Database Design Examples-1
22/03/2004
3 step design
Conceptual Design
Highest level design
Issues: data types, relationships, constraints
Uses ER model
Logical Design
Implementation of conceptual model
3 ways: hierarchical, network, relational
Apply relational model
Uses RA (relational algebra) as a formal query language
Physical Design
Actual computer implementation
Issues: mem. manag., storage, indexing
ER model
The most popular conceptual data modeling technique.
(Give an example of other conceptual design tool?)
Integrates with “relational model”.
The scenario is partitioned into “entities” which are
characterized with “attributes” and interrelated via
“relationships”. “Entity set” is a set of entities of
the same type.
Entity is an independent conceptual existence in the
scenario.
An attribute (or a set of attributes) that uniquely
identifies instance of an entity is called the key.
1-) Super key 2-) candidate key (minimal
superkey) 3-) primary key (candidate key chosen by DBA)
An attribute can be single-valued or multi-valued.
cont..
At least 2 entities participate in a relationship.
(give an example of recursive relationship)
Binary relationship, ternary relationship…
(give an example of ternary relationship)
Cardinality constraints: 1-1, 1-N, N-M
Relationships may have their own attributes
Example of an ER diagram:
m1
E1
a1
1
R
N
E2
a3
Can we migrate a3?
IF we can, to where?
a2
d
cont..
Weak Entity set: An entity set that does not have enough
attributes to form a primary key.
Think of Transaction entity set(transaction#, date, amount) assuming
that different accounts might have similar transactions.
We need a strong entity set (owner set) in order to
distinguish the entities of weak entity set.
Question: Is there a way to represent this kind of scenario
without using another entity set?
account
Acc#
1
R
date
N
transaction
amount
Tran#
Relational Model
Data representation model introduced by Codd,
1970.
E-R
RM
Relation  Table
Attributes  Columns
Table is an unordered collection of tuples(rows).
Degree of a relation is the # of columns.
Data types of attributes: DOMAINS
int, float, character,date, large_object (lob),
user-defined data types(only for ORDBs)
cont..
Logical consistency of data is ensured by
certain constraints:
key :: every relation must have PK key
entity integrity :: no PK can be NULL
referential integrity :: value of attributes of the foreign
key either must appear as a value in PK of another table (or the
same table, give an example) or must be null.
Definitions:
PrimaryKey is chosen among the candidate
key by DBA.
ForeignKey is set of attributes in a relation
which is duplicated in another relation.
ERRM Rules
such as Erwin, Oracle
Designer 2000, Rational Rose can translate ER
to RM.
4 steps for transformation:
S/w packages
(CASE tools)
1.) Map each entity set in ER into a separate table in RM
(Also, map the attributes, and PK)
2.) Weak entity set with attributes (a1,..an) and owner
set attributes (b1,b2,..bm): MAP it to a table with
{(a1,..an) U (b1,b2,..bm)} attributes. (b1,b2,..bm)
becomes the foreign key. {(a1,..an) U discriminator}
becomes the PK.
cont..
3.) Binary Relationship S between R1 and R2 entity
sets. Assume (a1,a2,…an) is the attributes of S.
If cardinality is 1-1: Chose either relations( say S)
and extend it with {PK(T) U (a1,a2,…an)}
If cardinality is 1-N: Chose N-side relation( say S)
and extend it with {PK(T) U (a1,a2,…an)}
If cardinality is N-M: Represent it with a new
relation with PK(T) U PK(S) U {(a1,a2,…an)}
4.) Multivalued Attribute ‘A’ of entity set R is
represented with a new relation with {A U PK(S)}.
What is the PK of new table?
Example 1-
(3-step design,SQL)
DB of a “Managing customer orders”
Scenario:
a customer has a unique customer number and
contact information
a customer can place many orders, but a given
purchase order is placed by one customer
a purchase order has a many-to-many relationship
with a stock item.
Here is the ER diagram.
Example-(Relational model)
CUSTOMER
PURCHASE_ORDER
Cust Cust Street City Zip
#
Name
Cust# Order Ship ToStre ToCity ToZip
Order#
Date Date et
Purchase
PK is (PurchaseOrder#)
FK is(Cust#)
CUST_PHONES
Cust
#
Phones
STOCK_ITEMS
CONTAINS
Purchase
Order#
Stock#
quantity
Corresponds to 1-N relationship
discount
PK is (Cust#, Phones)
PK is (PurchaseOrder#, Stock#)
Stock# Price
TaxRate
Corresponds to NM relationship
Physical Design-DDL
CREATE TABLE Customer (
CustNo
NUMBER NOT NULL,
CustName VARCHAR2(200) NOT NULL,
Street
VARCHAR2(200) NOT NULL,
City
VARCHAR2(200) NOT NULL,
State
CHAR(2) NOT NULL,
Zip
VARCHAR2(20) NOT NULL,
PRIMARY KEY (CustNo)
);
CREATE TABLE Contains (
PONo NUMBER REFERENCES PurchaseOrder,
StockNo
NUMBER REFERENCES Stock,
Quantity
NUMBER,
Discount
NUMBER,
PRIMARY KEY (PONo, StockNo)
);
CREATE TABLE PurchaseOrder (
PONo
NUMBER,
Custno
NUMBER REFERENCES
Customer,
/* purchase order no */
/* Foreign KEY referencing customer */
OrderDate DATE,
ShipDate
ToStreet
DATE,
/* date of order */
/* date to be shipped */
VARCHAR2(200), /* shipto address */
ToCity
VARCHAR2(200),
ToState
CHAR(2),
ToZip
VARCHAR2(20),
PRIMARY KEY(PONo)
);
cont..
