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Database Design Examples-1

Database Design Examples-1. 22/03/2004. 3 step design. Conceptual Design Highest level design Issues: data types, relationships, constraints Uses ER model Logical Design Implementation of conceptual model 3 ways: hierarchical, network, relational

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Database Design Examples-1

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  1. Database Design Examples-1 22/03/2004

  2. 3 step design • Conceptual Design Highest level design Issues: data types, relationships, constraints Uses ER model • Logical Design Implementation of conceptual model 3 ways: hierarchical, network, relational Apply relational model Uses RA (relational algebra) as a formal query language • Physical Design Actual computer implementation Issues: mem. manag., storage, indexing

  3. ER model • The 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.

  4. 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: R E1 E2 1 N d m1 a3 a1 Can we migrate a3? IF we can, to where? a2

  5. 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? date R account transaction 1 N amount Acc# Tran#

  6. Relational Model • Data representation model introduced by Codd, 1970. • E-R RM Relation  Table Attributes  Columns • Table 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)

  7. 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.

  8. ERRM Rules • S/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.

  9. 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?

  10. 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.

  11. Example-(Relational model) CUSTOMER PURCHASE_ORDER PK is (PurchaseOrder#) FK is(Cust#) Corresponds to 1-N relationship CUST_PHONES STOCK_ITEMS CONTAINS PK is (Cust#, Phones) Corresponds to N-M relationship PK is (PurchaseOrder#, Stock#)

  12. Physical Design-DDL 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 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 Contains ( PONo NUMBER REFERENCES PurchaseOrder, StockNo NUMBER REFERENCES Stock, Quantity NUMBER, Discount NUMBER, PRIMARY KEY (PONo, StockNo) ) ;

  13. 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 ) ;

  14. 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');

  15. 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..

  16. SQL • Q1: 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 ;

  17. SQL • Q3: 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)

  18. SQL • Q4: 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)

  19. DML • Delete 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..!!!!)

  20. area catch depth distance Overlap Park Lake M N Pname Lname area Pid Lid 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.

  21. Relations for Parking DB PARK LAKE PARK_LAKE RA (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)

  22. RA operations • select: <selection operation>(relation R) retrieves the subset of rows. • project: <list of attributes>(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

  23. Join and natural join operations • A 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(1attr1), oldname) means that the relation “oldname” becomes the “newname”. Also the first attribute of “newname” table is called the “attr1”

  24. Relational Algebra on Parking DB • Find 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)

  25. 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)

  26. 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

  27. 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(1Pid1,2Lid1),Pid,Lid ( ParkLake ¤ Park))  (t2(1Pid2,2Lid2),Pid,Lid ( ParkLake ¤ Park))  (t, t1 x t2) Pname (Pid1=Pid2)  (Lid1 Lid2) (t)

  28. 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

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