Transforming ER & EER diagrams into Relations (Chapter 9) Overview

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1 Transforming ER & EER diagrams into Relations (Chapter 9) "!$# %&%#('()* Overview A relatively straightforward process with a welldefined set of rules. Many CASE Tools can automatically perform many of the conversion steps. CASE tools often cannot model complex data relationships. There are sometimes legitimate alternatives where you will need to choose a particular solution. You must be prepared to perform a quality check on the results obtained with a CASE Tool. +, -,.,/01,2, ":$; <&5<;(=(>? 1

2 Map Simple Regular Entities Each regular entity type in an ER diagram is transformed into a relation. The name given to the relation is generally the same as the entity type. Each simple attribute of the entity type become an attribute of the relation. Choose one of the key(s) as primary key of the relation. How about composite attributes? How about multi-valued A B A C ADEFAGA HEIEGB JKLM M"O$P Q&JQP(R(ST Map Composite Attributes When a regular entity type has a composite attribute, only the simple component attributes of the composite attribute are included in the new relation. Street City ID COSTOMER Address State Zip CUSTOMER(ID,, Street, City, State, Zip) U V W V Ẍ VYZ[V\V ]Z^Z\W _`ab cb"d$ë f&_fe(g(hi 2

3 Map Multi-valued Attributes When a regular entity type contains a multivalued attribute, two new relations (rather than one) are created. The first relation contains all of the attributes of the entity type except the multi-valued attribute. The second relation contains two sets of attributes. The primary key from the first relation, which becomes a foreign key of the second relation. Multi-valued attribute itself. The primary key of the second of relation is the combination of all attributes. The name of the second relation should capture the meaning of the multi-valued attribute. j k l k m knopkqk rosoql tuvw xw"y$z {&t{z( (}~ Example SS Skills Address Street City State Zip (SS,, State, City, State, Zip) _SKILL(SS, Skill) ƒ ˆ Š Œ Œ"Ž$ & ( ( 3

4 Map Binary One-Many Relationship Create a relation for each of the two entity types participating in the relationship. Include the primary key attribute (or attributes) of the entity on the one-side of the relationship as a foreign key in the relation that is on the many-side of the relationship ( a mnemonic you can use to remember this rule is this: The primary key migrates to many side). Dept SS 1 STUDET Major_in DEPT STUDET(SS,, Dept) DPET(Dept) š œ žÿ " $ &ž ( ( Map Binary Many-Many Relationships Suppose that there is a binary relationship (M:) between two entity types A and B. For such a relationship, create a new relation C: Include as foreign key attributes in C the primary key for each of the two participating entity type. These attributes become the primary key of C. Any attributes that are associated with the relationship are included with the relation C. Grade CID SID STUDET M taking COURSE Text STUDET(SID, AME) COURSE(CID, Text) Taking(SID, CID, Grade) ª «ª ª ª ª ± ² «³ µ " $¹ º&³º¹(»(¼½ 4

5 Map Binary One-One Relationship Binary 1:1 relationship can be viewed as a special case of 1:m relationship. The process of mapping such a relationship to relations requires to steps: first, two relations are created, one for each of the participating entity type. Second, the primary key of one of the relations is included as foreign key in the other relation. Dept SS 1 1 Manages DEPT (SS,, Dept) (SS, ) DEPT(Dept) DEPT(Dept, ManagerSS) ¾ À Á ÂÃÄ Å ÆÃÇÃÅÀ ÈÉÊË ÌË"Í$Î Ï&ÈÏÎ(Ð(ÑÒ Map Unary One-Many Relationship The entity type in the unary relationship is mapped to a relation using the procedure described before. Then a foreign key attribute is added within the same relation that references the primary key values. ote that the foreign key attribute name should reflect the role name on the one-side. A recursive foreign key is a foreign key in a relation that references the primary key values of that same relation. Supervisee SS Manages Supervisor 1 (SS,, SupervisorSS) Ó Ô Õ Ô Ö Ô ØÙÔÚÔ ÛØÜØÚÕ ÝÞßà áà"â$ã ä&ýäã(å(æç 5