CREATE TABLE Cust_Phones (
CustNo
NUMBER REFERENCES Customer,
Phones VARCHAR2(20),
PRIMARY KEY (CustNo, Phones)
);
CREATE TABLE Stock (
StockNo NUMBER PRIMARY KEY,
Price
NUMBER,
TaxRate NUMBER
);
DML (data manipulation
language)
INSERT INTO Stock VALUES(1004, 6750.00, 2) ;
INSERT INTO Stock VALUES(1011, 4500.23, 2) ;
INSERT INTO Stock VALUES(1534, 2234.00, 2) ;
INSERT INTO Stock VALUES(1535, 3456.23, 2) ;
******************************************
INSERT INTO Customer VALUES (1, 'Jean Nance', '2 Avocet Drive',
'Redwood Shores', 'CA', '95054') ;
INSERT INTO Customer VALUES (2, 'John Nike', '323 College Drive',
'Edison', 'NJ', '08820') ;
******************************************
INSERT INTO Cust_Phones (1, '415-555-1212‘);
INSERT INTO Cust_Phones (2, '609-555-1212');
INSERT INTO Cust_Phones (2, '201-555-1212');
cont..
INSERT INTO PurchaseOrder VALUES (1001, 1, SYSDATE, '10-MAY1997',NULL, NULL, NULL, NULL) ;
INSERT INTO PurchaseOrder VALUES (2001, 2, SYSDATE, '20 MAY1997', '55 Madison Ave', 'Madison', 'WI', '53715') ;
**********************************************
INSERT INTO Contains VALUES( 1001, 1534, 12, 0) ;
INSERT INTO Contains VALUES(1001, 1535, 10, 10) ;
INSERT INTO Contains VALUES(2001, 1004, 1, 0) ;
INSERT INTO Contains VALUES(2001, 1011, 2, 1) ;
**********************************************
NOTE:
You can use bulk loading if the DB has this functionality. Example: Oracle has
SQL*Loader, sqlldr command for bulk loading..
SQL
Q1: Get Customer and Data Item Information for a Specific
Purchase Order
SELECT C.CustNo, C.CustName, C.Street, C.City, C.State, C.Zip,
P.PONo, P.OrderDate,
CO.StockNo, CO.Quantity, CO.Discount
FROM Customer C, PurchaseOrder P, Contains CO
WHERE C.CustNo = P.CustNo
AND P.PONo = CO.PONo
AND P.PONo = 1001 ;
Q2: Get the Total Value of Purchase Orders
SELECT
FROM
WHERE
P.PONo, SUM(S.Price * CO.Quantity)
PurchaseOrder P, Contains CO, Stock S
P.PONo = CO.PONo
AND CO.StockNo = S.StockNo
GROUP BY P.PONo ;
SQL
Q3: List the Purchase Orders whose total value is greater than that
of a specific Purchase Order.
SELECT
FROM
WHERE
P.PONo
PurchaseOrder P, Contains CO, Stock S
P.PONo = CO.PONo
AND CO.StockNo = S.StockNo
AND SUM(S.Price
* CO.Quantity) > SELECT
SUM(S.Price * CO.Quantity)
FROM
Contains CO, Stock S
WHERE
CO.PONo = 1001
AND CO.StockNo = S.StockNo
NOTE:
What if “>1” customers can have the same PurchaseOrderNumber?  Use “ANY” for a general solution..
SELECT
FROM
WHERE
P.PONo
PurchaseOrder P, Contains CO, Stock S
P.PONo = CO.PONo
AND CO.StockNo = S.StockNo
GROUP BY P.PONo ;
HAVING SUM(S.Price * CO.Quantity) > ALL(SELECT
FROM
WHERE
SUM(S.Price * CO.Quantity)
Contains CO, Stock S
CO.PONo = 1001
AND CO.StockNo = S.StockNo)
SQL
Q4: Find the Purchase Order that has the
maximum total value.
