CIW 1D0-541 問題集

質問 1：
Consider the Dept1_Parts and Dept2_Parts relations shown in the exhibit. Which of the following SQL statements would create a set difference of the two relations with the widest variety of Structured Query Language dialects?

A. SELECT * FROM Dept1_Parts DIFFERENCE (SELECT Part_ID FROM Dept2_Parts);
B. SELECT * FROM Dept1_Parts EXCEPT (SELECT Part_ID FROM Dept2_Parts);
C. SELECT * FROM Dept1_Parts MINUS (SELECT Part_ID FROM Dept2_Parts);
D. SELECT * FROM Dept1_Parts DIFFERENCE (SELECT Part_ID FROM Dept2_Parts);
正解：D

質問 2：
Which process is used to prevent the current database operation from reading or writing a data item while that data item is being accessed by another operation?
A. Deadlock
B. Transaction
C. Lock
D. Time stamp
正解：C

質問 3：
Consider the Project relation shown in the exhibit as well as the following SQL statement: Which of the following tables shows the Project relation after execution of this SQL statement?


A. D
B. A
C. B
D. C
正解：B

質問 4：
Consider the Dept1_Parts and Dept2_Parts relations shown in the exhibit. Which of the following SQL statements would create a set difference of the two relations with the widest variety of Structured Query Language dialects?

A. SELECT * FROM Dept1_Parts WHERE Part_ID NOT IN (SELECT Part_ID FROM Dept2_Parts);
B. SELECT * FROM Dept1_Parts DIFFERENCE (SELECT Part_ID FROM Dept2_Parts);
C. SELECT * FROM Dept1_Parts EXCEPT(SELECT PartJD FROM Dept2_Parts);
D. SELECT * FROM Dept1_Parts MINUS (SELECT Part_ID FROM Dept2_Parts);
正解：A

質問 5：
Consider the Information Engineering diagram shown in the exhibit. 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 RESIDENT relation shown in this diagram?

A. CREATE TABLE RESIDENT (
R_ID INTEGER NOT NULL PRIMARY KEY,
Room_NumINTEGER,
Res_NameVARCHAR (20),
Building_IDINTEGER NOT NULL,
FOREIGN KEY Building_ID REFERENCES BUILDING (Building_ID));
B. CREATE TABLE RESIDENT (
R_ID INTEGER NULL PRIMARY KEY,
Room_NumFLOAT,
Res_NameVARCHAR,
Building_IDINTEGER NULL,
FOREIGN KEY Building_ID REFERENCES BUILDING (Building_ID));
C. CREATE TABLE RESIDENT (
R_ID INTEGER NOT NULL PRIMARY KEY,
Room_NumINTEGER,
Res_NameVARCHAR (20),
Building_IDINTEGER NOT NULL);
D. CREATE TABLE RESIDENT (
R_ID INTEGER NOT NULL PRIMARY KEY,
Room_NumBINARY,
Res_NameVARCHAR (20),
Building_IDINTEGER NOT NULL,
FOREIGN KEY Building_ID REFERENCES BUILDING (Building_ID));
正解：A