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The Relational Model Chapter 3 Instructor: Xin Zhang Database Management Systems Raghu Ramakrishnan 1 Why Study the Relational Model? Most widely used model. – Vendors: IBM, Informix, Microsoft, Oracle, Sybase, etc. “Legacy systems” in older models – e.g., IBM’s IMS Recent competitor: Object-Oriented model – ObjectStore, Versant, Ontos – a synthesis emerging: object-relational model Informix Universal Server, UniSQL, O2, Oracle, DB2 Database Management Systems Raghu Ramakrishnan 2 Relational Database: Definitions Relational database: a set of relations. Relation: consist of 2 parts: – Instance : a table, with rows and columns. #rows = cardinality, #fields = degree – Schema : specifies name of relation, plus name, type of each column, and their domain. E.g. Students(sid: string, name: string, login: string, age: integer, gpa: real) Can think of a relation as a set of rows or tuples. (i.e., all rows are distinct) Database Management Systems Raghu Ramakrishnan 3 Example Instance of Students Relation sid 53666 53688 53650 name login Jones jones@cs Smith smith@eecs Smith smith@math age 18 18 19 gpa 3.4 3.2 3.8 Cardinality = 3, degree = 5 , all rows distinct Do all columns in a relation instance have to be distinct? Database Management Systems Raghu Ramakrishnan 4 Example Instance of Bank Account Acct 1005 1002 1003 1008 1010 1012 Name Address Jones 11 First St Smith 12 First St Smith 12 First St Green 14 Fourth St Balance $1023.22 $22.43 $11000.12 $1077.23 $5,000,000,000.00 Gates 20 Tenth St Smith 15 Second St $443.77 Cardinality = 6, degree = 4 , all rows distinct Database Management Systems Raghu Ramakrishnan 5 Advantages of Relational Database A major strength of the relational model: supports simple, powerful querying of data. Queries can be written intuitively, and the DBMS is responsible for efficient evaluation. Query Language (SQL) is well-defined. Database Management Systems Raghu Ramakrishnan 6 Integrity Constraints (ICs) Domain, primary key, foreign key constraints IC: condition that must be true for any instance of the database; e.g., domain constraints. – ICs are specified when schema is defined. – ICs are checked when relations are modified. A legal instance of a relation is one that satisfies all specified ICs. – DBMS should not allow illegal instances. If the DBMS checks ICs, stored data is more faithful to real-world meaning. – Avoids data entry errors, too! Database Management Systems Raghu Ramakrishnan 7 Where do ICs Come From? ICs are based upon the semantics of the realworld enterprise that is being described in the database relations. We can check a database instance to see if an IC is violated, but we can NEVER infer that an IC is true by looking at an instance. – An IC is a statement about all possible instances! Key and foreign key ICs are the most common; more general ICs supported too. Database Management Systems Raghu Ramakrishnan 8 Primary Key Constraints A set of fields is a key for a relation if : 1. No two distinct tuples can have same values in all key fields, and 2. This is not true for any subset of the key. – Part 2 false? A superkey. – If there’s >1 key for a relation, one of the keys is chosen (by DBA) to be the primary key. E.g., sid is a key for Students. (What about name?) The set {sid, gpa} is a superkey. Database Management Systems Raghu Ramakrishnan 9 Primary and Candidate Keys in SQL Possibly many candidate keys (specified using UNIQUE), one of which is chosen as the primary key. “For a given student and course, CREATE TABLE Enrolled (sid CHAR(20), there is a single grade.” vs. cid