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IST210 1
DATABASE DESIGN IIST 210: Organization of Data
IST210 2
I want a database to maintain departments in my company. Store information about my employees, their projects and assignments. I
want ….
$$$
User requirement
A database
YOUR JOB!
Data Modeling (Ch.4)
Database Design (Ch.5)
IST210 3
Two-Step Approach: Step 1
• Input: User requirement; Output: E-R Diagram• Use Entity-Relationship Diagram (E-R Diagram) to
capture all user requirements
Data Modeling (Ch.4)
Database Design (Ch.5)
IST210 4
Two-Step Approach: Step 2
Data Modeling (Ch.4)
Database Design (Ch.5)
• Input: E-R diagram; Output: A database• Transferring a data model to a relational database
• Entities Relations• Relationships Foreign keys and extra
IST210 5
Chapter Key Objective
Data Model (E-R Diagram) Relational Model (Tables)
1. Entity Table• Entity name Table name• Identifier Primary key• Attributes Attributes
2. Relationship Foreign key or a new table• Determined by the maximal cardinality
3. Specify the attribute properties• Data type• Key• Required
• Determined by the minimal cardinality• Remarks
IST210 6
E-R Diagram Tables
1. Entity Table• Entity name Table name• Identifier Primary key• Attributes Attributes
2. Relationship Foreign key or a new table3. Specify the attribute properties
IST210 7
Example
LockerID
LockerRoomLockerSize
EmployeeID
NameOfficeNumberOfficePhone
LOCKEREMPLOYEE1) Entity name Table name2) Identifier Primary key3) Attributes Attributes
EMPLOYEE(EmployeeID, Name, OfficeNumber, OfficePhone)LOCKER(LockerID, LockerRoom, LockerSize)
*primary key: use underline
IST210 8
E-R Diagram Tables
1. Entity Table• Entity name Table name• Identifier Primary key• Attributes Attributes
2. Relationship Foreign key or a new table• Determined by the maximal cardinality
3. Specify the attribute properties
IST210 9
Maximal Cardinality
• 1:1 (one-to-one) Relationship
• 1:N (one-to-many) Relationship
• N:M (many-to-many) RelationshipEntity A Entity B
Entity A Entity B
Entity A Entity B
IST210 10
1:1 Relationship
LockerID
LockerRoom
LockerSize
1 #2 10
5 #2 5
EMPLOYEE LOCKER
Relationship is not captured yet!
Add foreign key ?
Employee ID
Name OfficeNumber
OfficePhone
4 Tom 281 9182
EMPLOYEE(EmployeeID, Name, OfficeNumber, OfficePhone)LOCKER(LockerID, LockerRoom, LockerSize)
IST210 11
Employee ID
Name OfficeNumber
OfficePhone
LockerID
4 Tom 281 9182 1
LockerID
LockerRoom
LockerSize
1 #2 10
5 #2 5
EMPLOYEE LOCKER
Employee 4 owns locker 1; locker 1 belongs to employee 4.
Employee ID
Name OfficeNumber
OfficePhone
4 Tom 281 9182
LockerID
LockerRoom
LockerSize
EmployeeID
1 #2 10 4
5 #2 5
EMPLOYEE LOCKER
LockerID
LockerRoom
LockerSize
EmployeeID
1 #2 10 4
5 #2 5
EMPLOYEE LOCKER
Employee ID
Name OfficeNumber
OfficePhone
LockerID
4 Tom 281 9182 1
Option A
Option B
Option C
IST210 12
1:1 Relationship
EMPLOYEE(EmployeeID, Name, OfficeNumber, OfficePhone, LockerID)LOCKER(LockerID, LockerRoom, LockerSize)
EMPLOYEE(EmployeeID, Name, OfficeNumber, OfficePhone)LOCKER(LockerID, LockerRoom, LockerSize, EmployeeID)
Option A
Option B
Both options are correct. Option A is preferred.• If choosing Option A, since every employee has one locker, LockerID must
have a value in EMPLOYEE table.• If choosing Option B, since some lockers does not belong to anyone,
EmployeeID will be NULL value for some lockers and it takes extra space. (See the previous slide, Option B needs one more cell.)
*foreign key: italic
IST210 13
1:1 Relationship• The maximum cardinality determines how a relationship is
represented• 1:1 relationship
• The key from one relation is placed in the other as a foreign key• It does not matter which table receives the foreign key• We prefer the option taking less space
IST210 14
1:N Relationship
Employee ID
LastName
FirstName
OfficeNumber
OfficePhone
4 Green Tom 281 9182
LockerID
LockerRoom
LockerSize
1 #2 10
5 #2 5
EMPLOYEE LOCKER
Relationship is not captured yet! Add foreign key ?
