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Logical Schema Design: the Relational Data Model

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Logical Schema Design: the Relational Data Model LogicalLogical SchemaSchema Design:Design: TheThe RelationalRelational DataData ModelModel Basics of the Relational Model From Conceptual to Logical Schema LogicalLogical SchemaSchema DesignDesign Select data model Hierarchical data model: hierarchies of record types mainframe oldie, still in use, outdated Network data model: graph like data structures, still in use, outdated Relational data model: the most important one today Object-Oriented data model: more flexible data structures, less powerful data manipulation language Object-Relational: the best of two worlds? FU-Berlin, DBS I 2006, Transform conceptual model into logical schema of data model easy for Relational Data Model (RDM) Hinze / Scholz / Hinze can be performed automatically (e.g. by Oracle Designer) 1.2 LogicalLogical SchemaSchema Design:Design: RelationalRelational DataData ModelModel The Relational Data Model Important terms 1970 introduced by E.F. Codd, honoured by Turing award (Codd: A Relational Model of Data for Large Shared Data Banks. ACM Comm. 13( 6): 377- 387, 1970) Basic ideas: Database is collection of relations Relation R = set of n-tuples Relation schema R(A1,A2,…An) Attributes Ai = atomic values of domain Di=dom(Ai) Student relation (table) FName Name Email SID FU-Berlin, DBS I 2006, attribute Tina Müller mueller@... 13555 Anna Katz katz@... 12555 tuple Carla Maus piep@... 11222 relation schema: Hinze / Scholz / Hinze Student(Fname, Name, Email, SID) or Student(Fname:string, Name:string, Email:string, SID:number) 1.3 LogicalLogical SchemaSchema Design:Design: RelationalRelational DataData ModelModel Relation Schema R(A1, A2, …, An) notation sometimes R:{[A1, A2, …, An]} Relation R ⊆ dom(A1) x dom(A2) x … x dom(An) Attribute set R ={A1, A2, …, An} Degree of a relation: number of attributes Example: Student(Fname, Name, Email, SID) FU-Berlin, DBS I 2006, Student ⊆ string x string x string x number R (Student) ={Fname, Name, Email, SID} Hinze / Scholz / Hinze Database Schema is set of relation schemas 1.4 LogicalLogical SchemaSchema Design:Design: RelationalRelational DataData ModelModel Time-variant relations Relations have state at each point in time. Integrity constraints on state part of DB schema Tuples (rows, records) Not ordered No duplicate tuples (Relations are sets) Null-values for some attributes possible Distinguishable based on tuple values (key-concept) FU-Berlin, DBS I 2006, Different to object identification in o-o languages: there, each object has implicit identity, usually completely unrelated to its fields Hinze / Scholz / Hinze 1.5 LogicalLogical SchemaSchema Design:Design: RelationalRelational DataData ModelModel Superkey: Important terms subset of attributes of a relation schema R for which no two tuples have the same value. Every relation at least 1 superkey. Example: Student(Fname, Name, Email, SID) Superkeys: {Fname, Name, Email, SID}, {Name, Email, SID}, {Fname, Email, SID}, {Fname, Name, SID}, {Fname, SID}, {Name, SID}, {Fname, Name, Email}, FU-Berlin, DBS I 2006, {Fname, Email}, {Name, Email}, {Email, SID}, {Email}, {SID} Hinze / Scholz / Hinze 1.6 LogicalLogical SchemaSchema Design:Design: RelationalRelational DataData ModelModel Candidate Key: Important terms superkey K of relation R such that if any attribute A ∈ K is removed, the set of attributes K\A is not a superkey of R. Example: Student(Fname, Name, Email, SID) Candidate Keys: {Email}, {SID} Primary Key: FU-Berlin, DBS I 2006, arbitrarily designated candidate key Example: Student(Fname, Name, Email, SID) Hinze / Scholz / Hinze Primary Key: {SID} 1.7 LogicalLogical SchemaSchema Design:Design: RelationalRelational DataData ModelModel Foreign Key: Important terms set of attributes FK in relation schema R1 that references relation R2 if: Attributes of FK have the same domain as the attributes of primary key K2 of R2 A value of FK in tuple t1 in R1 either occurs as a value of K2 for some t2 in R2 or is null. Example: FU-Berlin, DBS I 2006, R1: Exercise(EID, Tutor, ExcerciseHours, Lecture) R2: Student(Fname, Name, Email, SID) Hinze / Scholz / Hinze K1={EID}, K2={SID} Tutor is foreign key of Student (from Exercise ). 