Question 1

A stylized presentation of data for a selected audience is known as a(n)_____________.

Accepted Answer

The answer of A stylized presentation of data for a...

Question 2

Using the materialization strategy,when a user submits a query using a view,the query defining the view is first executed creating a temporary table,and then the query using the view is executed on the temporary table.

Accepted Answer

A) True 
 B)False

Question 3

Only single-table views are updatable with SQL modification statements.

Accepted Answer

A) True 
 B)False

Question 4

The use of views in application development can provide the following benefit:

Accepted Answer

A)  data independence 
B)  simplification 
C)  security 
D)  all of the above 
A)  data independence 
B)  simplification 
C)  security 
D)  all of the above

Question 5

The major query formulation issue for hierarchical reports is the _______________ of output,i.e.whether the output should contain individual rows or groups of rows.

Accepted Answer

The answer of The major query formulation issue for hierarchical...

Question 6

Although there are many advantages to using views,the main drawback is that performance can sometimes be poor in processing some complex views.

Accepted Answer

A) True 
 B)False

Question 7

In a 1-M updatable query,to insert a row into the child table,you would provide only the values needed to insert a row into the child table.

Accepted Answer

A) True 
 B)False

Question 8

If a table name used in a view is changed,the view definition does not have to be changed but the applications using the view must be changed.

Accepted Answer

A) True 
 B)False

Question 9

In a 1-M updatable query,to insert a row into both the parent and child tables,you must supply the primary keys and all the required columns of both the parent and child tables.

Accepted Answer

A) True 
 B)False

Question 10

Typically in a hierarchical form,the data from the two 1-M tables shown on the main form and the subform would be the result of executing a join query on the two tables for a specific value of the foreign key.