CHAR(20), “Students can take only one grade CHAR(2), course, and receive a single grade PRIMARY KEY (sid,cid) ) for that course; further, no two CREATE TABLE Enrolled students in a course receive the (sid CHAR(20), same grade.” cid CHAR(20), grade CHAR(2), PRIMARY KEY (sid), UNIQUE (cid, grade) ) Database Management Systems Raghu Ramakrishnan 10 Primary and Candidate Keys in SQL (continued) For Students relation with SID as the primary key CREATE TABLE Students (sid CHAR(20), name CHAR(20), login CHAR(10), age INTEGER, gpa REAL, PRIMARY KEY (sid) ) Are there any separate fields or combinations of fields which also are candidates for primary key? – How about login? – How about age? – How about age & gpa? Database Management Systems Raghu Ramakrishnan 11 Foreign Keys, Referential Integrity Foreign key : Set of fields in one relation that is used to `refer’ to a tuple in another relation. (Must correspond to primary key of the second relation.) Like a `logical pointer’. E.g. sid is a foreign key referring to Students: – Enrolled(sid: string, cid: string, grade: string) – If all foreign key constraints are enforced, referential integrity is achieved, i.e., no dangling references. Database Management Systems Raghu Ramakrishnan 12 Foreign Keys in SQL Only students listed in the Students relation should be allowed to enroll for courses. CREATE TABLE Enrolled (sid CHAR(20), cid CHAR(20), grade CHAR(2), PRIMARY KEY (sid,cid), FOREIGN KEY (cid) REFERENCES Courses Enrolled FOREIGN KEY (sid) REFERENCES Students ) Students sid cid grade 53666 53666 53650 53666 Carnatic101 Reggae203 Topology112 History105 Database Management Systems C B A B sid 53666 53688 53650 Raghu Ramakrishnan name login Jones jones@cs Smith smith@eecs Smith smith@math age 18 18 19 gpa 3.4 3.2 3.8 13 Foreign Keys in SQL (continued) Bank Account Example - Only allow accounts opened for customers listed in the customer information relation First, an instance of customer information Name Address Phone Email 6032223456 Jones@cs Jones 11 First St 6033335567 Smith@cs Smith 12 First St Green 14 Fourth St 7814447890 Green@cs 1223336789 Gates@ms Gates 11 First St Smith 12 Second St 6039788765 Smith@ee Database Management Systems Raghu Ramakrishnan 14 Foreign Keys in SQL (continued) Creates the customer information relation CREATE TABLE Customer_Info (name CHAR(20), addr CHAR(40), phone CHAR(10), email char (40), PRIMARY KEY (name, addr)) Now create the bank account relation with a foreign key CREATE TABLE Bank_Acct (acct CHAR (4), name CHAR (20), address char (40), balance REAL, PRIMARY KEY (acct) , Foreign Key (name, address) references Customer_Info) Raghu Ramakrishnan Database Management Systems 15 Foreign Keys in SQL (continued) Acct 1005 1002 1003 1008 1010 1012 Name Address Jones 11 First St Smith 12 First St Smith 12 First St Green 14 Fourth St Gates 20 Tenth St Smith 15 Second St Name Address Jones 11 First St Smith 12 First St Green 14 Fourth St Gates 11 First St Smith 12 Second St Database Management Systems Balance $1023.22 $22.43 $11000.12 $1077.23 $5,000,000,000.00 $443.77 Phone 6032223456 6033335567 7814447890 1223336789 6039788765 Raghu Ramakrishnan Email Jones@cs Smith@cs Green@cs Gates@ms Smith@ee 16 Enforcing Referential Integrity Consider Students and Enrolled; sid in Enrolled is a foreign key that references Students. What should be done if an Enrolled tuple with a nonexistent student id is inserted? (Reject it!) What should be done if a Students tuple is deleted? – – – – Also delete all Enrolled tuples that refer to it. Disallow deletion of a Students tuple that is referred to. Set sid in Enrolled tuples that refer