Database 1

Database Design Examples-122/03/2004

3 step designConceptual Design Highest level design Issues: data types, relationships, constraints Uses ER modelLogical Design Implementation of conceptual model 3 ways: hierarchical, network, relational Apply relational model Uses RA (relational algebra) as a formal query languagePhysical Design Actual computer implementation Issues: mem. manag., storage, indexing

ER modelThe 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:

E1E2m1a2Ra11Na3dCan we migrate a3?IF we can, to where?

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?accounttransactionTran#R1NamountdateAcc#

Relational ModelData representation model introduced by Codd, 1970. E-R RM Relation Table Attributes ColumnsTable 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.

ERRM RulesS/w packages (CASE tools) such as Erwin, Oracle Designer 2000, Rational Rose can translate ER to RM. 4 steps for transformation: 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)CUSTOMERPURCHASE_ORDERCUST_PHONESSTOCK_ITEMSCONTAINSPK is (PurchaseOrder#) FK is(Cust#)PK is (PurchaseOrder#, Stock#)Corresponds to N-M relationshipCorresponds to 1-N relationshipPK is (Cust#, Phones)

Cust#CustNameStreetCityZip

PurchaseOrder#Cust#OrderDateShipDateToStreetToCityToZip

Cust#Phones

Stock#PriceTaxRate

PurchaseOrder#Stock#quantitydiscount

Physical Design-DDLCREATE 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 PurchaseOrder ( PONo NUMBER, /* purchase order no */ Custno NUMBER REFERENCES Customer, /* Foreign KEY referencing customer */ OrderDate DATE, /* date of order */ ShipDate DATE, /* date to be shipped */ ToStreet VARCHAR2(200), /* shipto address */ ToCity VARCHAR2(200), ToState CHAR(2), ToZip VARCHAR2(20), PRIMARY KEY(PONo) ) ; CREATE TABLE Contains ( PONo NUMBER REFERENCES PurchaseOrder, StockNo NUMBER REFERENCES Stock, Quantity NUMBER, Discount NUMBER, PRIMARY KEY (PONo, StockNo) ) ;

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-MAY-1997',NULL, NULL, NULL, NULL) ;INSERT INTO PurchaseOrder VALUES (2001, 2, SYSDATE, '20 MAY-1997', '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..

SQLQ1: 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 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 ;

SQLQ3: List the Purchase Orders whose total value is greater than that of a specific Purchase Order. SELECT P.PONo FROM PurchaseOrder P, Contains CO, Stock S WHERE 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 P.PONo FROM PurchaseOrder P, Contains CO, Stock S WHERE P.PONo = CO.PONo AND CO.StockNo = S.StockNo GROUP BY P.PONo ; HAVING SUM(S.Price * CO.Quantity) > ALL(SELECT SUM(S.Price * CO.Quantity) FROM Contains CO, Stock S WHERE CO.PONo = 1001 AND CO.StockNo = S.StockNo)

SQLQ4: 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)

DMLDelete 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 table Cust_Phones; drop table Contains; drop table Stock; drop table PurchaseOrder; drop 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.

Relations for Parking DBPARKLAKEPARK_LAKERA (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)

Pid Pname areadistance s1I15052s2V25575s3B175300

Lid Lname catch depth100 L w 20200 C t 30300 S v 45400 T b 28

Lid Pid area100 s1 100200 s1 255300 s3 175

RA operations select: (relation R) retrieves the subset of rows. project: (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 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

Requires compatibility

Join and natural join operationsA 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(1attr1), oldname) means that the relation oldname becomes the newname. Also the first attribute of newname table is called the attr1

Relational Algebra on Parking DBFind 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(1Pid1,2Lid1), Pid,Lid ( ParkLake Park)) (t2(1Pid2,2Lid2), 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