Huiswerk Lees delen 3 2 3 3 van

Huiswerk Lees delen 3. 2, 3. 3 van hoofdstuk 3. opgaven voor hoofdstuk 2: modelleeropgave 5 opgaven voor hoofdstuk 3: maak de queries voor de vragen uit 3. 5 in relationele algebra; maak de queries 1 -6 voor de bierdrinkerdatabase in tuppel calculus EN in relationele algebra. Database System Concepts 2. 1 ©Silberschatz, Korth and Sudarshan

Silberschatz 3. 5 a employee (person-name, street, city) works (person-name, company-name, salary) company (company-name, city) manages (person-name, manager_name) Find the names of all employees who work for FBC { t | w works ( t[person-name] =w[person-name] Λ w[company-name]=“FBC” )} Database System Concepts 2. 2 ©Silberschatz, Korth and Sudarshan

Silberschatz 3. 5 b employee (person-name, street, city) works (person-name, company-name, salary) company (company-name, city) manages (person-name, manager_name) Find the names and the cities of residence of all employees who work for FBC { t | e employee (t[person-name] = e[person-name] Λ t[city] = e[city] Λ w works ( w[person-name] =e[person-name] Λ w[company-name]=“FBC”))} Database System Concepts 2. 3 ©Silberschatz, Korth and Sudarshan

Silberschatz 3. 5 c employee (person-name, street, city) works (person-name, company-name, salary) company (company-name, city) manages (person-name, manager_name) Find the names, street address, and the cities of residence of all employees who work for First Bank Corporation and earn more that $10000 per annum. t employee w works ( w[person-name] =t[person-name] Λ { t | e employeeΛ (t[person-name] = e[person-name] Λ t[city] = e[city] Λ t[street] = e[street] Λ w[company-name]=“FBC” Λ w works ( w[person-name] =e[person-name] Λ w[salary] >10000))} w[company-name]=“FBC” Λ w[salary] >10000))} Database System Concepts 2. 4 ©Silberschatz, Korth and Sudarshan

Silberschatz 3. 5 d employee (person-name, street, city) works (person-name, company-name, salary) company (company-name, city) manages (person-name, manager_name) Find the names of all employees who live in the same city as the company for which they work. { t | w works ( t[person-name] =w[person-name] Λ c company ( c[company-name=w[company-name] Λ e employee ( e[person-name] = w[person-name] Λ e[city] = c[city] )))} Database System Concepts 2. 5 ©Silberschatz, Korth and Sudarshan

Silberschatz 3. 5 e employee (person-name, street, city) works (person-name, company-name, salary) company (company-name, city) manages (person-name, manager_name) Find the names of all employees who live in the same city and on the same street as do their managers. { t | e employee (t[person-name] = e[person-name] Λ m manages ( m[person-name] = e[person-name] Λ em employee (em[person-name] = m[person-name] Λ e[city] = em[city] Λ e[street] = em[street] )))} Database System Concepts 2. 6 ©Silberschatz, Korth and Sudarshan

Silberschatz 3. 5 f employee (person-name, street, city) works (person-name, company-name, salary) company (company-name, city) manages (person-name, manager_name) Find the names of all employees in this database who do not work for FBC. single company assumption { t | w works (t[person-name]=w[person-name] Λ [company-name]≠“FBC”)} Multiple company assumption { t | w works (t[person-name]=w[person-name] Λ w 1 works (w[person-name] = w 1[person-name] w 1[company-name]≠“FBC”))} Which assumption holds according to the definition of the database? Database System Concepts 2. 7 ©Silberschatz, Korth and Sudarshan

Silberschatz 3. 5 g employee (person-name, street, city) works (person-name, company-name, salary) company (company-name, city) manages (person-name, manager_name) Find the names of all employees who earn more than every employee of SBC. { t | w works (t[person-name]=w[person-name] Λ w 1 works (w 1[company-name] = “SBC” w[salary] > w 1[salary] ))} Database System Concepts 2. 8 ©Silberschatz, Korth and Sudarshan

Silberschatz 3. 5 h employee (person-name, street, city) works (person-name, company-name, salary) company (company-name, city) manages (person-name, manager_name) Assume the companies may be located in several cities. Find all companies located in every city in which SBC is located. { t | c company (t[company-name]=c[company-name] Λ c 1 company (c 1[company-name] = “SBC” c 2 company (c 2[company-name]=c[company-name] Λ c 2[city] = c 1[city] )))} Database System Concepts 2. 9 ©Silberschatz, Korth and Sudarshan