to it to a default sid. (In SQL, also: Set sid in Enrolled tuples that refer to it to a special value null, denoting `unknown’ or `inapplicable’.) Similar if primary key of Students tuple is updated. Database Management Systems Raghu Ramakrishnan 17 Referential Integrity in SQL/92 SQL/92 supports all 4 CREATE TABLE Enrolled options on deletes and (sid CHAR(20), updates. cid CHAR(20), grade CHAR(2), – Default is NO ACTION PRIMARY KEY (sid,cid), (delete/update is rejected) FOREIGN KEY (sid) – CASCADE (also delete REFERENCES Students all tuples that refer to ON DELETE CASCADE) deleted tuple) – SET NULL / SET DEFAULT (sets foreign key value of referencing tuple) Database Management Systems Raghu Ramakrishnan 18 The SQL Query Language Developed by IBM (system R) in the 1970s Need for a standard since it is used by many vendors Standards: – – – – SQL-86 SQL-89 (minor revision) SQL-92 (major revision, current standard) SQL-99 (major extensions) Database Management Systems Raghu Ramakrishnan 19 The SQL Query Language How to support database operations by using SQL? a. How to generate (create) a relation (table) in database? b. How to modify the table (relation) in database? delete and insert? or modify one attribute? c. How to realize the access (query) relation in database? Database Management Systems Raghu Ramakrishnan 20 Creating Relations in SQL Creates the Students relation. Observe that the type (domain) of each field is specified, and enforced by the DBMS whenever tuples are added or modified. sid 53666 53688 53650 Database Management Systems CREATE TABLE Students (sid CHAR(20), name CHAR(20), login CHAR(10), age INTEGER, gpa REAL) name login Jones jones@cs Smith smith@eecs Smith smith@math Raghu Ramakrishnan age 18 18 19 gpa 3.4 3.2 3.8 21 As another example, the Enrolled table holds information about courses that students take. sid 53831 53831 53650 53666 cid Carnatic101 Reggae203 Topology112 History105 Database Management Systems grade C B A B CREATE TABLE Enrolled (sid CHAR(20), cid CHAR(20), grade CHAR(2)) Raghu Ramakrishnan 22 Creating Relations in SQL (cont) Creates the Bank Account relation. CREATE TABLE Bank_Acct (acct CHAR(4), name CHAR(20), address char(40), balance REAL) Acct 1005 1002 1003 1008 Name Address Jones 11 First St Smith 12 First St Smith 12 First St Green 14 Fourth St 1010 1025 Gates 20 Tenth St Blue 14 Third St Database Management Systems Raghu Ramakrishnan Balance $1023.22 $22.43 $11000.12 $1077.23 $5,000,000,000.00 $2344.22 23 Destroying and Altering Relations DROP TABLE Students Destroys the relation Students. The schema information and the tuples are deleted. ALTER TABLE Students ADD COLUMN firstYear: integer The schema of Students is altered by adding a new field; every tuple in the current instance is extended with a null value in the new field. Database Management Systems Raghu Ramakrishnan 24 Adding and Deleting Tuples Can insert a single tuple using: INSERT INTO Students (sid, name, login, age, gpa) VALUES (53699, 'Green ', 'green@ee', 18, 3.5) More inserts: INSERT INTO Students (sid, name, login, age, gpa) VALUES (53666, 'Jones', 'jones@cs', 18, 3.4) INSERT INTO Students (sid, name, login, age, gpa) VALUES (53688, 'Smith ', 'smith@eecs', 18, 3.2) INSERT INTO Students (sid, name, login, age, gpa) VALUES (53650, 'Smith ', 'smith@math', 18, 3.5) Database Management Systems Raghu Ramakrishnan 25 Students relation before inserts: sid 53666 53688 53650 name Jones Smith Smith age 18 18 19 gpa 3.4 3.2 3.8 Students relation after inserts: sid 53666 53688 53650 53600 login jones@cs smith@eecs smith@math name Jones Smith Smith Green login jones@cs smith@eecs