Accepted Answer

A) True 
 B)False

Question 11

Figuer:
CUSTOMER (primary key = CID) $\begin{array}{|l|l|l|l|l|}
\hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { Cregion } \
\hline 10 & \text { BLACK } & 40 & \text { SD } & \text { SD } \
\hline 20 & \text { GREEN } & 25 & \text { SD } & \text { SD } \
\hline 30 & \text { JONES } & 30 & \text { La Mesa } & \text { SD } \
\hline 40 & \text { MARTIN } & 35 & \text { LA } & \text { LA } \
\hline 50 & \text { SIMON } & 22 & \text { Pomona } & \text { LA } \
\hline 60 & \text { VERNON } & 60 & \text { Poway } & \text { SD } \
\hline
\end{array}$ RENTALS (primary key = RTN) \[\begin{array} { | l | l | l | l | l | l | l | l | } 
\hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } & \text { VIN } & \text { DATE-IN } \
\hline 10 & \text { FORD } & \text { 10-Oct-00 } & \text { SD } & \text { SD } & 1 & \text { F1 } & \text { 01-Dec-00 } \
\hline 10 & \text { GM } & 01 - \text { Jan-01 } & \text { SD } & \text { LA } & 2 & & \
\hline 10 & \text { FORD } & 01 - \text { Feb-01 } & \text { LA } & \text { SD } & 3 & & \
\hline 20 & \text { NISSAN } & 07 - J u l - 00 & \text { SD } & & 4 & \text { N1 } & \
\hline 30 & \text { FORD } & 01 - J u l - 00 & \text { SD } & \text { SD } & 5 & \text { F2 } & 30 \text {-Nov-00 } \
\hline 30 & \text { GM } & 01 - \text {-Dec-00 } & \text { SD } & \text { SD } & 6 & \text { G1 } & \
\hline 40 & \text { FORD } & 01 - \text { Aug-00 } & \text { LA } & \text { LA } & 7 & \text { F3 } & \
\hline 50 & \text { GM } & 01 - \text { Sep-00 } & \text { LA } & \text { SD } & 8 & \text { G2 } & \
\hline 60 & \text { NISSAN } & 02 \text {-Jan-01 } & \text { SD } & \text { LA } & 9 & & \
\hline
\end{array}\] In the table RENTALS, RTN provides the rental number (it is the primary key), CID refers to the CID in CUSTOMER, PICKUP is the city where the car was (or will be) picked up and Return is the city where the car was (or will be) returned, VIN is a foreign key that references the table CAR. The rental company has two branches, one in San Diego (SD) and one in Los Angeles (LA). Cars are picked up and returned by the customers at these two locations.
RENTCOST $$\begin{array} { | l | l | } 
\hline \text { MAKE } & \text { COST } \
\hline \text { FORD } & 30 \
\hline \text { GM } & 40 \
\hline \text { NISSAN } & 30 \
\hline \text { TOYOTA } & 20 \
\hline \text { VOLVO } & 50 \
\hline
\end{array}$$ In RENTCOST, COST shows the base cost of renting a given MAKE for one day.
CAR $$\begin{array} { | l | l | l | l | } 
\hline \text { VIN } & \text { MAKE } & \text { RENTED } & \text { AT } \
\hline \text { F1 } & \text { FORD } & \text { NO } & \text { SD } \
\hline \text { F2 } & \text { FORD } & \text { NO } & \text { SD } \
\hline \text { F3 } & \text { FORD } & \text { YES } & \text { IR } \
\hline \text { F4 } & \text { FORD } & \text { NO } & \text { LA } \
\hline \text { G1 } & \text { GM } & \text { YES } & \text { SD } \
\hline \text { G2 } & \text { GM } & \text { YES } & \text { SD } \
\hline \text { N1 } & \text { NISSAN } & \text { YES } & \text { LA } \
\hline \text { N2 } & \text { NISSAN } & \text { NO } & \text { LA } \
\hline \text { T1 } & \text { TOYOTA } & \text { NO } & \text { IR } \
\hline \text { T2 } & \text { TOYOTA } & \text { NO } & \text { IR } \
\hline \text { V1 } & \text { VOLVO } & \text { NO } & \text { LA } \
\hline \text { V2 } & \text { VOLVO } & \text { NO } & \text { LA } \
\hline
\end{array}$$ The table CAR (primary key = VIN) provides information about each car, in particular if it is currently rented, and where its usual storage location is (attribute AT).
CREATE VIEW CUST_SD AS
SELECT * FROM CUSTOMER
WHERE Cregion = 'SD'
CREATE VIEW CUST_40D AS
SELECT CNAME, RENTALS.MAKE, RENTCOST.COST
FROM CUSTOMER, RENTALS, RENTCOST
WHERE CUSTOMER.CID = RENTALS.CID
AND RENTALS.MAKE = RENTCOST.MAKE
AND RENTCOST < 40
CREATE VIEW Make_View (Make, NumRentals, Cregion, Storageat) AS
SELECT RENTALS.MAKE, Count (RTN), Cregion, AT
FROM CUSTOMER, RENTALS, CAR
WHERE CUSTOMER.CID = RENTALS.CID
AND RENTALS.VIN = CAR.VIN
GROUP BY RENTALS.MAKE, Cregion, At
-The number of columns in the view CUST_SD is:

