A, Bowenpally, Hyderabad
Shankar #4-12/1/2, Malakpet, Hyderabad
Ashutosh #1-234, Chikkadpally, Hyderabad
Aruna #1-234, Chikkadpally, Hyderabad
Anvesh #2-89/A/1, Santoshnar, Hyderabad
4. Display the IFSC, bank name, branch details for all banks.MySQL> SELECT BANK_NAME,IFSC,BRANCH FROM BANK;
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BANK_NAME IFSC BRANCH
SBI 11007 Dilshuknagar
SBI 11557 Gaddiannaram
SBI 23239 BNReddy Nagar
SBI 45379 Uppal
5. Display the account number of custmers whose names start with a and end withh.
MySQL> SELECT CUST_ID,CUST_NAME FORM CUSTOMER WHERE NAME
LIKE A%H;
CUST_ID CUST_NAME
1010054214 Ashuthosh
1010054216 Anvesh
6 . Find names of customers whose age is between 30 and 45.MySQL> SELECT CUST_NAME FROM CUSTMER WHERE AGE BETWEEN 30
AND 45;
CUST_NAME
Nagaraju
Ashuthosh
Shankar
7. Display all customer names whose name starts with a.MySQL>SELECT CUST_NAME FROM CUSTOMER WHERE NAME LIKEA%;
CUST_NAME
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Ashutosh
Aruna
Anvesh
8. Display sorted list of customer names.MySQL> SELECT CUST_NAME FROM CUSTOMER ORDER BY CUST_NAME;
CUST_NAME
Anvesh
Aruna
Ashutosh
Nagaraju
Preetam
Shankar
Sreeram
Week 8 and 9: Querying (Continued)
1. Write a query to display the customer name ,customer id, IFSCMySQL>SELECT C.CUST_ID,C.CUST_NAME,A.IFSC FROM ACCOUNT A,
CUSTOMER C WHERE C.CUST_ID=A.ACCOUNT_NO;
CUSTOMER_ID CUST_NAME IFSC
1010054211 Nagaraju 23239
1010054212 Sreeram 23239
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1010054213 Shankar 11007
1010054214 Ashutosh 11557
1010054215 Aruna 45379
1010054216 Anvesh 11007
2. Write a query to display account number where amount is less than 20,000.MySQL> SELECT ACCOUNT_NO FROM ACCOUNT WHERE AMOUNT SELECT COUNT(ACCOUNT_NO) FROM ACCOUNT HAVING
IFSC=11007;
COUNT(ACCOUNT_NO)
2
4. Display the account number, name, amount of all customers.MySQL>SELECT C.CUST_ID ACCOUNT_NO,C.CUST_NAME NAME,A.AMOUNT
FROM CUSTOMER C, ACCOUNT A WHERE A.ACCOUNT_NO=C.CUST_ID;
ACCOUNT_NO NAME AMOUNT
1010054211 Nagaraju 15,000
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1010054212 Sreeram 14,000
1010054213 Shankar 23,000
1010054214 Ashutosh 11,000
1010054215 Aruna 29,000
1010054216 Anvesh 15,000
5. Write a query to display the account numbers and IFSC where brach name isdilshuknagar.
MySQL>SELECT A.ACCOUNT_NO, A.IFSC FROM ACCOUNT A, BANK B
WHERE A.IFSC=B.IFSC AND B.BRANCH=Dilshuknagar;
ACCOUNT_NO IFSC
1010015F413 11007
10100DS54216 11007
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