smith@math green@ee age 18 18 19 18 gpa 3.4 3.2 3.8 3.5 Question: what happens if you insert the same info at a second time? Database Management Systems Raghu Ramakrishnan 26 sid name login age gpa 53666 Jones jones@cs 18 3.4 53688 Smith smith@eecs 18 3.2 53650 Smith smith@math 19 3.8 53600 Green green@ee 18 3.5 Can delete all tuples satisfying some condition (e.g., name = Smith): DELETE FROM Students S WHERE S.name = 'Smith' Students instance after delete: sid name login 53666 Jones jones@cs 53600 Green green@ee Database Management Systems Raghu Ramakrishnan age 18 18 gpa 3.4 3.5 27 Adding and Deleting Tuples (continued) Insert a tuples into the Bank_Acct instance: INSERT INTO Bank_Acct (acct, name, address, balance) VALUES (1025, 'Blue', '14 Third St', 2344.22) INSERT INTO Bank_Acct (acct, name, address, balance) VALUES (1005, 'Jones', '11 First St', 1023.22) INSERT INTO Bank_Acct (acct, name, address, balance) VALUES (1002, 'Smith', '12 First St', 22.43) INSERT INTO Bank_Acct (acct, name, address, balance) VALUES (1003, 'Smith', '12 First St', 11000.12) INSERT INTO Bank_Acct (acct, name, address, balance) VALUES (1008, 'Green', '14 Fourth St', 1077.23) INSERT INTO Bank_Acct (acct, name, address, balance) VALUES (1010, 'Gates', '20 Tenth St', 5000000000.00) Database Management Systems Raghu Ramakrishnan 28 Adding and Deleting Tuples (continued) Bank_Acct instance after insert: Acct Name Address Balance $1023.22 1005 Jones 11 First St $22.43 1002 Smith 12 First St 1003 Smith 12 First St $11000.12 1008 Green 14 Fourth St $1077.23 $5,000,000,000.00 1010 Gates 20 Tenth St $2344.22 1025 Blue 14 Third St Database Management Systems Raghu Ramakrishnan 29 Adding and Deleting Tuples (continued) Delete all tuples satisfying some condition (e.g., acct = 1008) from Bank_Acct instance: DELETE FROM Bank_Acct B WHERE B.acct = 1008 Database Management Systems Raghu Ramakrishnan 30 Adding and Deleting Tuples (continued) Bank_Acct instance after delete: Acct 1005 1002 1003 1010 1012 1025 Name Address Jones 11 First St Smith 12 First St Smith 12 First St Gates 20 Tenth St Smith 12 Second St Blue 14 Third St Balance $1023.22 $22.43 $11000.12 $5,000,000,000.00 $443.77 $2344.22 Powerful variants of these commands are available; more later! Database Management Systems Raghu Ramakrishnan 31 The SQL Query Language To find all 18 year old students, we can write: SELECT * FROM Students S WHERE S.age=18 sid 53666 53688 53650 sid name login Jones jones@cs Smith smith@eecs Smith smith@math name 53666 Jones login jones@cs age gpa 18 3.4 18 3.2 19 3.8 age gpa 18 3.4 53688 Smith smith@ee 18 3.2 •To find just names and logins, replace the first line: SELECT S.name, S.login from Students S Database Management Systems Raghu Ramakrishnan 32 The SQL Query Language (continued) To find all bank customers whose name is Smith : SELECT * FROM Bank_Acct B WHERE B.name='Smith' Name Address Phone 6033335567 Smith 12 First St Smith 12 Second St 6039788765 Database Management Systems Raghu Ramakrishnan Email Smith@cs Smith@ee 33 Adding and Deleting Tuples (continued) Insert tuples into the Enrolled instance: INSERT INTO Enrolled (sid, cid, grade) VALUES ('53831', 'Carnatic 101', 'C') INSERT INTO Enrolled (sid, cid, grade) VALUES ('53831', 'Reggae 203', 'B') INSERT INTO Enrolled (sid, cid, grade) VALUES ('53650', 'Topology 112', 'A') sid cid grade INSERT INTO Enrolled (sid, cid, grade) 53831 Carnatic101 C VALUES ('53666', 'History 105', 'B') 53831 Reggae203 B 53650 Topology112 A 53666 History105 B Database Management Systems Raghu Ramakrishnan 34 Querying Multiple Relations What does the following query compute? SELECT