Concept explainers
Synonyms:
When more than one name is assigned to same attribute, the attribute names are referred as “synonyms’. It exists when the same attribute has more than one name.
Example:
Suppose in table STUDENT, one of the attribute names is “STU_NUM” which displays the student registration number. Also, another attribute is “STU_ID” which also displays the student registration number. Then the attribute names are named as “Synonyms”.
Primary Key:
A Primary Key in a
Example:
Students in Universities are assigned a unique registration number.
Therefore, in a STUDENT database table, the attribute “reg_no” acts as primary key.
Foreign Key:
Foreign Key is a column in a relational database table which provides a relation between two tables. It provides a cross reference between tables by pointing to primary key of another table.
Example:
In STUDENT database table, the attribute “reg_no” acts as primary key and in COURSE database table in which the student selects his or her course, the same “reg_no” acts as foreign key for the STUDENT table.
Many to One Relationship:
When more than one record in a database table is associated with only one record in another table, the relationship between the two tables is referred as many to one relationship. It is also represented as M: 1 relationship.
One to Many Relationship:
When one record in a database table is associated with more than one record in another table, the relationship between the two tables is referred as one to many relationship. It is also represented as1: M relationship. This is the opposite of many to one relationship.
One to One Relationship:
When one record in a database table is associated with one record in another table, the relationship between the two tables is referred as one to one relationship. It is also represented as1: 1relationship.
CROW FOOT ERD:
The Crow Foot ERD is an Entity Relationship Diagram which is used to represent the cardinalities present in the basic ER diagram. It is used to represent the relationships present between two tuples or tables present in the database.
Explanation of Solution
Given database tables:
Table Name: CHARTER
CHAR_TRIP | CHAR_DATE | CHAR_PILOT | CHAR_COPILOT | AC_NUMBER | CHAR_DESTINATION | CHAR_DISTANCE | CHAR_HOURS_FLOWN | CHA_HOURS_WAIT | CHAR_FUEL_GALLONS | CHAR_OIL_QTS | CUS_CODE |
10001 | 05-Feb-18 | 104 | 2289L | ATL | 936.0 | 5.1 | 2.2 | 354.1 | 1 | 10011 | |
10002 | 05-Feb-18 | 101 | 2778V | BNA | 320.0 | 1.6 | 0.0 | 72.6 | 0 | 10016 | |
10003 | 05-Feb-18 | 105 | 109 | 4278Y | GNV | 1574.0 | 7.8 | 0.0 | 339.8 | 2 | 10014 |
10004 | 06-Feb-18 | 106 | 1484P | STL | 472.0 | 2.9 | 4.9 | 97.2 | 1 | 10019 | |
10005 | 06-Feb-18 | 101 | 2289L | ATL | 1023.0 | 5.7 | 3.5 | 397.7 | 2 | 10011 | |
10006 | 06-Feb-18 | 109 | 4278Y | STL | 472.0 | 2.6 | 5.2 | 117.1 | 0 | 10017 | |
10007 | 06-Feb-18 | 104 | 105 | 2778V | GNV | 1574.0 | 7.9 | 0.0 | 348.4 | 2 | 10012 |
10008 | 07-Feb-18 | 106 | 1484P | TYS | 644.0 | 4.1 | 0.0 | 140.6 | 1 | 10014 | |