LockerID
LockerRoomLockerSize
EmployeeID
NameOfficeNumberOfficePhone
LOCKEREMPLOYEE
Assume one-to-many relationship
IST210 15
Employee ID
Name OfficeNumber
OfficePhone
LockerID
4 Tom 281 9182 1, 5
LockerID
LockerRoom
LockerSize
1 #2 10
5 #2 5
EMPLOYEE LOCKER
Employee 4 owns locker 1 and 5.
Employee ID
Name OfficeNumber
OfficePhone
4 Tom 281 9182
LockerID
LockerRoom
LockerSize
EmployeeID
1 #2 10 4
5 #2 5 4
EMPLOYEE LOCKER
EMPLOYEE LOCKER
EmployeeID
Name OfficeNumber
OfficePhone
LockerID
4 Tom 281 9182 1, 5
Option A
Option B
Option C
LockerID
LockerRoom
LockerSize
EmployeeID
1 #2 10 4
5 #2 5 4
IST210 16
1:N Relationship
EMPLOYEE(EmployeeID, Name, OfficeNumber, OfficePhone)LOCKER(LockerID, LockerRoom, LockerSize, EmployeeID)
IST210 17
1:N Relationship• Like a 1:1 relationship, a 1:N relationship is saved by
placing the key from one table into another as a foreign key
• However, in a 1:N the foreign key always goes into the many-side of the relationship• The 1 side is called the parent• The N side is called the child
IST210 18
N:M Relationship
Employee ID
Name OfficeNumber
OfficePhone
4 Tom 281 9182
10 John 777 1829
LockerID
LockerRoom
LockerSize
1 #2 10
5 #2 5
EMPLOYEE LOCKER
Relationship is not captured yet! Add foreign key ?
LockerID
LockerRoomLockerSize
EmployeeID
NameOfficeNumberOfficePhone
LOCKEREMPLOYEE
Assume many-to-many relationship
IST210 19
Employee ID
Name OfficeNumber
OfficePhone
LockerID
4 Tom 281 9182 1, 5
10 John 777 1829 1
LockerNumber
LockerRoom
LockerSize
1 #2 10
5 #2 5
EMPLOYEE LOCKER
Employee 4 owns locker 1 and 5. Employee 10 owns locker 1.
Employee ID
Name OfficeNumber
OfficePhone
4 Tom 281 9182
10 John 777 1829
LockerID
LockerRoom
LockerSize
EmployeeID
1 #2 10 4, 10
5 #2 5 4
EMPLOYEE LOCKER
Option A
Option B
IST210 20
Employee ID
Name OfficeNumber
OfficePhone
4 Tom 281 9182
10 John 777 1829
LockerID
LockerRoom
LockerSize
1 #2 10
5 #2 5
EMPLOYEE LOCKER
EmployeeID
LockerID
4 1
4 5
10 1
ASSIGNMENT
NOT SIMPLY ADDING FOREIGN KEYS! CREATE ANOTHER TABLE!
Employee 4 owns locker 1 and 5. Employee 10 owns locker 1.
IST210 21
N:M Relationship
EMPLOYEE(EmployeeID, LastName, FirstName, OfficeNumber, OfficePhone)LOCKER(LockerID, LockerRoom, LockerSize)ASSIGNMENT(EmployeeID, LockerID)
Note: EmployeeID and LockerID are both primary keys and foreign keys in ASSIGNMENT table
IST210 22
N:M Relationship• To represent a N:M relationship in relational design, a new
table must be created. • This table is called an intersection table
• An intersection table has a composite key consisting of the keys from each of the tables that it connects
IST210 23
Relationship Summary
EMPLOYEE(EmployeeID, Name, OfficeNumber, OfficePhone, LockerID)LOCKER(LockerID, LockerRoom, LockerSize)
EMPLOYEE(EmployeeID, Name, OfficeNumber, OfficePhone)LOCKER(LockerID, LockerRoom, LockerSize, EmployeeID)
N:M add an intersection table
EMPLOYEE(EmployeeID, Name, OfficeNumber, OfficePhone)LOCKER(LockerID, LockerRoom, LockerSize)ASSIGNMENT(EmployeeID, LockerID)
EMPLOYEE(EmployeeID, Name, OfficeNumber, OfficePhone)LOCKER(LockerID, LockerRoom, LockerSize, EmployeeID)
1:N add a foreign key to the many-side table
1:1 add a foreign key to either table
or
IST210 24
In-class Exercise 1Transform this diagram into tables
UserID
NameEmail
SpotID
LocationMonthlyCost
USERPARKING
IST210 25
In-class Exercise 1Transform this diagram into tables
UserID
NameEmail
SpotID
LocationMonthlyCost
USERPARKING
PARKING(SpotID, Location, MonthlyCost)USER(UserID, Name, Email, SpotID)
IST210 26
In-class Exercise 1: Notes
PARKING(SpotID, Location, MonthlyCost)USER(UserID, Name, Email, SpotID)
PARKING(SpotID, Location, MonthlyCost, UserID)USER(UserID, Name, Email)
Note: both options are correct, but the first one is preferred because of the minimal cardinality of SpotID is mandatory.