S.name, E.cid FROM Students S, Enrolled E WHERE S.sid=E.sid AND E.grade='B' sid 53666 53688 53650 53600 sid 53831 53831 53650 53666 name Jones Smith Smith Green cid grade Carnatic101 C Reggae203 B Topology112 A History105 B Database Management Systems login jones@cs smith@eecs smith@math green@ee age 18 18 19 18 gpa 3.4 3.2 3.8 3.5 we get: S.name E.cid Jones History 105 Raghu Ramakrishnan 35 Querying Multiple Relations (continued) What does the following query compute? SELECT B.name, C.email FROM Bank_Acct B, Customer_Info C WHERE B.name=C.name AND B.addr=C.addr AND B.balance>$1000.00 Given the following instances of Bank_Acct and Customer_Info Database Management Systems Raghu Ramakrishnan 36 Querying Multiple Relations (continued) Acct 1005 1002 1003 1008 1010 1012 Name Jones Smith Smith Green Gates Smith Address 11 First St 12 First St 12 First St 14 Fourth St 20 Tenth St 15 Second St Name Address Jones 11 First St Smith 12 First St Green 14 Fourth St Gates 11 First St Smith 12 Second St Database Management Systems Balance $1023.22 $22.43 $11000.12 $1077.23 $5,000,000,000.00 $443.77 Phone 6032223456 6033335567 7814447890 1223336789 6039788765 Raghu Ramakrishnan Email Jones@cs Smith@cs Green@cs Gates@ms Smith@ee 37 Querying Multiple Relations (continued) We get : Name Jones Smith Green Database Management Systems Email Jones@cs Smith@cs Green@cs Raghu Ramakrishnan 38 Views A view is just a relation, but we store a definition, rather than a set of tuples. CREATE VIEW YoungActiveStudents (name, grade) AS SELECT S.name, E.grade FROM Students S, Enrolled E WHERE S.sid = E.sid and S.age<21 Views can be dropped using the DROP VIEW command. How to handle DROP TABLE if there’s a view on the table? DROP TABLE command has options to let the user specify this. Database Management Systems Raghu Ramakrishnan 39 Views and Security Views can be used to present necessary information (or a summary), while hiding details in underlying relation(s). – Given YoungStudents, but not Students or Enrolled, we can find students s who have are enrolled, but not the cid’s of the courses they are enrolled in. Database Management Systems Raghu Ramakrishnan 40 Logical DB Design: ER to Relational Entity sets to tables. sss 312-24-234 324-56-678 ssn name Employees Database Management Systems name John Jack lot 18 2 lot CREATE TABLE Employees (ssn CHAR(11), name CHAR(20), lot INTEGER, PRIMARY KEY (ssn)) Raghu Ramakrishnan 41 Relationship Sets to Tables since name ssn dname lot Employees did Works_In budget Departments Relationship Set Database Management Systems Raghu Ramakrishnan 42 Relationship Sets to Tables In translating a relationship set to a relation, attributes of the relation must include: – Keys for each participating entity set (as foreign keys). This set of attributes forms a superkey for the relation. – All descriptive attributes. Database Management Systems CREATE TABLE Works_In( ssn CHAR(1), did INTEGER, since DATE, PRIMARY KEY (ssn, did), FOREIGN KEY (ssn) REFERENCES Employees, FOREIGN KEY (did) REFERENCES Departments) Raghu Ramakrishnan 43 Entity Sets and their Relationship Set CREATE TABLE Employees (ssn CHAR(11), name CHAR(20), lot INTEGER, PRIMARY KEY (ssn)) CREATE TABLE Departments (did CHAR(11), dname CHAR(20), budget INTEGER, PRIMARY KEY (did)) Database Management Systems CREATE TABLE Works_In( ssn CHAR(1), did INTEGER, since DATE, PRIMARY KEY (ssn, did), FOREIGN KEY (ssn) REFERENCES Employees, FOREIGN KEY (did) REFERENCES Departments) Raghu Ramakrishnan 44 since name ssn dname lot Employees address budget did Works_In Departments Locations capacity CREATE TABLE Works_In( ssn CHAR(11), did