CREATE VIEW X(Purchase,Total) AS
SELECT P.PONo, SUM(S.Price * CO.Quantity)
FROM
PurchaseOrder P, Contains CO, Stock S
WHERE P.PONo = CO.PONo
AND CO.StockNo = S.StockNo
GROUP BY P.PONO
---------------------------
SELECT
P.PONo
FROM
X
GROUP BY P.PONo ;
HAVING Total =( SELECT max(Total)
FROM X)
DML
Delete Purchase Order 1001
DELETE
FROM Contains
WHERE PONo = 1001 ;
DELETE
FROM PurchaseOrder
WHERE PONo = 1001 ;
(Important: The order of commands is important..!!!!)
Delete the database.
drop
drop
drop
drop
drop
table Cust_Phones;
table Contains;
table Stock;
table PurchaseOrder;
table Customer;
(Important: The order of commands is important..!!!!)
Example-2
(ER, relational algebra)
Scenario for Parking database:
We want to develop a database that has parks and lakes that
are overlapping with each other. Overlapping area is also stored.
 Parks have their name, area, distance and a unique id.
 Lakes have name, depth, catch and a unique id.
area
distance
Pname
Park
Pid
catch
M
Overlap
area
N
Lake
Lid
depth
Lname
Relations for Parking DB
PARK
Pid Pname area distance
LAKE
Lid
Lname
L
catch depth
w
20
s1
I
150
52
100
200
C
t
30
s2
V
255
75
300
S
v
45
400
T
b
28
s3
B
175
PARK_LAKE
Lid
100
200
300
Pid
s1
s1
s3
area
100
255
175
300
RA (relational algebra operations)
RA is a formal query language and the core of SQL. Not
implemented in commercial DBs.
RA consists of a set of operands (tables) and operations
(select, project, union,cross-product, difference,
intersection)
RA operations
select:
<selection operation>(relation R)
retrieves the subset of rows.
project: <list of attributes>
(relation R)
retrieves the subset of columns.
Assume R and S are tables:
union: R  S, all tuples that are R OR S
intersect: R  S, all tuples that are both R AND S Requires
compatibility
difference: R - S, all tuples that are in R but not in S
cross-product: R x S, all attributes of R followed by those of S
Join
and
natural join operations
A derived operation.
Is the cross-product followed by a select.
R
¤c S : c (R x S)
c:
the condition that usually refers to the attributes of both R and S.
If c is an equality condition and consists of one column (the
common column), then it is called natural join. (R
¤ S)
For complex queries, use the renaming operation, 
(newname(1attr1), oldname) means that
the relation “oldname” becomes the “newname”. Also the first attribute of
“newname” table is called the “attr1”
Relational Algebra on Parking DB
Find the name of the Park which contains
Lake with Lid=100.
1. solution:
Pname (Park ¤  Lid=100(ParkLake))
2. solution:
Pname ( Lid=100(ParkLake ¤ Park))
3. solution:
 (t1,  Lid=100(ParkLake))
 (t2, t1 ¤ Park)
Pname (t2)
cont..
Find the names of Parks with Lakes which have a
depth of above 25.
Pname (Park ¤ (ParkLake ¤ ( depth > 25 (Lake)))
Find the depth of lakes that overlap with ‘I’.
depth (Lake ¤ (ParkLake ¤ ( Pname=I (Park)))
Find the names of Parks with at least 1 lake.
Pname (Park ¤ (ParkLake))
Find the names of Parks with lakes whose catch is
either ‘b’ or ‘w’.
 (t1,  catch=’b’(Lake)   catch=’w’(Lake) )
Pname (Park ¤ ParkLake ¤ t1)
cont..
Find the names of Parks that have ‘b’ and
‘w’ as the catch in their lakes.
 (t1, Pname ( catch=’b’ (Lake) ¤ ParkLake ¤ Park))
 (t2, Pname ( catch=’w’ (Lake) ¤ ParkLake ¤ Park))
t1  t2
cont..
Find Pid of Parks that are 50 km away from the
city where catch is not ‘t’.
Pid ( distance>50 (Park)) - Pid ( catch=’t’ (Lake) ¤
ParkLake ¤ Park)
Find the names pf Parks that have at least 2
lakes.
 (t1(1Pid1,2Lid1), Pid,Lid ( ParkLake ¤ Park))
 (t2(1Pid2,2Lid2), Pid,Lid ( ParkLake ¤ Park))
 (t, t1 x t2)
Pname  (Pid1=Pid2)  (Lid1 Lid2) (t)
NEXT WEEK, 29/04/2004
MORE DB DESIGN EXAMPLES
(weak entity set, N-ary relationships,
EER model)
SQL examples
(set operations-union,intersect,minus
set comparison operations- contains, some, all
aggregate functions-count, some, avg,max,min
group by, having,order by
delete, update operations…)
1.vize: 5 nisan 2004