Modern Database Management (12th Edition)
12th Edition
ISBN: 9780133544619
Author: Jeffrey A. Hoffer, Ramesh Venkataraman, Heikki Topi
Publisher: PEARSON
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Chapter 6, Problem 6.37PAE
Are based on the class scheduling 3NF relations along with some sample data shown in Figure 6-11. Not shown in this figure are data for an ASSIGNMENT relation, which represents a many-to-many relationship between faculty and sections.
Note that values of the SectionNo column do not repeat across semesters.
Figure 6-11 Class scheduling relations (missing ASSIGNMENT)
Because of referential integrity, before any can be entered into the Section table, the course ID to be entered must already exist in the COURSE table. Write an SQL assertion that will enforce this constraint.
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For the following table for student grades, create a relational data model after normalization using MySQL.
PK/FK and cardinalities should be defined. If needed, you can make some assumptions.
All tables should be in the third normal form (3NF).
You need to submit a MySQL Workbench Model (Only files with the extension .mwb will be accepted).
[slide-19.png]
Q1. Please write SQL codes to create below reservation database table based on the previous slide (slide 19) Video E-R Modling Diagram
child table
RESERVATION
1. Reservation table is a child table, and the member_id column is a foreign key column which references member table’s member_id primary key column.
2. Reservation table is a child table, and the reserve_date and member_id columns are combined together as a composite primary key column.
DESC reservation
Name
Null?
Type
reserve_date
NOT NULL
DATE
member_ID
NOT NULL
NUMBER(4, 0)
title_ID
NOT NULL
NUMBER(3, 0)
Modify the MODEL table to add the attribute shown in the following table. MOD_WAIT_CHG should support up to 5 digits to the left of the decimal place and 2 digits to the right of the decimal place..
Chapter 6 Solutions
Modern Database Management (12th Edition)
Ch. 6 - Prob. 6.1RQCh. 6 - Prob. 6.2RQCh. 6 - Contrast the following terms: a. base table; view...Ch. 6 - Prob. 6.4RQCh. 6 - Prob. 6.5RQCh. 6 - Prob. 6.6RQCh. 6 - List six potential benefits of achieving an SQL...Ch. 6 - Prob. 6.8RQCh. 6 - Distinguish among data definition commands, data...Ch. 6 - Prob. 6.10RQ
Ch. 6 - Prob. 6.11RQCh. 6 - Prob. 6.12RQCh. 6 - Prob. 6.13RQCh. 6 - Drawing on material covered in prior chapters,...Ch. 6 - Explain and provide at least one example Of how to...Ch. 6 - Prob. 6.16RQCh. 6 - What is the difference between COUNT, COUNT...Ch. 6 - What is the evaluation order for the Boolean...Ch. 6 - If an SQL statement includes a GROUP BY clause,...Ch. 6 - Prob. 6.20RQCh. 6 - Prob. 6.21RQCh. 6 - Prob. 6.22RQCh. 6 - Explain why SQL is called a set-oriented language.Ch. 6 - When would the use of the LIKE key word with the...Ch. 6 - Prob. 6.25RQCh. 6 - Prob. 6.26RQCh. 6 - In what order are the clauses of an SQL statement...Ch. 6 - Within which clauses of an SQL statement can a...Ch. 6 - Prob. 6.29RQCh. 6 - Prob. 6.30RQCh. 6 - Prob. 6.31RQCh. 6 - Explain the difference between the WHERE and...Ch. 6 - Prob. 6.33RQCh. 6 - Prob. 6.34PAECh. 6 - Are based on the class scheduling 3NF relations...Ch. 6 - Prob. 6.36PAECh. 6 - Are based on the class scheduling 3NF relations...Ch. 6 - Are based on the class scheduling 3NF relations...Ch. 6 - Are based on the class scheduling 3NF relations...Ch. 6 - Prob. 6.40PAECh. 6 - Are based on the class scheduling 3NF relations...Ch. 6 - Prob. 6.42PAECh. 6 - Prob. 6.43PAECh. 6 - Are based on the class scheduling 3NF relations...Ch. 6 - Prob. 6.45PAECh. 6 - Prob. 6.46PAECh. 6 - Prob. 6.47PAECh. 6 - Prob. 6.48PAECh. 6 - Prob. 6.49PAECh. 6 - Prob. 6.50PAECh. 6 - Prob. 6.51PAECh. 6 - Prob. 6.52PAECh. 6 - Prob. 6.53PAECh. 6 - Modify the Product_T table by adding an attribute...Ch. 6 - Prob. 6.55PAECh. 6 - Add an order to the Order_T table and include a...Ch. 6 - Use the Pine Valley database to answer the...Ch. 6 - Prob. 6.58PAECh. 6 - Prob. 6.59PAECh. 6 - Prob. 6.60PAECh. 6 - Prob. 6.61PAECh. 6 - Prob. 6.62PAECh. 6 - Prob. 6.63PAECh. 6 - Prob. 6.64PAECh. 6 - Prob. 6.65PAECh. 6 - Prob. 6.66PAECh. 6 - Prob. 6.67PAECh. 6 - Prob. 6.68PAECh. 6 - Prob. 6.69PAECh. 6 - Prob. 6.70PAECh. 6 - Prob. 6.71PAECh. 6 - Prob. 6.72PAECh. 6 - Prob. 6.73PAECh. 6 - Prob. 6.74PAECh. 6 - Prob. 6.75PAECh. 6 - Prob. 6.76PAECh. 6 - Prob. 6.77PAECh. 6 - Prob. 6.78PAECh. 6 - Prob. 6.79PAECh. 6 - Prob. 6.80PAECh. 6 - Prob. 6.81PAECh. 6 - Prob. 6.82PAECh. 6 - Prob. 6.83PAECh. 6 - Prob. 6.84PAECh. 6 - Prob. 6.85PAE
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![[ 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](/questions-and-answers/slide-15-.png-q1.-please-write-sql-codes-to-create-below-member-database-table-based-on-the-previous/3b4446fb-5348-4dbb-944d-a99326b9259f){kind=link}
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