INTEGER, address CHAR(20) since DATE, PRIMARY KEY (ssn, did, address), FOREIGN KEY (ssn) REFERENCES Employees, FOREIGN KEY (did) REFERENCES Departments, FOREIGN KEY (address) REFERENCES Locations) Database Management Systems Raghu Ramakrishnan 45 CREATE TABLE Departments (did CHAR(11), dname CHAR(20), budget INTEGER, PRIMARY KEY (did)) CREATE TABLE Works_In( ssn CHAR(11), did INTEGER, CREATE TABLE Employees (ssn CHAR(11), name CHAR(20), lot INTEGER, PRIMARY KEY (ssn)) CREATE TABLE Locations (address CHAR(11), capacity CHAR(20), PRIMARY KEY (address)) address CHAR(20) since DATE, PRIMARY KEY (ssn, did, address), FOREIGN KEY (ssn) REFERENCES Employees, FOREIGN KEY (did) REFERENCES Departments) FOREIGN KEY (address) REFERENCES Locations) Database Management Systems Raghu Ramakrishnan 46 Describe Key Constraints since name dname ssn Each dept has at most one manager, according to the key constraint on Manages. lot Employees budget did Works_In Departments Relationship Set since name ssn dname lot Employees Database Management Systems Raghu Ramakrishnan did Manages budget Departments key constraint 47 Translating ER Diagrams with Key Constraints Map relationship to a table: – Note that did is the key now! – Separate tables for Employees and Departments. Database Management Systems since name ssn dname lot Employees did Manages budget Departments CREATE TABLE Manages( ssn CHAR(11), did INTEGER, since DATE, PRIMARY KEY (did), FOREIGN KEY (ssn) REFERENCES Employees, FOREIGN KEY (did) REFERENCES Departments) Raghu Ramakrishnan 48 since name ssn dname lot did Employees Manages budget Departments CREATE TABLE Manages( ssn CHAR(11), did INTEGER, since DATE, PRIMARY KEY (did), FOREIGN KEY (ssn) REFERENCES Employees, FOREIGN KEY (did) REFERENCES Departments) CREATE TABLE Employees (ssn CHAR(11), name CHAR(20), lot INTEGER, PRIMARY KEY (ssn)) Database Management Systems CREATE TABLE Departments (did CHAR(11), dname CHAR(20), budget INTEGER, PRIMARY KEY (did)) Raghu Ramakrishnan 49 Translating ER Diagrams with Key Constraints CREATE TABLE Dept_Mgr( Since each department has a unique manager, we could instead combine Manages and Departments. did INTEGER, dname CHAR(20), budget REAL, ssn CHAR(11), since DATE, PRIMARY KEY (did), FOREIGN KEY (ssn) REFERENCES Employees) since name ssn dname lot Employees Database Management Systems did Manages budget Departments Raghu Ramakrishnan 50 since name ssn dname lot Employees did Manages budget Departments CREATE TABLE Dept_Mgr( did INTEGER, dname CHAR(20), budget REAL, ssn CHAR(11), since DATE, PRIMARY KEY (did), FOREIGN KEY (ssn) REFERENCES Employees) Database Management Systems CREATE TABLE Employees (ssn CHAR(11), name CHAR(20), lot INTEGER, PRIMARY KEY (ssn)) 51 Raghu Ramakrishnan Can we do like this? since name ssn dname lot Employees did Works_In budget Departments Each object in database should have a primary key! Database Management Systems Raghu Ramakrishnan 52 Describe Participation Constraints Does every department have a manager? – If so, this is a participation constraint: the participation of Departments in Manages is said to be total (vs. partial). Every did value in Departments table must appear in a row of the Manages table (with a non-null ssn value!) since name ssn dname did lot Employees Manages budget Departments Works_In since Database Management Systems Raghu Ramakrishnan 53 Participation Constraints in SQL We can capture participation constraints involving one entity set in a binary relationship, but little else (without resorting to CHECK constraints). CREATE TABLE Dept_Mgr( did INTEGER, dname CHAR(20), budget REAL, ssn CHAR(11) NOT