IBM Solutions Expert 000-512 Exam

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Exam : IBM 000-512
Title : db2 udb v7.1 family fundamentals


1. Given the following SQL statements:
CREATE TABLE tab1 (col1 INT)
CREATE TABLE tab2 (col1 INT)
INSERT INTO tab1 VALUES (NULL),(1)
INSERT INTO tab2 VALUES (NULL),(1)
SELECT COUNT(*) FROM tab1
WHERE col1 IN
(SELECT col1 FROM tab2)
Which of the following is the result of the SELECT COUNT(*) statement?
A. 1
B. 2
C. 3
D. 4
E. 0
Answer: A

2. Given the two following tables:
Names
Name Number
Wayne Gretzky 99
Jaromir Jagr 68
Bobby Orr 4
Bobby Hull 23
Brett Hull 16
Mario Lemieux 66
Steve Yzerman 19
Claude Lemieux 19
Mark Messier 11
Mats Sundin 13
Points
Name Points
Wayne Gretzky 244
Jaromir Jagr 68
Bobby Orr 129
Bobby Hull 93
Brett Hull 121
Mario Lemieux 189
Joe Sakic 94
Which of the following statements will display the player Names, numbers and points for all players with an entry in both tables? Which of the following statements will display the player? Names, numbers and points for all players with an entry in both tables?
A. SELECT names.name, names.number, points.points FROM names INNER JOIN points ON names.name=points.name
B. SELECT names.name, names.number, points.points FROM names FULL OUTER JOIN points ON names.name=points.name
C. SELECT names.name, names.number, points.points FROM names LEFT OUTER JOIN points ON names.name=points.name
D. SELECT names.name, names.number, points.points FROM names RIGHT OUTER JOIN points ON names.name=points.name
E. SELECT names.name, names.number, points.points FROM names FULL OUTER JOIN points ON names.name=points.name
F. SELECT names.name, names.number, points.points FROM names LEFT OUTER JOIN points ON names.name=points.name
G. SELECT names.name, names.number, points.points FROM names RIGHT OUTER JOIN points ON names.name=points.name
H. SELECT names.name, names.number, points.points FROM names LEFT OUTER JOIN points ON names.name=points.name
I. SELECT names.name, names.number, points.points FROM names RIGHT OUTER JOIN points ON names.name=points.name
J. SELECT names.name, names.number, points.points FROM names LEFT OUTER JOIN points ON names.name=points.name
K. SELECT names.name, names.number, points.points FROM names RIGHT OUTER JOIN points ON names.name=points.name
Answer: A

3. Which of the following describes why savepoints are NOT allowed inside an atomic unit of work?
A. Atomic units of work span multiple databases, but savepoints are limited to units of work which operate on a single database.
B. A savepoint implies that a subset of the work may be allowed to succeed, while atomic operations must succeed or fail as a unit.
C. A savepoint requires an explicit commit to be released, and commit statements are not allowed in atomic operations such as compound SQL.
D. A savepoint cannot be created without an active connection to a database, but atomic operations can contain a CONNECT as a sub-statement.
Answer: B

4. Given a table T1, with a column C1 char(3), that contains strings in upper and lower case letters, which of the following queries will find all rows where C1 is the string 'ABC' in any case?
A. SELECT * FROM t1 WHERE c1 = 'ABC'
B. SELECT * FROM t1 WHERE UCASE(c1) = 'ABC'
C. SELECT * FROM t1 WHERE IGNORE_CASE(c1 = 'ABC')
D. SELECT * FROM t1 WHERE c1 = 'ABC' WITH OPTION CASE INSENSITIVE
Answer: B

5. Given the tables:
COUNTRY
ID NAME PERSON CITIES
1 Argentina 1 10
2 Canada 2 20
3 Cuba 2 10
4 Germany 1 0
5 France 7 5
STAFF
ID LASTNAME
1 Jones
2 Smith
The statement:
SELECT * FROM staff, country
will return how many rows?
A. 2
B. 4
C. 5
D. 7
E. 10
Answer: E