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2018 Jan Oracle Official New Released 1z0-882
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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?
-
SELECT * FROM tableA. tableB
-
SELECT * FROM tableA JOIN tableB
-
SELECT * FROM table A INNER JOIN tableB
-
SELECT * FROM tableA NATURAL JOIN tableB
-
SELECT amp; FROM tableA STRAIGHT JOIN tableB
Answer: D
Question No: 42
The contents of the parent and child tables are:
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?
-
It does not delete anything from any table but returns an error.
-
It deletes one row from the parent table but does not affect the child table.
-
It deletes one row from the parent table and two rows from the child table.
-
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:
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?
-
An error results as the SELECT must assign the return values to a user variable.
-
An error results as the count variable is not initialized with a value.
-
An error result as the function must be defined with the CONTAINS SQL clause.
-
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?
-
CREATE DATABASE $test
-
CREATE DATABASE 1$
-
CREATE DATABASE $
-
CREATE DATABASE _
-
CREATE DATABASE 12
Answer: A
Question No: 45
A statement exists that can duplicate the definition of the ‘world’table.
What is missing?
CREATE TABLE t1 world
-
FROM
-
USING
-
COPY
-
LIKE
Answer: D
Question No: 46
The data from t1 table is:
Assuming You want to see this output:
Which query achieves the preceding result?
-
SELECT name FROM t1 WHERE name LIKE ,_e%
-
SELECT name FROM t1 WHERE name LIKE,e%.;
-
SELECT name FROM t1 GROUP BY name ORDER by name LIMIT 1,1;
-
SELECT name FROM t1 GROUP BY name HAVING sun ( marks)=176 ORDER BY
name;
Answer: C
Question No: 47
Which statement describes the process of normalizing databases?
-
All text is trimmed to fit into the appropriate fields. Capitalization and spelling errors are corrected.
-
Redundant tables are combined into one larger table to simplify the schema design.
-
Numeric values are checked against upper and lower accepted bounds. All text is purged of illegal characters.
-
Columns that contain repeating data values are split into separate tables to reduce item duplication.
-
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.
Which query lists all names in colors1 and how many total matches are there in colors2?
-
SELECT colors1 .name.count (colors2.name) FROM colors1. Colors2
WHERE
Colors1. Name = (SELECT DISTINCT name FROM colors2 WHERE colors1.name=colors2.name)
GROUP BY colorse1.name,
-
SELECT colors1.name, count(colorse2. Name) FROM colorse1 .name =colors2.name
WHERE colors1. Name =colors2.name GROUP BY colors1.name,
-
SELECT colors1. Name count (colors2.name) FROM colors1
INNER JOIN colors2
on colors1. Name =colors2. Name GROUP BY colors1 .name;
-
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;’;
->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;’;
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?
-
Because comic_title is not the primary key
-
Because a LIKE statement always requires a full tables scan
-
Because comic _title is part of acovering index
-
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?
-
Table
-
Column
-
Index
-
Stored procedure
-
Trigger
Answer: B,C,D
Explanation: https://dev.mysql.com/doc/refman/5.6/en/identifier-case-sensitivity.html
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