IST210 27
In-class Exercise 2
BuildingNameApartmentID
NumberOfBedNumberOfBathRent
BuildingName
Address
APARTMENTBUILDING
Transform this diagram into tables
IST210 28
In-class Exercise 2Transform this diagram into tables
BUILDING(BuildingName, Address)APARTMENT(BuildingName, ApartmentID, NumberOfBedrooms, NumberofBaths, Rent)
IST210 29
In-class Exercise 2: Notes
BUILDING(BuildingName, Address)APARTMENT(BuildingName, ApartmentID, NumberOfBedrooms, NumberofBaths, Rent)
Notes:* A common mistake is making ApartmentID as the only primary key. (BuildingName, ApartmentID) should a composite key for APARTMENT * Another common mistake is to create a duplicate attribute BuildingName in APARTMENT. Or forget to make BuildingName as the foreign key
IST210 30
In-class Exercise 3
CourseID
CourseNameInstructor
StudentID
StudentNameEmail
COURSESTUDENT
Transform this diagram into tables
IST210 31
In-class Exercise 3
CourseID
CourseNameInstructor
StudentID
StudentNameEmail
COURSESTUDENT
Transform this diagram into tables
STUDENT(StudentID, StudentName, Email)COURSE(CourseID, CourseName, Instructor)REGISTRATION(StudentID, CourseID)
IST210 32
In-class Exercise 3: Notes
STUDENT(StudentID, StudentName, Email)COURSE(CourseID, CourseName, Instructor)REGISTRATION(StudentID, CourseID)
Notes:* In REGISTRATION table, StudetID and CourseID are both primary key and foreign key. So you should give both underline and italic (wave underline in hand-written format) to them.
IST210 33
E-R Diagram Tables
1. Entity Table• Entity name Table name• Identifier Primary key• Attributes Attributes
2. Relationship Foreign key or a new table• Determined by the maximal cardinality• 1:1 add a foreign key to either table• 1:N add a foreign key to the many-side table• N:M add an intersection table
3. Specify the attribute properties
IST210 34
E-R Diagram Tables
1. Entity Table• Entity name Table name• Identifier Primary key• Attributes Attributes
2. Relationship Foreign key or a new table• Determined by the maximal cardinality
3. Specify the attribute properties• Data type• Key• Required
• Determined by the minimal cardinality• Remarks
IST210 35
EMPLOYEE(EmployeeID, Name, OfficeNumber, OfficePhone, LockerID)LOCKER(LockerID, LockerRoom, LockerSize)
EMPLOYEE table
Column Name Data Type Key Required Remarks
LockerID
LockerRoom
LockerSize
LOCKER table
Column Name Data Type Key Required Remarks
EmployeeID
Name
OfficeNumber
OfficePhone
LockerID
IST210 36
EMPLOYEE(EmployeeID, Name, OfficeNumber, OfficePhone, LockerID)LOCKER(LockerID, LockerRoom, LockerSize)
EMPLOYEE table
Column Name Data Type
LockerID int
LockerRoom char(10)
LockerSize float
LOCKER table
Column Name Data Type
EmployeeID int
Name char(50)
OfficeNumber char(20)
OfficePhone char(12)
LockerID int
Data Type• Determine the data types based on the type
of these attributes• Data types must be the types defined in
SQL• http://technet.microsoft.com/en-us/library/
ms187752.aspx
• Data type of a foreign key must be the same type as its referred primary key
• LockerID in EMPLOYEE table must have the same data type as LockerID in LOCKER because LockerID is a foreign key in EMPLOYEE
IST210 37
EMPLOYEE(EmployeeID, Name, OfficeNumber, OfficePhone, LockerID)LOCKER(LockerID, LockerRoom, LockerSize)
EMPLOYEE table
Column Name Data Type
LockerID Primary key
LockerRoom
LockerSize
LOCKER table
Column Name Key
EmployeeID Primary key
Name
OfficeNumber
OfficePhone
LockerID Foreign key
Key• Specify primary key and foreign key(s) in the
table
IST210 38
EMPLOYEE(EmployeeID, Name, OfficeNumber, OfficePhone, LockerID)LOCKER(LockerID, LockerRoom, LockerSize)
EMPLOYEE table
Column Name Required
LockerID Yes