Accepted Answer

A)  5 
B)  4 
C)  3 
D)  2 
A)  5 
B)  4 
C)  3 
D)  2

Question 12

Figuer:
CUSTOMER (primary key = CID)
CNAME is NOT NULL $$\begin{array} { | l | l | l | l | l | } 
\hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { Cregion } \
\hline 10 & \text { BLACK } & 40 & \text { SD } & \text { SD } \
\hline 20 & \text { GREEN } & 25 & \text { SD } & \text { SD } \
\hline 30 & \text { JONES } & 30 & \text { La Mesa } & \text { SD } \
\hline 40 & \text { MARTIN } & 35 & \text { LA } & \text { LA } \
\hline 50 & \text { SIMON } & 22 & \text { Pomona } & \text { LA } \
\hline 60 & \text { VERNON } & 60 & \text { Poway } & \text { SD } \
\hline
\end{array}$$ RENTALS (primary key = RTN)
CID is a foreign key referencing CUSTOMER
CID is NOT NULL
MAKE is NOT NULL
MAKE is a foreign key referencing RENTCOST $$\begin{array} { | l | l | l | l | l | l | } 
\hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \
\hline 10 & \text { FORD } & 10 - \text { Oct-00 } & \text { SD } & \text { SD } & 1 \
\hline 10 & \text { GM } & 01 - \text {-Jan-01 } & \text { SD } & \text { LA } & 2 \
\hline 10 & \text { FORD } & 01 - \text { Feb-01 } & \text { LA } & \text { SD } & 3 \
\hline 20 & \text { NISSAN } & 07 - J u l - 00 & \text { SD } & & 4 \
\hline 30 & \text { FORD } & 01 - J u l - 00 & \text { SD } & \text { SD } & 5 \
\hline 30 & \text { GM } & 01 - \text {-Dec-00 } & \text { SD } & \text { SD } & 6 \
\hline 40 & \text { FORD } & 01 - \text { Aug-00 } & \text { LA } & \text { LA } & 7 \
\hline 50 & \text { GM } & 01 - \text { Sep-00 } & \text { LA } & \text { SD } & 8 \
\hline 60 & \text { NISSAN } & 02 \text {-Jan-01 } & \text { SD } & \text { LA } & 9 \
\hline
\end{array}$$ RENTCOST
MAKE is the primary key $$\begin{array} { | l | l | } 
\hline \text { MAKE } & \text { COST } \
\hline \text { FORD } & 30 \
\hline \text { GM } & 40 \
\hline \text { NISSAN } & 30 \
\hline \text { TOYOTA } & 20 \
\hline \text { VOLVO } & 50 \
\hline
\end{array}$$ (Access) View1:
SELECT RTN, MAKE, PICKUP, RENTALS.CID,
CUSTOMER.CID, CNAME, AGE, Cregion
FROM CUSTOMER INNER JOIN RENTALS
ON CUSTOMER.CID = RENTALS.CID
(Access) View2:
SELECT RTN, MAKE, PICKUP, RENTALS.CID,
CNAME, AGE, Cregion
FROM CUSTOMER INNER JOIN RENTALS
ON CUSTOMER.CID = RENTALS.CID
(Access) View3:
SELECT RTN, RENTALS.MAKE, PICKUP, RENTALS.CID,
CNAME, AGE, Cregion, COST
FROM (CUSTOMER INNER JOIN RENTALS
ON CUSTOMER.CID = RENTALS.CID.
INNER JOIN RENTCOST
ON RENTALS.MAKE = RENTCOST.MAKE
-(Access)INSERT INTO View2 (RTN,MAKE,PICKUP,RENTALS.CID,CNAME)
VALUES (11,'GM','SD',70,'JORDAN' )
Which of the following is true?

Accepted Answer

A)  This command inserts 1 row into RENTALS 
B)  This command inserts 1 row into CUSTOMER 
C)  After execution of this command, SELECT MAKE FROM View2 WHERE CNAME='JORDAN' will show 1 row 
D)  This command cannot execute 
A)  This command inserts 1 row into RENTALS 
B)  This command inserts 1 row into CUSTOMER 
C)  After execution of this command, SELECT MAKE FROM View2 WHERE CNAME='JORDAN' will show 1 row 
D)  This command cannot execute

Question 13

Which of the following operations is typical in queries for hierarchical reports?

Accepted Answer

A)  Join 
B)  Outer Join 
C)  Count 
D)  All of the above 
A)  Join 
B)  Outer Join 
C)  Count 
D)  All of the above