10009 | 07-Feb-18 | 105 | 2289L | GNV | 1574.0 | 6.6 | 23.4 | 459.9 | 0 | 10017 | |
10010 | 07-Feb-18 | 109 | 4278Y | ATL | 998.0 | 6.2 | 3.2 | 279.7 | 0 | 10016 | |
10011 | 07-Feb-18 | 101 | 104 | 1484P | BNA | 352.0 | 1.9 | 5.3 | 66.4 | 1 | 10012 |
10012 | 08-Feb-18 | 101 | 2289L | MOB | 884.0 | 4.8 | 4.2 | 215.1 | 0 | 10010 | |
10013 | 08-Feb-18 | 105 | 4278Y | TYS | 644.0 | 3.9 | 4.5 | 174.3 | 1 | 10011 | |
10014 | 09-Feb-18 | 106 | 4278V | ATL | 936.0 | 6.1 | 2.1 | 302.6 | 0 | 10017 | |
10015 | 09-Feb-18 | 104 | 101 | 2289L | GNV | 1645.0 | 6.7 | 0.0 | 459.5 | 2 | 10016 |
10016 | 09-Feb-18 | 109 | 105 | 2778V | MQY | 312.0 | 1.5 | 0.0 | 67.2 | 0 | 10011 |
10017 | 10-Feb-18 | 101 | 1484P | STL | 508.0 | 3.1 | 0.0 | 105.5 | 0 | 10014 | |
10018 | 10-Feb-18 | 105 | 104 | 4278Y | TYS | 644.0 | 3.8 | 4.5 | 167.4 | 0 | 10017 |
Table Name: AIRCRAFT
AC_NUMBER | MODE-CODE | AC_TTAF | AC_TTEL | AC_TTER |
1484P | PA23-250 | 1833.1 | 1833.1 | 101.8 |
2289L | C-90A | 4243.8 | 768.9 | 1123.4 |
2778V | PA31-350 | 7992.9 | 1513.1 | 789.5 |
4278Y | PA31-350 | 2147.3 | 622.1 | 243.2 |
Table Name: MODEL
MOD_CODE | MOD_MANUFACTER | MOD_NAME | MOD_SEATS | MOD_CHG_MILE |
B200 | Beechcraft | Super KingAir | 10 | 1.93 |
C-90A | Beechcraft | KingAir | 8 | 2.67 |
PA23-250 | Piper | Aztec | 6 | 1.93 |
PA31-350 | Piper | Navajao Chiettan | 10 | 2.35 |
Table Name: PILOT
EMP_NUM | PIL_LICENSE | PIL_RATINGS | PIL_MED_TYPE | PIL_MED_DATE | PIL_PTI35_DATE |
101 | ATP | ATP/SEL/MEL/Instr/CFII | 1 | 20-Jan-18 | 11-Jan-18 |
104 | ATP | ATP/SEL/MEL/Instr | 1 | 18-Dec-17 | 17-Jan-18 |
105 | COM | COMM/SEL/MEL/Instr/CFI | 2 | 05-Jan-18 | 02-Jan-18 |
106 | COM | COMM/SEL/MEL/Instr | 2 | 10-Dec-17 | 02-Feb-18 |
109 | COM | ATP/SEL/MEL/SES/Instr/CFII | 1 | 22-Jan-18 | 15-Jan-18 |
Table Name: EMPLOYEE
EMP_NUM | EMP_TITLE | EMP-LNAME | EMP_FNAME | EMP_INITIAL | EMP_CODE | EMP_HIRE_DATE |
100 | Mr. | Kolrnycz | George | D | 15-Jun-62 | 15-Mar-08 |
101 | Ms. | Lewis | Rhonda | G | 19-Mar-85 | 25-Apr-06 |
102 | Mr. | Vandam | Rhett | 14-Nov-78 | 18-May-13 | |
103 | Ms. | Jones | Anne | M | 11-May-94 | 26-Jul-17 |
104 | Mr. | Lange | John | P | 12-Jul-91 | 20-Aug-10 |
105 | Mr. | Williams | Robert | D | 14-Mar-95 | 19-Jun-17 |
106 | Mrs. | Duzak | Jeanine | K | 12-Feb-88 | 13-Mar-18 |
107 | Mr. | Deante | George | D | 01-May-95 | 02-Jul-16 |
108 | Mr. | Wiesanbach | Paul | R | 14-Feb-86 | 03-Jun-13 |
109 | Ms. | Travis | Elizabeth | K | 18-Jun-81 | 14-Feb-16 |
110 | Mrs. | Genkazi | Lieghla | W | 19-May-90 | 29-Jun-10 |
Table Name: EMPLOYEE
CUS_CODE | CUS_LNAME | CUS_FNAME | CUS_INITIAL | CUS_AREACODE | CUS_PHONE | CUS_BALANCE |
10010 | Ramas | Alfred | A | 615 | 844-2573 | 0.00 |
10011 | Dunne | Leona | K | 713 | 894-1293 | 0.00 |
10012 | Smith | Kathy | W | 615 | 894-2285 | 896.54 |
10013 | Owolski | Paul | F | 615 | 894-2180 | 1285.19 |
10014 | Orlando | Myron | 615 | 222-1672 | 673.21 | |
10015 | OBrian | Amy | B | 713 | 442-3381 | 1014.86 |
10016 | Brown | James | G | 615 | 297-1228 | 0.00 |
10017 | Williams | George | 615 | 290-2556 | 0.00 | |