LockerRoom Yes
LockerSize Yes
LOCKER table
Column Name Required
EmployeeID Yes
Name Yes
OfficeNumber No
OfficePhone No
LockerID Yes
Required• Whether an attribute is required or not
determines whether we allow NULL value for this attribute
• Primary key must be required• Normal attributes are determined based on
requirements or common sense• “not required” is preferred
• Foreign key is determined by minimum cardinality
• An employee must have at least one locker. So LockerID is required in EMPLOYEE
IST210 39
EMPLOYEE(EmployeeID, Name, OfficeNumber, OfficePhone, LockerID)LOCKER(LockerID, LockerRoom, LockerSize)
EMPLOYEE table
Column Name
Remarks
LockerID Surrogate key: initial value = 1 Increment = 1
LockerRoom
LockerSize Default value: 10
LOCKER table
Column Name
Remarks
EmployeeID Surrogate key: initial value = 1 Increment = 1
Name
OfficeNumber
OfficePhone Format: ###-###-####
LockerID Reference: LOCKER
Remarks• If a primary key is a surrogate key,
specify the initial value and increment
• If an attribute is a foreign key, specify which table it refers to
• If an attribute has a required format, specify the format
• If an attribute has a default value, specify the value
IST210 40
EMPLOYEE(EmployeeID, Name, OfficeNumber, OfficePhone, LockerID)LOCKER(LockerID, LockerRoom, LockerSize)
EMPLOYEE table
Column Name
Data Type
Key Required
Remarks
LockerID int Primary key
Yes Surrogate key: initial value = 1 Increment = 1
LockerRoom char(10) Yes
LockerSize float Yes Default value: 10
LOCKER table
Column Name
Data Type
Key Required Remarks
EmployeeID int Primary key
Yes Surrogate key: initial value = 1 Increment = 1
Name char(50) Yes
OfficeNumber char(20) No
OfficePhone char(12) No Format: ###-###-####
LockerID int Foreign key
Yes Reference: LOCKER
Now we complete converting an E-R diagram to tablesNext, we can use SQL to create a real database
IST210 41
In-class Exercise 1
PARKING(SpotID, Location, MonthlyCost)USER(UserID, Name, Email, SpotID)
IST210 42
PARKING(SpotID, Location, MonthlyCost)USER(UserID, Name, Email, SpotID)PARKING table
Column Name
Data Type
Key Required Remarks
UserID
Name
SpotID
USER table
Column Name
Data Type
Key Required Remarks
SpotID
Location
MonthlyCost
In-class Exercise 1
IST210 43
PARKING table
Column Name
Data Type
Key Required Remarks
UserID int Primary key
Yes Surrogate key: initial value = 1 Increment = 1
Name char(100) Yes
Email char(50) No
SpotID int Foreign key
Yes Reference: PARKING
USER table
Column Name
Data Type
Key Required Remarks
SpotID int Primary key
Yes Surrogate key: initial value = 1 Increment = 1
Location char(50) Yes
MonthlyCost int Yes Default: 50
PARKING(SpotID, Location, MonthlyCost)USER(UserID, Name, Email, SpotID)
IST210 44
PARKING table
Column Name
Data Type
Key Required
Remarks
UserID int Primary key
Yes Surrogate key: initial value = 1 Increment = 1
Name char(100) Yes
Email char(50) No
SpotID int Foreign key
Yes Reference: PARKING
USER table
Column Name
Data Type
Key Required
Remarks
SpotID int Primary key
Yes Surrogate key: initial value = 1 Increment = 1
Location char(50) Yes
MonthlyCost int Yes Default: 50
Notes: • See the red parts• Char is not a data type, char(50) is a data type. If you use char, you need
to specify the max length• When you can use other data types, do not use char. For example, it is
better to use int for MonthlyCost instead of using char
PARKING(SpotID, Location, MonthlyCost)USER(UserID, Name, Email, SpotID)
IST210 45
In-class Exercise 2Transform this diagram into tables
BUILDING(BuildingName, Address)APARTMENT(BuildingName, ApartmentID, NumberOfBedrooms, NumberofBaths, Rent)
IST210 46
BUILDING table
Column Name
Data Type