6 Map Unary Many-Many relationship With this type of relationship: Two relations are created: one to represent the entity type in the relationship and the other an associative relation to represent the M: relationship itself. The primary key of the associative relation consists of two parts: both take their values from the primary key of the other relation. Any attribute of the relationship is included in the associative relation. ID Item M PART Contains Quantity Unit_cost Components COMPOET(ItemID, ComponentID, Quantity) PART(ID,, Unit_Cost) è é ê é ë éìíîéïé ðíñíïê òóôõ öõ" $ø ù&òùø(ú(ûü Map Weak Entities For each weak entity type, create a new relation and include all of the simple attributes (or simple components of composite attributes) as attributes of this relation. Then, include the primary key of the owner relation as a foreign key attribute in this relation. The primary key of the new relation is the combination of this primary key of the owner and the partial key of the weak entity type. salary sex SS name addr. Employees birthdate 1 DEPEDAT_ OF name birthdate Dependants sex ý þ ÿ þ þ þ þ ÿ!!" # relationship 6

7 Example Results Employee( SS, name, addr, salary sex, birthdate) Dependants(name, birthdate, sex, relationship, empss) The relation for the weak entity not only has the attributes of itself, but also has the key attributes of the other entity sets. Do not construct a relation for a double-diamond relationship. $&%'(% )% * +,-%. %/+0+.1' !:!; < Map Ternary (n-ary) Relationship It is recommended that you convert the ternary (n-ary) relationship to a number of binary relationships, and then transform the diagram into relationships. S Quantity Proj SUPPLIER SUPPLY PROJECT S Quantity Proj PART Parto SUPPLIER 1 1 SS SUPPLY SPJ PROJECT SP 1 PART Parto A B C-> D >EBFBD1? GHIJK JLMGM!O!P Q 7

8 Example Result SUPPLIER(S) PROJECT(Proj) SUPPLY(S, Proj, PartID, Quantity) PART(PartID) R&ST(S US V W X-S Y SZW[WY1T \]^_` _abc\cb!d!e f Map Supertype/Subtype Relationships The relational data model does not yet directly support/subtype relationships. There are various strategies that database designer can use to represent these relationships. In this lecture, we introduce the most commonly employed strategy. g&hi(h jh k l m-h n holpln1i qrstu tvwxqxw!y!z { 8

9 Mapping Strategy Create a separate relation for the supertype and for each of its subtypes. Assign to the relation created for the supertype the attributes that are common to all memebers of the supertype, including the primary key. Assign to the relation for each subtype the primary key of the supertype, and this primary key is also a foreign key that references the primary key in relation representing the supertype. Assign to the relation for each subtype the attributes that are unique to that subtype. &}~(} } -} ƒ } ƒ1~ ˆ Š Œ Œ!Ž! Example Address SS Date_hired d HOURLY SALARED COSULTAT Hourly_rate Annual_salary Stock_option Contract_o. Billing_rate ( SS,, Address, Date_hired) HOURLY_(SS, Hourly_Rate) SALARIED_(SS, Annual_salary, Stock_Option) COSULTAT(SS, Contract_o, Billing_rate) & ( - š 1 œ žÿ ž!! 9

10 Problem Map Categories Multiple supertypes may have different primary keys. Solution The concept of surrogate key, a specified new key attribute for the subtype. & ( ª «- ««1 ±²³ ³µ!!¹ º Example C Address SS COMPAY PERSO U Total_Balance B ACCOUT HOLDER M Has_Acct BAK»&¼½(¼ ¾¼ À Á-¼  ¼ÃÀÄÀÂ1½ ÅÆÇÈÉ ÈÊËÌÅÌË!Í!Î Ï 10

11 Example Results COMPAY(C, Address, OwnerID) PERSO(SS,, OwnerID) ACCOUTHOLDER(OwnerID) HASACCOUT(OwnerID, B, Total_Balance) BAK(B) Ð&ÑÒ(Ñ ÓÑ Ô Õ Ö-Ñ ÑØÕÙÕ 1Ò ÚÛÜÝÞ ÝßàáÚáà!â!ã ä 11

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