Reduction of an E-R Schema to Tables Primary keys allow entity sets and relationship sets to be expressed uniformly as tables which represent the contents of the database. A database which conforms to an E-R diagram can be represented by a collection of tables. For each entity set and relationship set there is a unique table which is assigned the name of the corresponding entity set or relationship set. Each table has a number of columns (generally corresponding to attributes), which have unique names. Converting an E-R diagram to a table format is the basis for deriving a relational database design from an E-R diagram. Database System Concepts 2. 10 ©Silberschatz, Korth and Sudarshan

Representing Entity Sets as Tables A strong entity set reduces to a table with the same attributes. Database System Concepts 2. 11 ©Silberschatz, Korth and Sudarshan

Composite and Multivalued Attributes Composite attributes are flattened out by creating a separate attribute for each component attribute E. g. given entity set customer with composite attribute name with component attributes first-name and last-name the table corresponding to the entity set has two attributes name. first-name and name. last-name A multivalued attribute M of an entity E is represented by a separate table EM Table EM has attributes corresponding to the primary key of E and an attribute corresponding to multivalued attribute M E. g. Multivalued attribute dependent-names of employee is represented by a table employee-dependent-names( employee-id, dname) Each value of the multivalued attribute maps to a separate row of the table EM E. g. , an employee entity with primary key John and dependents Johnson and Johndotir maps to two rows: (John, Johnson) and (John, Johndotir) Database System Concepts 2. 12 ©Silberschatz, Korth and Sudarshan

Representing Weak Entity Sets A weak entity set becomes a table that includes a column for the primary key of the identifying strong entity set Database System Concepts 2. 13 ©Silberschatz, Korth and Sudarshan

Representing Relationship Sets as Tables A many-to-many relationship set is represented as a table with columns for the primary keys of the two participating entity sets, and any descriptive attributes of the relationship set. E. g. : table for relationship set borrower Database System Concepts 2. 14 ©Silberschatz, Korth and Sudarshan

Redundancy of Tables Many-to-one and one-to-many relationship sets that are total on the many-side can be represented by adding an extra attribute to the many side, containing the primary key of the one side E. g. : Instead of creating a table for relationship account- branch, add an attribute branch to the entity set account Database System Concepts 2. 15 ©Silberschatz, Korth and Sudarshan

Redundancy of Tables (Cont. ) For one-to-one relationship sets, either side can be chosen to act as the “many” side That is, extra attribute can be added to either of the tables corresponding to the two entity sets If participation is partial on the many side, replacing a table by an extra attribute in the relation corresponding to the “many” side could result in null values The table corresponding to a relationship set linking a weak entity set to its identifying strong entity set is redundant. E. g. The payment table already contains the information that would appear in the loan-payment table (i. e. , the columns loan-number and payment-number). Database System Concepts 2. 16 ©Silberschatz, Korth and Sudarshan

Representing Specialization as Tables Method 1: Form a table for the higher level entity Form a table for each lower level entity set, include primary key of higher level entity set and local attributes table attributes person name, street, city customer name, credit-rating employee name, salary Drawback: getting information about, e. g. , employee requires accessing two tables Database System Concepts 2. 17 ©Silberschatz, Korth and Sudarshan

Representing Specialization as Tables (Cont. ) Method 2: Form a table for each entity set with all local and inherited attributes table person customer employee table name, attributes street, city, credit-rating street, city, salary If specialization is total, table for generalized entity (person) not required to store information Can be defined as a “view” relation containing union of specialization tables But explicit table may still be needed foreign key constraints Drawback: street and city may be stored redundantly for persons who are both customers and employees Database System Concepts 2. 18 ©Silberschatz, Korth and Sudarshan

Relations Corresponding to Aggregation To represent aggregation, create a table containing primary key of the aggregated relationship, the primary key of the associated entity set Any descriptive attributes Database System Concepts 2. 19 ©Silberschatz, Korth and Sudarshan

Relations Corresponding to Aggregation (Cont. ) E. g. to represent aggregation manages between relationship works-on and entity set manager, create a table manages(employee-id, branch-name, title, manager-name) Table works-on is redundant provided we are willing to store null values for attribute manager-name in table manages Database System Concepts 2. 20 ©Silberschatz, Korth and Sudarshan

Example Click to add Text

Translate into the relational model Database System Concepts 2. 22 ©Silberschatz, Korth and Sudarshan