Oracle 1Z0-882 Online Practice Questions and Exam Preparation

Oracle 1Z0-882 Online Questions & Answers

• Question 1:

  Assume the user has just connected to the MySQL server. What is the result of the query SELECT @ a?

  A. An error that @ a is undefined
  B. A single NULL
  C. An empty string
  D. The value of GLOBAL variable @ a
  D. The value of GLOBAL variable @ a

• Question 2:

  A MySQL command- line client is started with safe updates disabled. Mysql - -safe updates=0

  What happens when you execute an UPDATE statement without a WHERE clause?

  A. Results in an error
  B. Updates every row in the specified table(s)
  C. Results in - -safe-updates being enabled automatically
  D. Causes a syntax error
  A. Results in an error

• Question 3:

  Assume your connection uses SQL mode ANSI_QUOTES. Which two statements cause a syntax error?

  A. CREATE TABLE FRIENDS (NAME CHAR (10))
  B. CREATE TABLE BINARY (PRIMARY SMALLINT)
  C. CREATE TABLE `TABLE' (COLUMN' INTEGER)
  D. CREATE TABLE "CONDITION" ("DESCRIBE" TEXT)
  E. CREATE TABLE INTERVAL (ELAPSED_TIME TIME)
  A. CREATE TABLE FRIENDS (NAME CHAR (10))
  B. CREATE TABLE BINARY (PRIMARY SMALLINT)

• Question 4:

  When executing a stored routine, how is the SQL_MODE determined?

  A. By the default SQL_MODE of the server
  B. By the current SQL _MODE of the session
  C. By the SQL_MODE that was set when the routine was defined
  D. By using TRADITTIONAL regardless of any other settings
  B. By the current SQL _MODE of the session

• Question 5:

  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?

  A. SELECT * FROM tableA. tableB
  B. SELECT * FROM tableA JOIN tableB
  C. SELECT * FROM table A INNER JOIN tableB
  D. SELECT * FROM tableA NATURAL JOIN tableB
  E. SELECT and FROM tableA STRAIGHT JOIN tableB
  C. SELECT * FROM table A INNER JOIN tableB

• Question 6:

  You want to query the VARCHAR column ` code' values that match:

  

  Start with "p"

  

  End with "_"

  

  Contain more than 3 characters

  Assume that sql_mode is blank.

  Which two queries select only those rows?

  A. SELECT code FROM operations WHERE code LIKE "pandand_";
  B. SELECT code FROM operations WHERE code LIKE " 'and'and_' "ESCAPE " ` '';
  C. SELECT code FROM operations WHERE code LIKE "pand_\_";
  D. SELECT code FROM operations WHERE code LIKE "p_\and\_";
  E. SELECT code FROM operations WHERE code LIKE "p_7_and;_"ESCAPE "/";
  C. SELECT code FROM operations WHERE code LIKE "pand_\_";
  E. SELECT code FROM operations WHERE code LIKE "p_7_and;_"ESCAPE "/";

• Question 7:

  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 8:

  The friends table has the columns and contents as shown:

  Mysql> SELECT * FROM friends;

  

  This statement was executed: SELECT AVG (age) FROM friends What value is returned?

  A. NULL
  B. 10
  C. 20
  A. NULL

• Question 9:

  Which three statement types can be prepared?

  A. LOAD DATA INFILE
  B. CREATE TABLE
  C. CREATE VIEW
  D. ALTER VIEW
  E. CALL
  A. LOAD DATA INFILE
  B. CREATE TABLE

• Question 10:

  You have two lists of values to correlate.

  

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

  A. 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,
  B. SELECT colors1.name, count(colorse2. Name) FROM colorse1 .name =colors2.name WHERE colors1. Name =colors2.name GROUP BY colors1.name,
  C. SELECT colors1. Name count (colors2.name) FROM colors1 INNER JOIN colors2 on colors1. Name =colors2. Name GROUP BY colors1 .name;
  D. SELECT colors1.name, count (colors2.name) FROM JOIN colors2 on colors1 .name =colors2.name GROUP BY colors1.name;
  E. SELECT colors1.name, count (colors2.name) FROM colors1 RIGHT JOIN colors1 on colors1 .name =colors2.name GROUP BY colors1.name;
  D. SELECT colors1.name, count (colors2.name) FROM JOIN colors2 on colors1 .name =colors2.name GROUP BY colors1.name;