10018 | Fariss | Anne | G | 713 | 382-7185 | 0.00 |
10019 | Smith | Olette | K | 615 | 297-3809 | 453.98 |
PRIMARY KEY in the above tables:
For Table Name: CHARTER:
Primary Key: CHAR_TRIP
“CHAR_TRIP” acts as primary key of the table because the attribute “CHAR_TRIP” is a unique ID that is assigned to every individual trip by the charter plane. It also uniquely identifies every other row present in the database table.
For Table Name: AIRCRAFT:
Primary Key: AC_NUMBER
“AC_NUMBER” acts as primary key of the table because the attribute “AC_NUMBER” is a unique number that is assigned to every individual charter plane and is used to distinguish among them. It also uniquely identifies every other row present in the database table.
For Table Name: MODEL:
Primary Key: MOD_CODE
“MOD_CODE” acts as primary key of the table because the attribute “MOC_CODE” is a unique number that is assigned to every individual model of the charter plane and is used to distinguish models among them. It also uniquely identifies every other row present in the database table.
For Table Name: PILOT:
Primary Key: EMP_NUM
“EMP_NUM” acts as primary key of the table because the attribute “EMP_NUM” is a unique number that is assigned to every pilot that flies an aircraft. It also uniquely identifies every other row present in the database table.
For Table Name: EMPLOYEE:
Primary Key: EMP_NUM
“EMP_NUM” acts as primary key of the table because the attribute “EMP_NUM” is a unique number or ID that is assigned to every employee that works in the airline. It also uniquely identifies every other row present in the database table.
For Table Name: CUSTOMER:
Primary Key: CUS_CODE
“CUS_CODE” acts as primary key of the table because the attribute “CUS_CODE” is a unique code that is assigned to every customer that books a flight with the airline. It also uniquely identifies every other row present in the database table.
FOREIGN KEY in the above tables:
For Table Name: CHARTER:
Primary Key: CHAR_PILOT,CHAR_COPILOT,AC_NUMBER,CUS_CODE
“CHAR_PILOT” acts as foreign key of the table because the attribute “CHAR_PILOT” is also present in the table PILOT and it references PILOT and hence it forms a link between the two tables.
“CHAR_COPILOT” acts as foreign key of the table because the attribute “CHAR_COPILOT” is also present in the table PILOT and it references PILOT and hence it forms a link between the two tables.
“AC_NUMBER” acts as foreign key of the table because the attribute “AC_NUMBER” is also present in the table AIRCRAFT and it references AIRCRAFT and hence it forms a link between the two tables.
“CUS_CODE” acts as foreign key of the table because the attribute “CUS_CODE” is also present in the table CUSTOMER and it references CUSTOMER and hence it forms a link between the two tables.
For Table Name: AIRCRAFT:
Foreign Key: MOD_CODE
“MOD_CODE” acts as foreign key of the table because the attribute “MOD_CODE” is also present in the table MODEL and it references MODEL and hence it forms a link between the two tables.
