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Mapping from ER Model to Relational Model

Last Updated : 24 Sep, 2024
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Converting an Entity-Relationship (ER) diagram to a Relational Model is a crucial step in database design. The ER model represents the conceptual structure of a database, while the Relational Model is a physical representation that can be directly implemented using a Relational Database Management System (RDBMS) like Oracle or MySQL. In this article, we will explore how to convert an ER diagram to a Relational Model for different scenarios, including binary relationships with various cardinalities and participation constraints.

Case 1:  Binary Relationship with 1:1 cardinality with total participation of an entity 

erm1

A person has 0 or 1 passport number and Passport is always owned by 1 person. So it is 1:1 cardinality with full participation constraint from Passport. 

First Convert each entity and relationship to tables.  Person table corresponds to Person Entity with key as Per-Id. Similarly Passport table corresponds to Passport Entity with key as Pass-No. Has Table represents relationship between Person and Passport (Which person has which passport). So it will take attribute Per-Id from Person and Pass-No from Passport. 

Person Has Passport
Per-IdOther Person AttributePer-IdPass-NoPass-NoOther PassportAttribute
PR1-PR1PS1PS1-
PR2-PR2PS2PS2-
PR3-      

                                                                     Table 1                                                                                  

As we can see from Table 1, each Per-Id and Pass-No has only one entry in  Has Table. So we can merge all three tables into 1 with attributes shown in Table 2. Each Per-Id will be unique and not null. So it will be the key. Pass-No can’t be key because for some person, it can be NULL.

Per-IdOther Person AttributePass-NoOther PassportAttribute

Table 2

Case 2: Binary Relationship with 1:1 cardinality and partial participation of both entities 

erm2A male marries 0 or 1 female and vice versa as well. So it is 1:1 cardinality with partial participation constraint from both. First Convert each entity and relationship to tables.  Male table corresponds to Male Entity with key as M-Id. Similarly Female table corresponds to Female Entity with key as F-Id. Marry Table represents relationship between Male and Female (Which Male marries which female). So it will take attribute M-Id from Male and F-Id from Female. 

Male Marry Female
M-IdOther Male AttributeM-IdF-IdF-IdOther FemaleAttribute
M1-M1F2F1-
M2-M2F1F2-
M3-    F3-

Table 3

As we can see from Table 3, some males and some females do not marry. If we merge 3 tables into 1, for some M-Id, F-Id will be NULL. So there is no attribute which is always not NULL. So we can’t merge all three tables into 1. We can convert into 2 tables. In table 4, M-Id who are married will have F-Id associated. For others, it will be NULL. Table 5 will have information of all females. Primary Keys have been underlined. 

M-IdOther Male AttributeF-Id

 Table 4   

F-IdOther FemaleAttribute

                                                                Table 5                                                                 

Note: Binary relationship with 1:1 cardinality will have 2 table if partial participation of both entities in the relationship. If atleast 1 entity has total participation, number of tables required will be 1. 

Case 3: Binary Relationship with n: 1 cardinality 

erm3

In this scenario, every student can enroll only in one elective course but for an elective course there can be more than one student. First Convert each entity and relationship to tables.  Student table corresponds to Student Entity with key as S-Id. Similarly Elective_Course table corresponds to Elective_Course Entity with key as E-Id. Enrolls Table represents relationship between Student and Elective_Course (Which student enrolls in which course). So it will take attribute S-Id from Student and E-Id from Elective_Course. 

Student Enrolls Elective_Course
S-IdOther Student AttributeS-IdE-IdE-IdOther Elective CourseAttribute
S1-S1E1E1-
S2-S2E2E2-
S3- S3E1 E3-
S4- S4E1   

Table 6

As we can see from Table 6, S-Id is not repeating in Enrolls Table. So it can be considered as a key of Enrolls table. Both Student and Enrolls Table’s key is same. We can merge it as a single table. The resultant tables are shown in Table 7 and Table 8. Primary Keys have been underlined. 

S-IdOther Student AttributeE-Id

Table 7 

E-IdOther Elective CourseAttribute

Table 8

Case 4: Binary Relationship with m: n cardinality

erm4

In this scenario, every student can enroll in more than 1 compulsory course and for a compulsory course there can be more than 1 student. First Convert each entity and relationship to tables.  Student table corresponds to Student Entity with key as S-Id. Similarly Compulsory_Courses table corresponds to Compulsory Courses Entity with key as C-Id. Enrolls Table represents relationship between Student and Compulsory_Courses (Which student enrolls in which course). So it will take attribute S-Id from Person and C-Id from Compulsory_Courses. 

Student Enrolls Compulsory_Courses
S-IdOther Student AttributeS-IdC-IdC-IdOther Compulsory CourseAttribute
S1-S1C1C1-
S2-S1C2C2-
S3- S3C1 C3-
S4- S4C3 C4-
   S4C2   
   S3C3   

Table 9

As we can see from Table 9, S-Id and C-Id both are repeating in Enrolls Table. But its combination is unique; so it can be considered as a key of Enrolls table. All tables’ keys are different, these can’t be merged.  Primary Keys of all tables have been underlined. 

Case 5: Binary Relationship with weak entity

erm5

In this scenario, an employee can have many dependents and one dependent can depend on one employee. A dependent does not have any existence without an employee (e.g; you as a child can be dependent of your father in his company). So it will be a weak entity and its participation will always be total. Weak Entity does not have key of its own. So its key will be combination of key of its identifying entity (E-Id of Employee in this case) and its partial key (D-Name). 

First Convert each entity and relationship to tables.  Employee table corresponds to Employee Entity with key as E-Id. Similarly Dependents table corresponds to Dependent Entity with key as  D-Name and E-Id. Has Table represents relationship between Employee and Dependents (Which employee has which dependents). So it will take attribute E-Id from Employee and D-Name from Dependents.  

Employee Has Dependents
E-IdOther Employee AttributeE-IdD-NameD-NameE-IdOther DependentsAttribute
E1-E1RAMRAME1-
E2-E1SRINISRINIE1-
E3-E2RAMRAME2-
  E3ASHISHASHISHE3-

 Table 10

As we can see from Table 10, E-Id, D-Name is key for Has as well as Dependents Table. So we can merge these two into 1. So the resultant tables are shown in Tables 11 and 12. Primary Keys of all tables have been underlined. 

E-IdOther Employee Attribute

Table 11

D-NameE-IdOther DependentsAttribute

Conclusion

Converting an ER diagram to a Relational Model is a crucial step in database design. The ER model represents the conceptual structure, while the Relational Model is a physical representation that can be directly implemented using a Relational Database Management System (RDBMS) like Oracle or MySQL. We've explored how to convert ER diagrams to Relational Models for different scenarios, including binary relationships with various cardinalities and participation constraints. We've covered five cases, highlighting key considerations and resulting table structures. By understanding these scenarios, database designers and developers can effectively translate conceptual ER models into physical Relational Models, ensuring successful database implementation using RDBMS. So, mapping from ER Model to Relational Model is a vital skill, and we hope this article has been helpful.


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