Oracle 1Z0-882 Online Practice Questions and Exam Preparation

Oracle 1Z0-882 Online Questions & Answers

• Question 81:

  You have a database `dev' that contains 15 tables, all of which use the CHARACTER SET `utfs' and the COLLATION `utfs_general_ci'.

  You perform the command:

  ALTER DATABASE `dev' CHARACTER SET ='latin' COLLATION='latin1'_swedish_ci'

  What is the result?

  A. You get an error because database are not allowed to have CHARACTER SET or COLLATION attributes.
  B. You get an error because the settings for CHARACTER SET and COLLATION attributes do not match the settings for the tables inside the database.
  C. You get an error while trying to change from a more inclusive CHARACTER SET like `utfs to a less' inclusive CHARACTER SET like `latin'.
  D. You get an error because changes to the CHARACTER SET or COLLATION attribute can happen only for empty databases.
  E. The statement succeeds and new tables created in this database use the new settings as their default values.
  F. The statement succeeds and all of the tables inside the database are converted to user the new settings.
  E. The statement succeeds and new tables created in this database use the new settings as their default values.

• Question 82:

  Your MYSQL server was successfully running for a days, and then suddenly stopped .You are sure that no mysqld process is running. Which two may provide diagnostic information to help determine why the MYSQL server stopped?

  A. The general query log file
  B. The syslog on Linux/UNIX or the Event view on windows
  C. The slow query log file
  D. The MYSQL server error log file
  E. The binary log file
  D. The MYSQL server error log file
  E. The binary log file

• Question 83:

  Which statement correctly demonstrates using a subquery as a scalar expression?

  A. SELECT ( SELECT SUM (population) FROM Country SELECT SUM (population) FROM Country WHERE Code ='CAN' )
  B. SELECT SUM (population) FROM Country WHERE Code ="USA" + (SELECT SUM (population) FROM Country WHERE Code =.'CAN' )
  C. SELECT (SELECT SUM (population) FROM Country WHERE Code ='USA' FROM country WHERE Code= `CAN' )
  D. (SELECT SUM (population) FROM Country WHERE Code ="USA'
  D. (SELECT SUM (population) FROM Country WHERE Code ="USA'

• Question 84:

  The city table has the following structure:

  

  Consider the statement with an incorrect field name:

  PREPARE countryBYID FROM `SELECT country FROM city WHERE ID=?, What happens if a prepared statement named countryByID already exists when the above statement is executed?

  A. A duplicate name error will result because a prepared statement with the same name already exists.
  B. An unknown column error will result and the old prepared statement definition will remain in effect.
  C. An unknown column error will result and no prepared statement named countryByID will exist.
  D. A warning will result and the old prepared statement definition will remain in effect.
  A. A duplicate name error will result because a prepared statement with the same name already exists.

• Question 85:

  Which there statements describe valid reasons why queries that use "SELECT" construct are discouraged?

  A. SELECT * may cause more data than you need to be read from disk if your application needs only some columns.
  B. SELECT * causes more data than you need to be sent via the client/server protocol if your application needs only some columns.
  C. SELECT * prevents the use of indexes, so a full table scan for every query.
  D. SELECT *causes your application to depend on the columns present when you wrote it , so your application could break if the table structure changes.
  E. SELECT * causes the statements to return all rows from the table.
  D. SELECT *causes your application to depend on the columns present when you wrote it , so your application could break if the table structure changes.
  E. SELECT * causes the statements to return all rows from the table.

• Question 86:

  Consider the stored procedure

  CREATE PROCEDURE param_test (

  IN P_in INT,

  OUT P_out INT,

  INPUT P_inout INT)

  BEGIN

  SELECT P_in, P_out, P_ inout;

  SET P_in, P_inout

  END

  You execute a series of commands:

  

  What is the output of the CALL and SELECT?

  A. (0,0,0) and (0,0,0)
  B. (0,0,0,) and (0,200,300)
  C. (0,NULL,0) and(0,200,300)
  D. (0,NULL,0) and (100,200,300)
  A. (0,0,0) and (0,0,0)

• Question 87:

  You execute this EXPLAIN statement for a SELECT statement on the table named comics.which contains 1183 rows:

  Mysql> explain select comic_ title, publisher from comics where comic_title like `and Actionand';

  

  1 row in set (0.00 sec)

  You create the following index:

  CREATE INDEX cimic_title_idx ON comics (comic_title, publisher); You run the same EXPLAIN statement again;

  Mysql > explain select comic_title ,publisher from comics where comic_title like `and Actionand';

  

  1 row in the second SELECT statement need to read all 1183 rows in the index comic_title_idx?

  A. Because comic_title is not the primary key
  B. Because a LIKE statement always requires a full tables scan
  C. Because comic _title is part of acovering index
  D. Because a wildcard character is at the beginning of the search word
  B. Because a LIKE statement always requires a full tables scan

• Question 88:

  Consider a table my_table , with contents shown:

  

  You execute:

  SELECT a b, b a FROM my_table WHERE a < s ORDER BY b;

  What does this statement return?

  

  A. Option A
  B. Option B
  C. Option C
  D. Option D
  A. Option A

• Question 89:

  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, )
  A. CREATE TABLE `manager' `manager' varchar (50) DEFAULT NULL, `employee2' varchar (50) DEFAULT NULL, `employee' varchar (50) DEFAULT NULL, UNIQUE ( `manager `, `employee1', `employee2, `employee3') )

• Question 90:

  When working with stored routines, these details are available:

  

  The affected rows count

  

  The number of conditions that occurred

  

  The condition information, such as the error code and message

  Where can you find these default?

  A. In the Handler area, defined in the DECLARE handler_action HANDLER block in a stored routine
  B. In the Signal area, which is set with the help of the SIGNAL statement in a stored routine
  C. In the Diagnostics area, part, of which can be stored in user-defined or routine variables
  D. In the Error area, which can be accessed with the help of the SHOW ERRORS statement
  C. In the Diagnostics area, part, of which can be stored in user-defined or routine variables