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2016 Jun 1Z0-882 test preparation
Q11. What are two ways in which normalizing your tables helps improve performance In MySQL?
A. Smaller table sizes and row lengths improve sorting operations.
B. Separate tables allow indexing more columns.
C. Fewer nullable column improve index usage.
D. Normalizing Improves the performance of innodb_file_per _table.
Q12. A complex query consists of eight populated tables that are all connected via INNER JOIN operands as shown:
You modify the query and replace the SELECT operand with SELECT STRAIGHT JOIN.
What is the effect of adding STRAIGHT JOINs to the query?
A. The optimizer processes only the JOINs in the sequence listed in the query.
B. The optimizer will only JOIN the tables by using their PRIMARY KEYS or UNIQUE constraints.
C. The optimizer will only JOIN the tables in sequence from smallest to largest.
D. The optimizer ignores all terms in the WHERE clause until all JOINs have been completed.
Q13. You are connected to a MySQL server and using a prepared statement. You accidentally exit your session.
What will happen if you log back in to use your prepared statement?
A. The statement exists, but will need to be deallocated and re-created.
B. The statement exists, but the user variables need to be redefined.
C. The statement can be used, if the MySQL server hasn’t been restarted.
D. The statement no longer exists.
Q14. Inspect the CREATE TABLE below:
Mysql> CREATE TABLE foo (a INT, PRIMARY KEY (a)) ENGINE =InnoDB;
Query Ok, 0 rows affected, 2 warnings (0.11 sec)
Mysql> SHOW WARNINGS;
Which two is true connecting the meaning of the warnings?
A. The InnoDB storage engine was disabled during server startup.
B. Global variable skip _innodb was set to ON after the server had started.
C. The default storage engine MYISAM was used for the table created.
D. MYSQL server was not started with the option default –storage –engine=InnoDB
E. Needed to specify TYPE = InnoDB instead of ENGINE=InnoDB
Q15. The tab-delimited file”/tmp/people,txt contains:
1636 Carsten Pederson Denmark 4672 Kai Voigt Germany 4628 Max Mether France
This is the structure of the people table: Mysq1> DESCRIBE people;
Which statement will load the first and last names into the Names column and the country into the country column?
A. LOAD DATA INFILE ‘/tmp/people.txt’INTO TABLE PEOPLE@First=$2. @Last=$3
(CONCAT (@First, ‘ ‘,@Last) , @ Country)
B. LOAD DATA INFILE ‘/tmp/people.txt ‘ INTO TABLE People
@Skip=$1 , @ First=$2, @Last=$3, @ Country=4,
(CONCAT (@First, ‘ ‘ .@ Last) , @ Country)
C. LOAD DATA INFILE ‘/tmp/people.txt ‘INTO TABLE People
(@ Skip, @First , @Last, @Country
SET Name=CONCAT (@First, ‘ ‘,@Last)
D. LOAD DATA INFILE ‘/tmp/people,txt, INTO TABLE People.
(@Skip. @First, @Last, @Country)
E. It is not possible to load the data from the file/tmp/people.txt into the people table,as
Updated 1Z0-882 practice exam:
Q16. You have two tables: news_source and news_feed. Here is some sample data from the news _feed table:
A. Option A
B. Option B
C. Option C
D. Option D
Q17. The people table contains the data as shown:
Which two statements return two rows each?
A. SELECT DISTINCT last_name, first_name FROM people
B. SELECT 1,2 FROM people GROUP BY last_name
C. SELECT first_name, last _name FROM people WHERE age LIKE ‘2’
D. SELECT 1, 2 FROM people WHERE last _name =’smith’
E. SELECT first _name, last_name FROM people LIMIT 1, 2
Q18. Consider the structures of the country and countrylanguage tables. mysql >DESCRIBE country;
mysql> DESCRIBE countrylanguage;
Which query will give you the list of all European countries where German is spoken?
A. SELECT Code AS c, Name FROM Country WHERE Continent = ‘Europe’ AND EXISTS ( SELECT * FROM CountryLanguage WHERE CountryCode = Code And Language= ‘German’ )
B. SELECT Code AS c, Name FROM Country WHERE Continent = ‘Europe’ AND Name IN ( SELECT * FROM CountryLanguage WHERE CountryCode = Code AND Language =’German’ )
C. SELECT Code AS c, Name FROM Country WHERE Continent = ‘ Europe’ AND EXIST ANY ( SELECT Language, CountryCode FROM CountryLanguage WHERE CountryCode =Code AND Language = ‘German’ ) D. SELECT Code AS c, Name FROM Country WHERE Continent = ‘Europe’ AND ( SELECT * FROM CountryLanguage WHERE CountryCode =Code AND Language =’German’ )
Q19. A table (t1) contains 1000 random integer values in the first column (col1). The random values range from 1 to 1000.
You execute this query: SELECT col1 FROM t1 WHERE col1< 100 UNION SELECT col1 FROM t1 WHERE col1 BETWEEN 100 and 200 UNION ALL
SELECT col1 FROM t1 WHERE col 1 >=900
What is the output?
A. A list of unique values within the ranges of 1-200 and 900-1000
B. A list of unique values within the range of 1-200 and a list of all values, including duplicates, on the table within the range of 900-1000
C. A list of all values , including duplicates, in the range of 1-200 and a list of unique values in the range of 900-1000
D. A list of all values, including duplicates, in the ranges of 1-200 and 900-1000
E. An error, because mixing UNION and UNION ALL in the same query is not permitted
Q20. You have been tasked to create a database that will store a list of all managers and the employees who report directly to them. The following is stipulated:
. No manage is managing more than three people.
. No employee can work for more than one manage.
Which of these designs represents a normalized schema that meets the project requirements?
A. CREATE TABLE ‘manager’
‘manager’ varchar (50) DEFAULT NULL,
‘employee2’ varchar (50) DEFAULT NULL,
‘employee’ varchar (50) DEFAULT NULL,
UNIQUE ( ‘manager ‘, ‘employee1’, ‘employee2, ‘employee3’)
B. CREATE TABLE ‘managers’ (
“id’ int(11) NOT NULL AUTO_INCREMENT,
‘manager’ varchar (50) DEFAULT NULL ,
PRIMARY KEY (‘id’)
CREATE TABLE “employees’ (
‘id’ int(11) NOT NULL AUTO _INCREMENT,
‘manager_id’ int(11) DEFAULT NULL,
‘employee varchar (25) DEFAULT NULL,
PRIMARY KEY (‘id’)
C. CREATE TABLE ‘manager’ (
‘manager’ varchar (50) DEFAULT NULL,
‘employee_list’varchar (150) DEFAULT NULL,
D. CREATE TABLE ‘message’ (
‘id’ int(11) NOT NULL AUTO_INCREMENT,
‘manager’ varchar(50) DEFAULT NULL,
PRIMARY KEY (“id’)
CREATE TABLE ‘employees’ (
‘id int (11) NOT NULL AUTO _INCREMENT,
‘ employees’ varchar(25) DEFAULT NULL,