Oracle 1z0-882 valid study dumps : Oracle Certified Professional, MySQL 5.6 Developer

Every 1z0-882 candidate eventually asks the same question: where is the study material that is actually valid? ValidDumps answers it with 100 Oracle Certified Professional, MySQL 5.6 Developer practice questions, each answer edited and re-verified by Oracle experts.

Oracle 1z0-882 Exam Overview:

Certification Vendor:	Oracle
Exam Name:	Oracle Certified Professional, MySQL 5.6 Developer
Exam Number:	1Z0-882
Certificate Validity Period:	Oracle does not specify a fixed expiration period for this certification
Related Certifications:	Oracle Certified Professional, MySQL 5.6 Developer Oracle Certified Professional, MySQL 5.7 Developer
Exam Price:	USD 245
Exam Duration:	120 minutes
Real Exam Qty:	70
Available Languages:	English
Passing Score:	60%
Exam Format:	Multiple Choice
Sample Questions:
Exam Way:	Delivered through Oracle's authorized testing partner (Pearson VUE) at test centers or via online proctored delivery where available.
Pre Condition:	No mandatory prerequisite exam; practical experience with MySQL 5.6 development is recommended.
Official Syllabus URL:	https://education.oracle.com/mysql-certification

Oracle 1z0-882 Exam Syllabus Topics:

Everything You Wanted to Ask About the 1z0-882 Exam

Oracle Certified Professional, MySQL 5.6 Developer Sample Questions:

Question 1

Consider the CREATE FUNCTION statement:
CREATE FUNCTION countrycount ()
BEGIN
DECLARE count INT;
SELECT COUNT (*) INTO count FROM country;
RETURN count ;
END
What is the outcome when you try to create the function?

A. An error result as the function must be defined with the CONTAINS SQL clause.
B. An error results as the SELECT must assign the return values to a user variable.
C. An error result as the variable type returned by the function must be defined with a RETURNS clause.
D. An error results as the count variable is not initialized with a value.

Question 2

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 all of the tables inside the database are converted to user the new settings.
F. The statement succeeds and new tables created in this database use the new settings as their default values.

Question 3

You have two tables: news_source and news_feed.

Here is some sample data from the news _feed table:

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

Question 4

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 will only JOIN the tables in sequence from smallest to largest.
B. The optimizer will only JOIN the tables by using their PRIMARY KEYS or UNIQUE constraints.
C. The optimizer ignores all terms in the WHERE clause until all JOINs have been completed.
D. The optimizer processes only the JOINs in the sequence listed in the query.

Question 5

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 EXIST ANY (
SELECT Language, CountryCode
FROM CountryLanguage
WHERE CountryCode =Code
AND Language = 'German'
)
B. SELECT Code AS c, Name
FROM Country
WHERE Continent = 'Europe'
AND EXISTS (
SELECT *
FROM CountryLanguage
WHERE CountryCode = Code
And Language= 'German'
)
C. SELECT Code AS c, Name
FROM Country
WHERE Continent = 'Europe'
AND (
SELECT *
FROM CountryLanguage
WHERE CountryCode =Code
AND Language ='German'
)
D. SELECT Code AS c, Name
FROM Country
WHERE Continent = 'Europe'
AND Name IN (
SELECT *
FROM CountryLanguage
WHERE CountryCode = Code
AND Language ='German'
)

Solutions: