Database Concepts (7th Edition)
7th Edition
ISBN: 9780133544626
Author: David M. Kroenke, David J. Auer
Publisher: PEARSON
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Chapter 5, Problem 5.2RQ
Explanation of Solution
Attribute:
In
- Generally columns in the table are called as an attributes.
Steps for transforming attributes into columns:
- Initially create a table for each entity.
- Each attributes in an entity must have a column in the corresponding table.
- Specify one column as the primary key for that table.
Column properties:
While transforming attributes into columns, it should specify some column properties. They are:
- Null status
- Data type
- Specify data constraints
- Default value
Example:
Consider the example of transforming attributes into columns is as follows:
Attribute:
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Use Mysql code
The database has three tables for tracking horse-riding lessons:
Horse with columns:
ID - primary key
RegisteredName
Breed
Height
BirthDate
Student with columns:
ID - primary key
FirstName
LastName
Street
City
State
Zip
Phone
EmailAddress
LessonSchedule with columns:
HorseID - partial primary key, foreign key references Horse(ID)
StudentID - foreign key references Student(ID)
LessonDateTime - partial primary key
Write a SELECT statement to create a lesson schedule for Feb 1, 2020 with the lesson date/time, student's first and last names, and the horse's registered name. Order the results in ascending order by lesson date/time, then by the horse's registered name. Make sure unassigned lesson times (student ID is NULL) appear in the results.
Hint: Perform a join on the LessonSchedule, Student, and Horse tables, matching the student IDs and horse IDs.
Alternate keys: Identify at least five keys (not already listed as PK or FK) needed by end users.
These indexes would be considered Alternate or Secondary keys and are mostly used for queries
and quick reporting. They may contain multiple columns.
The database has three tables for tracking horse-riding lessons:
Horse with columns:
ID - primary key
RegisteredName
Breed
Height
BirthDate
Student with columns:
ID - primary key
FirstName
LastName
Street
City
State
Zip
Phone
EmailAddress
LessonSchedule with columns:
HorseID - partial primary key, foreign key references Horse(ID)
StudentID - foreign key references Student(ID)
LessonDateTime - partial primary key
Write a SELECT statement to create a lesson schedule with the lesson date/time, horse ID, and the student's first and last names. Order the results in ascending order by lesson date/time, then by horse ID. Unassigned lesson times (student ID is NULL) should not appear in the schedule.
Hint: Perform a join on the Student and LessonSchedule tables, matching the student IDs
Chapter 5 Solutions
Database Concepts (7th Edition)
Ch. 5 - Explain how entities are transformed into tables.Ch. 5 - Prob. 5.2RQCh. 5 - Prob. 5.3RQCh. 5 - What is denormalization?Ch. 5 - Prob. 5.5RQCh. 5 - Explain the problems that denormalized tables may...Ch. 5 - Explain how the representation of weak entities...Ch. 5 - Explain how supertype and subtype entities are...Ch. 5 - Prob. 5.9RQCh. 5 - Prob. 5.10RQ
Ch. 5 - Show two different ways to represent the 1:1...Ch. 5 - Prob. 5.12RQCh. 5 - Prob. 5.13RQCh. 5 - Prob. 5.14RQCh. 5 - Prob. 5.15RQCh. 5 - Prob. 5.16RQCh. 5 - For your answer to question 5.15, code an SQL...Ch. 5 - Prob. 5.18RQCh. 5 - Prob. 5.20RQCh. 5 - Prob. 5.21RQCh. 5 - Explain how the terms parent table and child table...Ch. 5 - For your answers to questions 5.20, 5.21, and...Ch. 5 - Prob. 5.24RQCh. 5 - Prob. 5.25RQCh. 5 - Prob. 5.26RQCh. 5 - Prob. 5.27RQCh. 5 - Prob. 5.28RQCh. 5 - Define the three types of recursive binary...Ch. 5 - Prob. 5.30RQCh. 5 - Prob. 5.31RQCh. 5 - Prob. 5.32RQCh. 5 - Prob. 5.33RQCh. 5 - Prob. 5.34RQCh. 5 - Code an SQL statement that creates a table with...
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