“For Table Name: MODEL:
Foreign Key: None
There is no Foreign Key attribute present in the table because there is no attribute in the table except the primary key which is present in any other database table.
For Table Name: PILOT:
Primary Key: EMP_NUM
“EMP_NUM” acts as foreign key of the table because the attribute “EMP_NUM” is also present in the table EMPLOYEE and it references EMPLOYEE and hence it forms a link between the two tables.
For Table Name: EMPLOYEE:
Foreign Key: None
There is no Foreign Key attribute present in the table because there is no attribute in the table except the primary key which is present in any other database table.
For Table Name: CUSTOMER:
Foreign Key: None
There is no Foreign Key attribute present in the table because there is no attribute in the table except the primary key which is present in any other database table.
Relationship among the tables:
A CUSTOMER requests many CHARTER trips and more than one CHARTER trip can be requested by a single customer. Hence, the relationship between CUSTOMER and CHARTER is one to many or 1: M.
An AIRCRAFT can fly many CHARTER trips but that each CHARTER trip is flown by one AIRCRAFT. Hence, the relationship between AIRCRAFT and CHARTER is one to many or 1: M.
Each AIRCRAFT references a single MODEL but a MODEL references many AIRCRAFT. Hence, the relationship between AIRCRAFT and MODEL is many to one or M: 1.
Many CHARTER trips are flown by a single PILOT and with a single COPILOT but a PILOT can fly only one charter trip at a time. Hence, the relationship between CHARTER and PILOT is many to one or M: 1.
All PILOTS are EMPLOYEES, but not all EMPLOYEES are PILOTS – some are
There is an optional (default) 1:1 relationship between EMPLOYEE and PILOT. It can be represented that EMPLOYEE is the “parent” of PILOT.
Elimination of Homonyms:
In the above tables, there are two attributes that are homonyms. The attributes are CHAR_PILOTS and CHAR_COPILOTS.
The two homonyms attributes are eliminated by modifying the CHARTER table and deleting the CHAR_PILOTS and CHAR_COPILOTS attributes.
A new table CREW is added, which is a composite table and it acts as a link between the CHARTER and EMPLOYEE tables. One CHARTER requires many CREW members and hence there is a one to many relation between them. Many CREWS are employees but one EMPLOYEE can be a part of one crew and hence it represents a many to one relationship between them.
CROW FOOT diagram to represent relationship between CHARTER, MODEL, AIRCRAFT, CREW, EMPLOYEE, PILOT and CUSTOMER:
The CROW FOOT diagram to represent relationship between CHARTER, MODEL, AIRCRAFT, CREW, EMPLOYEE, PILOT and CUSTOMER is shown below:
The above diagram represents the one to many relationship between CUSTOMER and CHARTER, one to many relationship between AIRCRAFT and CHARTER, many to one relationship between AIRCRAFT and MODEL, many to one relation between CHARTER and PILOT and an optional one to one relationship between PILOT and EMPLOYEE, one to many relationship between CHARTER and CREW, many to one relationship between CREW and EMPLOYEE.
Want to see more full solutions like this?
