CIW 1D0-541 問題集

質問 1：
Consider the Recreation relation shown in the exhibit. You need to apply a SQL statement to the
Recreation relation that will return the following data:
Which SQL statement applied to the Recreation relation will return this data?

A. SELECT DISTINCT Activity FROM Recreation;
B. SELECT Activity FROM Recreation WHERE NOT LIKE Activity;
C. SELECT Activity FROM Recreation;
D. SELECT Activity FROM Recreation WHERE DISTINCT Activity;
正解：A

質問 2：
The exhibit shows a relation for a company projects. Which candidate key(s) would best serve as the primary key for this relation?

A. S_Date and E_Date
B. ltem_Num and E_Date
C. ProjJD
D. Proj_ID and Item_Num
正解：D

質問 3：
Which of the following best describes the two-tier database architecture?
A. The user accesses a database server using a terminal.
B. The user interface, data-processing logic, database access and data validation functions are performed on a mainframe server.
C. The user interface and data-processing logic are performed by the client whereas the server handles database access and data validation functions.
D. The user interface and data validation functions are performed by the client whereas the data-processing logic is performed on a server.
正解：C

質問 4：
Which subset of Structured Query Language (SQL) is used to create and name database entities?
A. Data Definition Language
B. Data Manipulation Language
C. Database Entity Language
D. Data Query Language
正解：A

質問 5：
Consider the Information Engineering diagram shown in the exhibit for a building management company. Referential integrity must be maintained such that a building cannot be deleted when it has residents. Building_ID, R_ID, Room_Count and Room_Num are integer numbers, whereas Bldg_Name, Location and Res_Name are all represented by variable-length strings with a maximum of 20 characters. Which SQL statement best implements the relations shown in this diagram?

A. CREATE TABLE BUILDING (
Building_ID INTEGER NOT NULL PRIMARY KEY,
Bldg_Name VARCHAR (20),
Location VARCHAR (20),
Room_Count INTEGER );
CREATE TABLE RESIDENT (
R_ID NOT NULL PRIMARY KEY,
Room_Num INTEGER,
Res_Name VARCHAR (20),
Building_ID INTEGER NOT NULL,
FOREIGN KEY Building_ID REFERENCES BUILDING (Building_ID)
ON DELETE NO CHECK
ON UPDATE CASCADE);
B. CREATE TABLE BUILDING (
Building_ID INTEGER NOT NULL PRIMARY KEY,
Bldg_Name VARCHAR (20),
Location VARCHAR (20),
Room_Count INTEGER );
CREATE TABLE RESIDENT (
R_ID NOT NULL PRIMARY KEY,
Room_Num INTEGER,
Res_Name VARCHAR (20),
Building_ID INTEGER NOT NULL,
FOREIGN KEY Building_ID REFERENCES BUILDING (Building_ID)
ON DELETE NO CHECK
ON UPDATE CASCADE);
C. CREATE TABLE BUILDING (
Building_ID INTEGER NOT NULL PRIMARY KEY,
Bldg_Name VARCHAR (20),
Location VARCHAR (20),
Room_Count INTEGER );
CREATE TABLE RESIDENT (
R_ID NOT NULL PRIMARY KEY,
Room_Num INTEGER,
Res_Name VARCHAR (20),
Building_ID INTEGER NOT NULL,
FOREIGN KEY Building_ID REFERENCES BUILDING (Building_ID)
ON DELETE NO CHECK
ON UPDATE CASCADE);
D. CREATE TABLE BUILDING (
Building_ID INTEGER NOT NULL PRIMARY KEY,
Bldg_Name VARCHAR (20),
Location VARCHAR (20),
Room_Count INTEGER );
CREATE TABLE RESIDENT (
R_ID NOT NULL PRIMARY KEY,
Room_Num INTEGER,
Res_Name VARCHAR (20),
Building_ID INTEGER NOT NULL,
FOREIGN KEY Building_ID REFERENCES RESIDENT (Building_ID)
ON DELETE NO CHECK);
正解：B