NULL, since DATE, PRIMARY KEY (did), FOREIGN KEY (ssn) REFERENCES Employees, ON DELETE NO ACTION) Database Management Systems Raghu Ramakrishnan 54 CREATE TABLE Dept_Mgr( Different? Why? did INTEGER, dname CHAR(20), budget REAL, ssn CHAR(11), since DATE, PRIMARY KEY (did), FOREIGN KEY (ssn) REFERENCES Employees) CREATE TABLE Dept_Mgr( did INTEGER, dname CHAR(20), budget REAL, ssn CHAR(11) NOT NULL, since DATE, PRIMARY KEY (did), FOREIGN KEY (ssn) REFERENCES Employees, ON DELETE NO ACTION) Database Management Systems Raghu Ramakrishnan 55 Different? since name ssn dname lot Employees did Manages budget Departments Some departments may not have manager currently! since name ssn dname did lot Employees Manages budget Departments Each department should always have manager! Database Management Systems Raghu Ramakrishnan 56 Describe Weak Entities A weak entity can be identified uniquely only by considering the primary key of another (owner) entity. – – Owner entity set and weak entity set must participate in a one-to-many relationship set (1 owner, many weak entities). Weak entity set must have total participation in this identifying relationship set. name ssn lot Employees Database Management Systems cost Policy Raghu Ramakrishnan pname age Dependents 57 Translating Weak Entity Sets Weak entity set and identifying relationship set are translated into a single table. – When the owner entity is deleted, all owned weak entities must also be deleted. CREATE TABLE Dep_Policy ( pname CHAR(20), age INTEGER, cost REAL, ssn CHAR(11) NOT NULL, PRIMARY KEY (pname, ssn), FOREIGN KEY (ssn) REFERENCES Employees, ON DELETE CASCADE) Database Management Systems Raghu Ramakrishnan 58 Binary vs. Ternary Relationships name ssn If each policy is owned by just 1 employee: – Key constraint on Policies would mean policy can only cover 1 dependent! What are the additional constraints in the 2nd diagram? Database Management Systems pname lot Employees Dependents Covers Bad design Policies policyid cost name ssn age pname lot age Dependents Employees Purchaser Better design policyid Raghu Ramakrishnan Beneficiary Policies cost 59 Review: Binary vs. Ternary Relationships Why? name ssn pname lot age Dependents Employees Purchaser Beneficiary Policies policyid Database Management Systems Raghu Ramakrishnan cost 60 Binary vs. Ternary Relationships (Contd.) CREATE TABLE Policies ( The key policyid INTEGER, constraints allow cost REAL, us to combine ssn CHAR(11) NOT NULL, Purchaser with PRIMARY KEY (policyid). Policies and FOREIGN KEY (ssn) REFERENCES Employees, Beneficiary with ON DELETE CASCADE) Dependents. Participation CREATE TABLE Dependents ( constraints lead to pname CHAR(20), NOT NULL age INTEGER, constraints. policyid INTEGER, PRIMARY KEY (pname, policyid). FOREIGN KEY (policyid) REFERENCES Policies, ON DELETE CASCADE) Database Management Systems Raghu Ramakrishnan 61 since name dname ssn did lot Employees Manages budget Departments Works_In since pname Policy age Dependents cost Database Management Systems Raghu Ramakrishnan 62 Relational Model: Summary A tabular representation of data. Simple and intuitive, currently the most widely used. Integrity constraints can be specified by the DBA, based on application semantics. DBMS checks for violations. – – Two important ICs: primary and foreign keys In addition, we always have domain constraints. Powerful and natural query languages exist. Rules to translate ER to relational model Database Management Systems Raghu Ramakrishnan 63 How to obtain your Oracle account? https://servlets.uncc.edu/oracle-account/ Database Management Systems Raghu Ramakrishnan 64