Question 14

Figuer:
CUSTOMER (primary key = CID)
CNAME is NOT NULL $$\begin{array} { | l | l | l | l | l | } 
\hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { Cregion } \
\hline 10 & \text { BLACK } & 40 & \text { SD } & \text { SD } \
\hline 20 & \text { GREEN } & 25 & \text { SD } & \text { SD } \
\hline 30 & \text { JONES } & 30 & \text { La Mesa } & \text { SD } \
\hline 40 & \text { MARTIN } & 35 & \text { LA } & \text { LA } \
\hline 50 & \text { SIMON } & 22 & \text { Pomona } & \text { LA } \
\hline 60 & \text { VERNON } & 60 & \text { Poway } & \text { SD } \
\hline
\end{array}$$ RENTALS (primary key = RTN)
CID is a foreign key referencing CUSTOMER
CID is NOT NULL
MAKE is NOT NULL
MAKE is a foreign key referencing RENTCOST $$\begin{array} { | l | l | l | l | l | l | } 
\hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \
\hline 10 & \text { FORD } & 10 - \text { Oct-00 } & \text { SD } & \text { SD } & 1 \
\hline 10 & \text { GM } & 01 - \text {-Jan-01 } & \text { SD } & \text { LA } & 2 \
\hline 10 & \text { FORD } & 01 - \text { Feb-01 } & \text { LA } & \text { SD } & 3 \
\hline 20 & \text { NISSAN } & 07 - J u l - 00 & \text { SD } & & 4 \
\hline 30 & \text { FORD } & 01 - J u l - 00 & \text { SD } & \text { SD } & 5 \
\hline 30 & \text { GM } & 01 - \text {-Dec-00 } & \text { SD } & \text { SD } & 6 \
\hline 40 & \text { FORD } & 01 - \text { Aug-00 } & \text { LA } & \text { LA } & 7 \
\hline 50 & \text { GM } & 01 - \text { Sep-00 } & \text { LA } & \text { SD } & 8 \
\hline 60 & \text { NISSAN } & 02 \text {-Jan-01 } & \text { SD } & \text { LA } & 9 \
\hline
\end{array}$$ RENTCOST
MAKE is the primary key $$\begin{array} { | l | l | } 
\hline \text { MAKE } & \text { COST } \
\hline \text { FORD } & 30 \
\hline \text { GM } & 40 \
\hline \text { NISSAN } & 30 \
\hline \text { TOYOTA } & 20 \
\hline \text { VOLVO } & 50 \
\hline
\end{array}$$ (Access) View1:
SELECT RTN, MAKE, PICKUP, RENTALS.CID,
CUSTOMER.CID, CNAME, AGE, Cregion
FROM CUSTOMER INNER JOIN RENTALS
ON CUSTOMER.CID = RENTALS.CID
(Access) View2:
SELECT RTN, MAKE, PICKUP, RENTALS.CID,
CNAME, AGE, Cregion
FROM CUSTOMER INNER JOIN RENTALS
ON CUSTOMER.CID = RENTALS.CID
(Access) View3:
SELECT RTN, RENTALS.MAKE, PICKUP, RENTALS.CID,
CNAME, AGE, Cregion, COST
FROM (CUSTOMER INNER JOIN RENTALS
ON CUSTOMER.CID = RENTALS.CID.
INNER JOIN RENTCOST
ON RENTALS.MAKE = RENTCOST.MAKE
-(Access)INSERT INTO View1 (RTN,MAKE,PICKUP,RENTALS.CID.
VALUES (10,'GM','SD',10)
Which of the following is not true?

Accepted Answer

A)  This command inserts 1 row into RENTALS 
B)  This command inserts 1 row into CUSTOMER 
C)  After execution of this command, SELECT MAKE FROM View1 will show 5 rows where MAKE has 'GM' for value 
D)  The CUSTOMER table is not updated 
A)  This command inserts 1 row into RENTALS 
B)  This command inserts 1 row into CUSTOMER 
C)  After execution of this command, SELECT MAKE FROM View1 will show 5 rows where MAKE has 'GM' for value 
D)  The CUSTOMER table is not updated

Question 15

A(n)__________________ is a document designed to support a business task,such as processing an order.

Accepted Answer

The answer of A(n)__________________ is a document designed to support...

Question 16

Modifications to views,such as deleting rows or changing the primary key column are subject to the integrity rules of the underlying base table.

Accepted Answer

A) True 
 B)False

Question 17

Like a form,a report can change the underlying data in the base tables.

Accepted Answer

A) True 
 B)False

Question 18

(Access)When implementing a hierarchical report involving two tables in a 1-M relationship,a SQL query using a join on the 1-table and the M-table is formulated,but the linking column is not identified in the query.

Accepted Answer

A) True 
 B)False

Question 19

The WITH RESTRICTIONS OPTION clause of the CREATE VIEW statement prevents side effects from occurring when a view is updated.

Accepted Answer

A) True 
 B)False

Question 20

Request to execute the following SQL2 command: DELETE FROM CUST_SD
WHERE RESID_CITY ='SD'
Which of the following statements is not true?

Accepted Answer

A)  After execution of this query, SELECT * from CUST_SD will show that two rows were deleted from CUST_SD 
B)  After execution of this query, SELECT * from CUST_SD will show that two rows were deleted from CUSTOMER 
C)  Both of the above 
D)  The DELETE statement cannot execute on CUST_SD 
A)  After execution of this query, SELECT * from CUST_SD will show that two rows were deleted from CUST_SD 
B)  After execution of this query, SELECT * from CUST_SD will show that two rows were deleted from CUSTOMER 
C)  Both of the above 
D)  The DELETE statement cannot execute on CUST_SD

A stylized presentation of data for a selected audience is known as a(n)_____________.

