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Roadmap for OO Design What is a Relational Database? Database = - PDF document

Roadmap for OO Design What is a Relational Database? Database = collection of tables RDBMS principles Table = collections of rows Row = set of attributes(fields) Implementing OO models in RDBMS Normalization and good models


  1. � Roadmap for OO Design What is a Relational Database? Database = collection of tables • RDBMS principles Table = collections of rows Row = set of attributes(fields) • Implementing OO models in RDBMS – Normalization and good models Students • More OO design Student ID Student Name Faculty ID 123 French KS – polymorphism, enhancing inheritance 256 Smith WM • Architecture 301 Jones GS – Alternate Data management – External control attributes – Reuse Domains and Keys Key Nomenclature Each attribute defines a Domain: Set of values for attribute Candidate Key : Uniquely identifies a row. Student ID={x | x is a valid student ID} Primary Key: a Candidate Key used for main access Student Name={x | x is a student name} Foreign Key: An attribute from one table that is a Primary Key in another table. Faculty ID={x | x is faculty ID initials} A Key is a is a set of attributes that identifies a row. Keys Example Why is it called Relational? Recall: Prof ID Name Dept Office M21 McUmber CSE 3100 , , , ( , , ) A B C A × B × C = a b c Given sets such that P01 Potter EEC 3200 Primary key , . . a ∈ A b ∈ B c ∈ C A × B × C is called a Cross Product S56 Stirewalt CSE 12 - 321 relation R ⊆ A × B × C A is a subset of the cross product : C00 Crewe CEP 45 - 001 The sets A,B,C are the domains (attribute values) Primary key Dept Chair Fax# CSE Stockman 355 - 1111 R A B ⊂ × A relation is a function if Foreign key EEC Rover 353 - 1212 a A , b , c B ( a , b ) R ( a , c ) R c b ∈ ∈ ∈ ∧ ∈ ⇒ = Note: referential integrity CEP Leheay 432 - 1234 ( P 01 ) f name = Potter PHY Izzo 355 - 4444 1

  2. Operations on Tables Join Example All the RDBMS operators act on tables to produce tables. Table A Table B Faculty ID Section Location Project operator = all rows, but selected columns Student ID Name Faculty ID 124 French A A 101 EB 201 Select operator = select rows based on criteria (such as 256 Smith B B 201 BH 100 key match) 301 Jones C C 001 EB 304 Join A, B over column x from A, y from B = (two steps): = × C A B 1. Form A.Student ID A.Name A.Faculty B.Faculty B.Section B.Location 124 French A A 101 EB 201 2. Keep only rows where x=y 256 Smith B B 201 BH 100 301 Jones C C 001 EB 304 Support for Existence-Based Implementing a Class Model Identity • RDBMS generates unique ID (usually • Objects have identity number) – Value based (keys are domain values) • MS-Access uses type counter (really a long – Existence based (keys are made-up Ids) value) • New Domains • Oracle uses type sequence – Identifier: contains existence Ids – Enumeration: discrete set of values • Other possibilities • Class table – Date/Time (if fine grained) • Assoc & Generalization table or relation – pointer connected with DB (Ingres) Implementing 1-1 Implementing Many to Many A A B B A-Id B-Id ……. B-Id ……. A-Id ………. A-Id B-Id B-Id ……. Foreign key can be on either side 2

  3. Example with Foreign Keys Implementing 1 to Many or A B City 1..* Airport 1..* serves City Table Serves Table Airport Table cityID cityName cityID airportCod e airportCod e airportNam e A-Id B-Id ……. B-Id ……. 01005 IAH 01005 Houston IAH Interconti nential 01005 HOU In the “0-1” case, flipping these requires 01006 Atlanta HOU Hobby 01008 TEW 01007 Albany a possible null key ATL Hartsfield 01009 LAN 01009 Lansing LAN Lansing 01006 ATL A-Id ……. B-Id A-Id ……. Exercise Review: Implementing 1-1 A B authors Person Citation birthDate edits title deathDate note name 1. Use existence-based identity to design tables for classes Person and Citation. 2. Design tables for associations authors and edits . A-Id B-Id ……. B-Id ……. Foreign key can be on either side Review: Implementing Many Review: Implementing 1 to to Many Many or A B A B A-Id B-Id ……. B-Id ……. In the “0-1” case, flipping these requires A-Id ………. A-Id B-Id B-Id ……. a possible null key A-Id ……. B-Id A-Id ……. 3

