[Free] 2017(Jan) EnsurePass Dumpsleader Oracle 1z0-882 Dumps with VCE and PDF Download 41-50

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Oracle Certified Professional, MySQL 5.6 Developer

Question No: 41

You want to compare all columns of table A to columns with matching names in table B. You want to select the rows where those have the same values on both tables.

Which query accomplishes this?

  1. SELECT * FROM tableA. tableB

  2. SELECT * FROM tableA JOIN tableB

  3. SELECT * FROM table A INNER JOIN tableB


  5. SELECT amp; FROM tableA STRAIGHT JOIN tableB

Answer: D

Question No: 42

The contents of the parent and child tables are:

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The child table has the parent_id column that has a foreign key constraint to the id column of the parent table with ON DELETE CASCADE clause.

Consider the command WHERE id =1; What is the effect of the above command?

  1. It does not delete anything from any table but returns an error.

  2. It deletes one row from the parent table but does not affect the child table.

  3. It deletes one row from the parent table and two rows from the child table.

  4. It deletes one row from the parent table and sets the parent _id column to NULL in the child.

Answer: C

Question No: 43

Consider the CREATE FUNCTION statement:



SELECT COUNT (*) INTO count FROM country; RETURN count ;


What is the outcome when you try to create the function?

  1. An error results as the SELECT must assign the return values to a user variable.

  2. An error results as the count variable is not initialized with a value.

  3. An error result as the function must be defined with the CONTAINS SQL clause.

  4. An error result as the variable type returned by the function must be defined with a RETURNS clause.

Answer: D Explanation:

Routine Functions must provide a RETURNS clause noting data-type just after func_name and parameters, before characteristics.

Question No: 44

Assume that none of the databases exist. Which statement results in an error?






Answer: A

Question No: 45

A statement exists that can duplicate the definition of the ‘world’table.

What is missing?


  1. FROM

  2. USING

  3. COPY

  4. LIKE

Answer: D

Question No: 46

The data from t1 table is:

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Assuming You want to see this output:

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Which query achieves the preceding result?

  1. SELECT name FROM t1 WHERE name LIKE ,_e%

  2. SELECT name FROM t1 WHERE name LIKE,e%.;

  3. SELECT name FROM t1 GROUP BY name ORDER by name LIMIT 1,1;

  4. SELECT name FROM t1 GROUP BY name HAVING sun ( marks)=176 ORDER BY


Answer: C

Question No: 47

Which statement describes the process of normalizing databases?

  1. All text is trimmed to fit into the appropriate fields. Capitalization and spelling errors are corrected.

  2. Redundant tables are combined into one larger table to simplify the schema design.

  3. Numeric values are checked against upper and lower accepted bounds. All text is purged of illegal characters.

  4. Columns that contain repeating data values are split into separate tables to reduce item duplication.

  5. Indexes are created to improve query performance. The data of types of columns are adjusted to use the smallest allocation.

Answer: D

Question No: 48

You have two lists of values to correlate.

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Which query lists all names in colors1 and how many total matches are there in colors2?

  1. SELECT colors1 .name.count (colors2.name) FROM colors1. Colors2


    Colors1. Name = (SELECT DISTINCT name FROM colors2 WHERE colors1.name=colors2.name)

    GROUP BY colorse1.name,

  2. SELECT colors1.name, count(colorse2. Name) FROM colorse1 .name =colors2.name

    WHERE colors1. Name =colors2.name GROUP BY colors1.name,

  3. SELECT colors1. Name count (colors2.name) FROM colors1

    INNER JOIN colors2

    on colors1. Name =colors2. Name GROUP BY colors1 .name;

  4. SELECT colors1.name, count (colors2.name)

FROM JOIN colors2

on colors1 .name =colors2.name GROUP BY colors1.name;

SELECT colors1.name, count (colors2.name) FROM colors1

RIGHT JOIN colors1

on colors1 .name =colors2.name GROUP BY colors1.name;

Answer: D

Question No: 49

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

Mysqlgt; explain select comic_ title, publisher from comics where comic_title like ‘amp; Actionamp;’;

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->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 gt; explain select comic_title ,publisher from comics where comic_title like ‘amp; Actionamp;’;

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1 row in set (0.00 sec)

Why did the second SELECT statement need to read all 1183 rows in the index comic_title_idx?

  1. Because comic_title is not the primary key

  2. Because a LIKE statement always requires a full tables scan

  3. Because comic _title is part of acovering index

  4. Because a wildcard character is at the beginning of the search word

Answer: C

Question No: 50

Which three database objects have non-case-sensitive names on all operating system?

  1. Table

  2. Column

  3. Index

  4. Stored procedure

  5. Trigger

Answer: B,C,D

Explanation: https://dev.mysql.com/doc/refman/5.6/en/identifier-case-sensitivity.html

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