Chapter 3 Solutions
Database Systems: Design, Implementation, & Management
- Modify the PILOT table to add the attribute shown in the following table, with a precision and scale of (8,2).arrow_forwardA social media application stored information about each registered user in the table create table user (id int primary key auto_increment, name varchar(100) not null) A user can have many "friends" and can also be a "friend" to many other users. Specify how this relationship would be implemented by coding the create table statement(s) for the tables.arrow_forwardWhat three join types are included in the outer join classification?arrow_forward
- Implement a new independent entity phone in the Sakila database. Attributes and relationships are shown in the diagram below The diagram uses Sakila naming conventions. Follow the Sakila conventions for your table and column names: All lower case Underscore separator between root and suffix Foreign keys have the same name as referenced primary key Write CREATE TABLE and ALTER TABLE statements that: Implement the entity as a new phone table. Implement the has relationships as foreign keys in the Sakila customer, staff, and store tables. Remove the existing phone column from the Sakila address table. Step 2 requires adding a foreign key constraint to an existing table. Ex: ALTER TABLE customer ADD FOREIGN KEY (phone_id) REFERENCES phone(phone_id) ON DELETE SET NULL ON UPDATE CASCADE; Specify data types as follows: phone_id, phone_number, and country_code have data type INT. phone_type has date type VARCHAR(12) and contains strings like 'Home', 'Mobile', and 'Other'. Apply these…arrow_forwardThe Driver Relationship team realized that maintaining driver IDs is difficult and requested an automatic way of incrementing the value when a new driver is added. You need to make the changes on the table to automatically increment the DRIVER_ID. After the change, you need to insert the following driver: First Name: Nursin Last Name: Yilmaz Driving License ID: 4141447 Start Date: 2021-12-28 Driving License Checked: True Rating: 4.0arrow_forwardImplement the above schema with its attributes and relation.• Drop the column PAdress from the above Person table.• Change the column PDesignation from above PersonOfficeDetails table VARCHAR (20) toVARCHAR (30).• Change the data type/definition of column POfficeAdress INT to TEXT from above PersonOfficeDetails table.• Rename Person Table to Person_AlterTable.• Insert one meaning full data on each table of your choicearrow_forward
- For 1 through 4, state whether the statements are True or False (Do not provide any explanation or comment but only an answer of either True or False) A composite attribute is an attribute that is composed of several attributes. A derived attribute is an attribute whose value is calculated and not permanently stored in adatabase. Every table is a relation. Referential integrity constraint is a rule stating that no primary key column can be optional.arrow_forward1-Write the syntax to create a relational table called staff that references an object table called people that was created using an people_type. Relational Table Name staff attribute staff_id attribute department attribute personarrow_forwardCreate a view vLab2_xxx in the class database listing the number of staff and the maximum and minimum salary for each branch city. Your view header names should be: (city, count, max_salary, min_salary), in ascending order of count.arrow_forward
- Create an MVC web application having the index view given below. The given view will be populated from the model book having data members (Id, Title). Write only the code for the model class with the action SubmitImg() to submit the data in the database table book(id, title). Use suitable data annotations attributes to perform empty field validation also. No need to write view code.arrow_forwardThink about the foreign-key constraint on the instructor's department name property. Please provide examples of additions and deletions that could go against the foreign-key restriction.arrow_forward[ slide-15 .png ] Q1. Please write SQL codes to create below member database table based on the previous slide (slide 15) Video E-R Modeling Diagram. child table parent table RENTAL MEMBER 1. The rental table is a child table, and the member table is a parent table. 2. We need to create member parent table first before creating rental child table. 3. The member_ID column is the primary key column, and rest columns are NOT NULL constraints. DESC MEMBER Name Null? Type member NOT NULL NUMBER(4, 0) first_name NOT NULL VARCHAR2(25) last_name NOT NULL VARCHAR2(25) street NOT NULL VARCHAR2(25) city NOT NULL VARCHAR2(15) phone NOT NULL CHAR(11) valid_date NOT NULL DATE balance NOT NULL NUMBERarrow_forward
- Database Systems: Design, Implementation, & Manag...Computer ScienceISBN:9781305627482Author:Carlos Coronel, Steven MorrisPublisher:Cengage LearningDatabase Systems: Design, Implementation, & Manag...Computer ScienceISBN:9781285196145Author:Steven, Steven Morris, Carlos Coronel, Carlos, Coronel, Carlos; Morris, Carlos Coronel and Steven Morris, Carlos Coronel; Steven Morris, Steven Morris; Carlos CoronelPublisher:Cengage Learning