  4. Normal Forms and OO Normal Forms Implementations (1) Schedule • “Normal Form” old RDBMS concept Course Section Desc Time Credits CSE 260 1 Math 8 : 00 MTF 3 • Concerned with redundancies and CSE 470 1 Software 11 : 30MWF 4 dependencies between attributes. CSE 260 2 Math 4 : 10MWF 3 CSE 101 3 Intro 9 : 10 MTF 3 • With Object Models we can (almost) + + CSE 231 2 C 10 : 20 TTh 3 ignore normalization Primary key • Requires good model and good If the description of CSE260 changes, have to change it implementation. in multiple places. Called modification anomaly . Normal Forms and OO Normal Forms and OO Implementations (2) Implementations (3) Schedule Schedule Course Section Desc Time Credits Course Section Desc Time Credits CSE 260 1 Math 8 : 00 MTF 3 CSE 260 1 Math 8 : 00 MTF 3 CSE 470 1 Software 11 : 30MWF 4 CSE 470 1 Software 11 : 30MWF 4 CSE 260 2 Math 4 : 10MWF 3 CSE 260 2 Math 4 : 10MWF 3 CSE 101 3 Intro 9 : 10 MTF 3 CSE 101 3 Intro 9 : 10 MTF 3 + + + + CSE 231 2 C 10 : 20 TTh 3 CSE 231 2 C 10 : 20 TTh 3 If I want to add CSE870, and have a description and credits, If no sections of CSE101 are offered, and I delete it, I lose I still need a section and time before it can be added. the description and credits. This is a deletion anomaly . This is an insertion anomaly. Normal Forms & Second Normal Form Correct Models 0..* Schedule Better Model Course Schedule Course Section Desc Time Credits CSE 260 1 Math 8 : 00 MTF 3 Course Schedule CSE 470 1 Software 11 : 30MWF 4 CSE 260 2 Math 4 : 10MWF 3 Course Desc Credits Course Section Time CSE 101 3 Intro 9 : 10 MTF 3 CSE 260 Math 3 CSE 260 1 8 : 00 MTF + + CSE 231 2 C 10 : 20 TTh 3 CSE 470 Software 4 CSE 470 1 11 : 30 MWF This table has problems because it has partial dependencies, and is CSE 101 Intro 3 CSE 101 3 9 : 10 MTF not in in second normal form . It’s also the implementation of an incorrect + + CSE 231 C 3 CSE 231 2 10 : 20 TTh object model. Description and Credits only depend on Course. Second normal form means there are no partial dependencies. 4

  5. Third Normal Form (I) Third Normal Form (II) Prof Prof ID Name Dept Office Chair FAX 0..* Key Dept Professor B1 Brown CSE 263 Stockman 555 - 7441 J2 Jones EEC 241 Fui 555 - 7318 M3 Morris CSE 127 Stockman 555 - 7441 Professor W4 West PSY 257 Freud 505 - 0012 Dept Prof ID Name Dept Office Dept Chair FAX B1 Brown CSE 263 This “professor” class is modeled incorrectly. The table is in CSE Stockman 555 - 7441 J2 Jones EEC 241 second normal form, but has anomalies: Modification : Changing dept chair requires multiple modifications PSY Freud 505 - 0012 M3 Morris CSE 127 Insertion : A new Dept requires at least one prof even if we have EEC Fui 555 - 7318 W4 West PSY 257 a chair and FAX. Why? Chair and FAX depend on only Dept. Called Transitive Now there are no partial or transitive dependencies. Dependency Things NOT To Do Conclusions So Far…. Company Board • Classes are tables Name Founded boardSize Name boardSize Founded • Associations are relations or tables and relations Don’t combine optional classes • Proper class model produces normalized A B A-Id B-Id forms B-Id A-Id • Normalized forms eliminate anomalies Don’t double up on foreign keys • Concentrate on class model -- proper classes and multiplicities and good relational model results. Generalization Association Classes Super Super-ID Common fields grantor User grantee owner Authorization Super-ID Special fields Table Sub 1 Sub 2 Requires a join to get the object User Table Authorization Alternatives: Push subclass attributes up and use a “type” field userID …… tableID owner grantee tableID grantor userID tableID userID Push superclass attributes down and replicate tables FK key 5

  6. Link Attributes Like any other association table, but add the extra attributes. A B attr B-Id ……. A-Id ………. A-Id B-Id Link-attributes 6

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