Using the materialization strategy,when a user submits a query using a view,the query defining the view is first executed creating a temporary table,and then the query using the view is executed on the temporary table.

Only single-table views are updatable with SQL modification statements.

The use of views in application development can provide the following benefit:

The major query formulation issue for hierarchical reports is the _______________ of output,i.e.whether the output should contain individual rows or groups of rows.

Although there are many advantages to using views,the main drawback is that performance can sometimes be poor in processing some complex views.

In a 1-M updatable query,to insert a row into the child table,you would provide only the values needed to insert a row into the child table.

If a table name used in a view is changed,the view definition does not have to be changed but the applications using the view must be changed.

In a 1-M updatable query,to insert a row into both the parent and child tables,you must supply the primary keys and all the required columns of both the parent and child tables.

Typically in a hierarchical form,the data from the two 1-M tables shown on the main form and the subform would be the result of executing a join query on the two tables for a specific value of the foreign key.

Which of the following operations is typical in queries for hierarchical reports?

A(n)__________________ is a document designed to support a business task,such as processing an order.

Modifications to views,such as deleting rows or changing the primary key column are subject to the integrity rules of the underlying base table.

Like a form,a report can change the underlying data in the base tables.

(Access)When implementing a hierarchical report involving two tables in a 1-M relationship,a SQL query using a join on the 1-table and the M-table is formulated,but the linking column is not identified in the query.

The WITH RESTRICTIONS OPTION clause of the CREATE VIEW statement prevents side effects from occurring when a view is updated.

Request to execute the following SQL2 command: DELETE FROM CUST_SD WHERE RESID_CITY ='SD' Which of the following statements is not true?

Introduction to Database Management

Introduction to Database Development

The Relational Data Model

Query Formulation With SQL

Understanding Entity Relations

Developing Data Models for Business Databases

Normalization of Relational Tables

Physical Database Design

Advanced Query Formulation With SQL

Stored Procedures and Triggers

View Design and Integration

Data and Database Administration

Transaction Management

Data Warehouse Technology and Management

Client-Server Processing and Distributed Databases

Object Database Management Systems

Filters

Exam 10: Application Development With Views

A stylized presentation of data for a selected audience is known as a(n)_____________.

Using the materialization strategy,when a user submits a query using a view,the query defining the view is first executed creating a temporary table,and then the query using the view is executed on the temporary table.

Only single-table views are updatable with SQL modification statements.

The use of views in application development can provide the following benefit:

The major query formulation issue for hierarchical reports is the _______________ of output,i.e.whether the output should contain individual rows or groups of rows.

Although there are many advantages to using views,the main drawback is that performance can sometimes be poor in processing some complex views.

In a 1-M updatable query,to insert a row into the child table,you would provide only the values needed to insert a row into the child table.

If a table name used in a view is changed,the view definition does not have to be changed but the applications using the view must be changed.

In a 1-M updatable query,to insert a row into both the parent and child tables,you must supply the primary keys and all the required columns of both the parent and child tables.

Typically in a hierarchical form,the data from the two 1-M tables shown on the main form and the subform would be the result of executing a join query on the two tables for a specific value of the foreign key.

Which of the following operations is typical in queries for hierarchical reports?

A(n)__________________ is a document designed to support a business task,such as processing an order.

Modifications to views,such as deleting rows or changing the primary key column are subject to the integrity rules of the underlying base table.

Like a form,a report can change the underlying data in the base tables.

(Access)When implementing a hierarchical report involving two tables in a 1-M relationship,a SQL query using a join on the 1-table and the M-table is formulated,but the linking column is not identified in the query.

The WITH RESTRICTIONS OPTION clause of the CREATE VIEW statement prevents side effects from occurring when a view is updated.

Request to execute the following SQL2 command: DELETE FROM CUST_SD WHERE RESID_CITY ='SD' Which of the following statements is not true?

Introduction to Database Management

Introduction to Database Development

The Relational Data Model

Query Formulation With SQL

Understanding Entity Relations

Developing Data Models for Business Databases

Normalization of Relational Tables

Physical Database Design

Advanced Query Formulation With SQL

Stored Procedures and Triggers

View Design and Integration

Data and Database Administration

Transaction Management

Data Warehouse Technology and Management

Client-Server Processing and Distributed Databases

Object Database Management Systems

Filters