Key Required Remarks
BuildingName
ApartmentID
NumberOfBed
NumberOfBath
Rent
APARTMENT table
Column Name
Data Type
Key Required Remarks
BuildingName
Address
BUILDING(BuildingName, Address)APARTMENT(BuildingName, ApartmentID, NumberOfBedrooms, NumberofBaths, Rent)
In-class Exercise 2
IST210 47
BUILDING table
Column Name
Data Type
Key Required Remarks
BuildingName Char(20) Primary key, foreign key
Yes Reference: BUILDING
ApartmentID int Primary key Yes
NumberOfBed float Yes
NumberOfBath float Yes
Rent float Yes
APARTMENT table
Column Name
Data Type
Key Required Remarks
BuildingName
Char(20) Primary key
Yes
Address Char(100) Yes Format: street, city, state, zip code
BUILDING(BuildingName, Address)APARTMENT(BuildingName, ApartmentID, NumberOfBedrooms, NumberofBaths, Rent)
IST210 48
BUILDING table
Column Name
Data Type
Key Required Remarks
BuildingName Char(20) Primary key, foreign key
Yes Reference: BUILDING
ApartmentID int Primary key Yes
NumberOfBed float Yes
NumberOfBath
float Yes
Rent float Yes
APARTMENT table
Column Name
Data Type
Key Required Remarks
BuildingName
Char(20) Primary key
Yes
Address Char(100) Yes Format: street, city, state, zip code
Notes: • ApartmentID should NOT be a surrogate key. In real scenarios, we are using
some meaningful ApartmentIDs, such as 100 or 201 instead of a meaningless system-generated id.
• Use float for #ofBed, #ofBath, and Rent. Because #ofBath could be 1.5 and rent could be $890.50, which are not integers.
IST210 49
In-class Exercise 3Transform this diagram into tables
STUDENT(StudentID, StudentName, Email)COURSE(CourseID, CourseName, Instructor)REGISTRATION(StudentID, CourseID)
IST210 50
STUDENT(StudentID, StudentName, Email)COURSE(CourseID, CourseName, Instructor)REGISTRATION(StudentID, CourseID)
STUDENT table
Column Name
Data Type Key Required Remarks
CourseID
CourseName
Instructor
COURSE table
Column Name
Data Type
Key Required Remarks
StudentID
StudentName
Column Name
Data Type
Key Required Remarks
StudentID
CourseID
REGISTRATION table
In-class Exercise 3
IST210 51
STUDENT(StudentID, StudentName, Email)COURSE(CourseID, CourseName, Instructor)REGISTRATION(StudentID, CourseID)
STUDENT table
Column Name
Data Type Key Required Remarks
CourseID Char(20) Primary key
Yes Format: DepartmentName + CourseNumber
CourseName Char(50) Yes
Instructor Char(100) No
COURSE table
Column Name
Data Type Key Required Remarks
StudentID int Primary key
Yes Surrogate key: initial value = 1 Increment = 1
StudentName Char(100) Yes
Email Char(50) No
Column Name
Data Type
Key Required Remarks
StudentID int Primary key, foreign key Yes Reference: STUDENT
CourseID Char(20) Primary key, foreign key Yes Reference: COURSE
REGISTRATION table
IST210 52
STUDENT table
Column Name
Data Type Key Required Remarks
CourseID Char(20) Primary key
Yes Format: DepartmentName + CourseNumber
CourseName Char(50) Yes
Instructor Char(100) No
COURSE table
Column Name
Data Type
Key Required Remarks
StudentID int Primary key
Yes Surrogate key: initial value = 1 Increment = 1
StudentName Char(100) Yes
Email Char(50) No
Column Name
Data Type
Key Required Remarks
StudentID int Primary key, foreign key
Yes Reference: STUDENT
CourseID Char(20) Primary key, foreign key
Yes Reference: COURSE
REGISTRATION table
Notes:• A CourseID should not be integer, for example, “IST210” is a courseID.
CourseName for IST210 is “Organization of the data”.
IST210 53
E-R Diagram Tables
1. Entity Table• Entity name Table name• Identifier Primary key• Attributes Attributes
2. Relationship Foreign key or a new table3. Specify the attribute properties
IST210 54
Attendance check?
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