1.8 LogicalLogical SchemaSchema Design:Design: TransformationTransformation 1. Select data model → relational data model 2. Transform conceptual model into logical schema of relational data model Define relational schema, table names, attributes and types, invariants Design steps: Translate entities into relations Translate relationships into relations Simplify the design FU-Berlin, DBS I 2006, Define tables in SQL Define additional invariants (Formal analysis of the schema) Hinze / Scholz / Hinze 1.9 LogicalLogical SchemaSchema Design:Design: EntitiesEntities Step 1: Transform entities For each entity E create relation R with all attributes Key attributes of E transform to keys of the relation director year id category Price_per_day length title Movie FU-Berlin, DBS I 2006, Relational Schema: Movie(id, title, category, year, director, Price_per_day, length) Hinze / Scholz / Hinze 1.10 LogicalLogical SchemaSchema Design:Design: WeakWeak EntitiesEntities Step 2: Transform weak entities For each weak entity WE create relation R Include all attributes of WE Add key of identifying entity to weak entities key Part of key is foreign key eid cid (0,*) (1,1) Employee has Child FU-Berlin, DBS I 2006, Relational Schema: Employee(eid) Child(cid, eid) Hinze / Scholz / Hinze Note: weak entity and defining relationship transformed together! 1.11 LogicalLogical SchemaSchema Design:Design: WeakWeak EntitiesEntities Step 2: Transform weak entities Item iid Example: (1,1) has date (1,*) (1,*) (1,1) Delivery contains Order did oid FU-Berlin, DBS I 2006, Relational Schema: contact date Delivery(did, date) Order(oid, did, date, contact) Item(iid, oid, did) Hinze / Scholz / Hinze 1.12 LogicalLogical SchemaSchema Design:Design: RelationshipsRelationships Step 3: Transform Relationships For each 1:1-relationship between E1, E2 create relation R Include all key attributes Define as key: key of entity E1 or entity E2 Choose entity with total participation for key (driving licence) (1,1) (1,1) Country has President CName PName FU-Berlin, DBS I 2006, Relational Schema: Country(CName) President(PName) Hinze / Scholz / Hinze has(CName, PName) or has(Cname, PName) 1.13 LogicalLogical SchemaSchema Design:Design: RelationshipsRelationships Step 3: Transform Relationships For each 1:N-relationship between E1, E2 create relation R Include all key attributes Define as key: key of N-side of relationship lid Lecture hold Lecturer (1,1) (0,*) lnr name FU-Berlin, DBS I 2006, Relational Schema: Lecture(lnr) Lecturer(lid, name) hold(lnr,lid) Hinze / Scholz / Hinze 1.14 LogicalLogical SchemaSchema Design:Design: RelationshipsRelationships Step 3: Transform Relationships For each N:M-relationship create relation R Include all key attributes Define as key: keys from both entities User_account has emailMessage (0,*) (1,*) id username from FU-Berlin, DBS I 2006, Relational Schema: User_account(username) emailMessage(id, from) has(username, messageid) Hinze / Scholz / Hinze 1.15 LogicalLogical SchemaSchema Design:Design: RelationshipsRelationships Step 3: Transform Relationships Include all relationship attributes in relation R lid Lecture hold Lecturer (1,1) (0,N) lnr date name Relational Schema: Lecture(lnr) Lecturer(lid, name) hold(lnr,lid, date) lnr FU-Berlin, DBS I 2006, Rename keys in recursive relationships Lecture Relational Schema: predecessor successor require(preLNR, succLNR) (0,1) (0,*) Hinze / Scholz / Hinze require 1.16 LogicalLogical SchemaSchema Design:Design: RelationshipsRelationships Step 3: Transform Relationships For each n-ary relationship (n>2) create relation R Include all key attributes Define as key: keys from all numerously involved entities rating (0,*) (1,*) Lecturer recommend Textbook ISBN (0,*) title lid name lnr FU-Berlin, DBS I 2006, Lecture author Relational Schema: Lecture(lnr) Lecturer(lid, name) Hinze / Scholz / Hinze Textbook(ISBN, title, author) recommend(lid, lnr, ISBN, rating) 1.17 LogicalLogical SchemaSchema Design:Design: GeneralizationGeneralization Step 4: Generalization FName 3 Variants for Transformation Person Name Email is-a is-a pid SID Student Lecturer Contract Person(pid, name, fname, email) Student(pid, sid) Lecturer(pid, contract) FU-Berlin, DBS I 2006, Student(pid, name, fname, email, sid) Lecturer(pid, name, fname, email, contract) Hinze / Scholz / Hinze Person(pid, name, fname, email, sid, contract) 1.18 LogicalLogical SchemaSchema Design:Design: GeneralizationGeneralization aid (1) Separate relation for each entity A a Key in specialized relations: is-a is-a foreign key to general relation b B C c A(aid, a) B: B(aid, b) A: C(aid, c) C: Gathering data from different tables time consuming FU-Berlin, DBS I 2006, Appropriate specialization prevents unnecessary data access Hinze / Scholz / Hinze 1.19 LogicalLogical SchemaSchema Design:Design: GeneralizationGeneralization aid (2) Relations for specialization A a Separate tables which include is-a is-a A’s attributes B C Separate keys for relations b c AB(aid, a, b) AB: AC(aid, a, c) AC: FU-Berlin, DBS I 2006, Only valid for exhaustive specialization (no data in A only) Time consuming data retrieval for non-distinct specializations Hinze / Scholz / Hinze 1.20 LogicalLogical SchemaSchema Design:Design: GeneralizationGeneralization aid (3) one relation for all entities A a is-a is-a b B C c ABC(aid, a, b, c) ABC: NULLs, if subtypes are disjoint
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