It is impossible to pass Oracle 1z0 051 pdf exam without any help in the short term. Come to Examcollection soon and find the most advanced, correct and guaranteed Oracle oracle 1z0 051 practice questions. You will get a surprising result by our Up to the minute Oracle Database: SQL Fundamentals I practice guides.


♥♥ 2018 NEW RECOMMEND ♥♥

Free VCE & PDF File for Oracle 1Z0-051 Real Exam (Full Version!)

★ Pass on Your First TRY ★ 100% Money Back Guarantee ★ Realistic Practice Exam Questions

Free Instant Download NEW 1Z0-051 Exam Dumps (PDF & VCE):
Available on: http://www.surepassexam.com/1Z0-051-exam-dumps.html

Q11. - (Topic 1) 

View the Exhibit and examine the structure of the CUSTOMERS table. Evaluate the query statement: 

What would be the outcome of the above statement? 

A. It executes successfully. 

B. It produces an error because the condition on CUST_LAST_NAME is invalid. 

C. It executes successfully only if the CUST_CREDIT_LIMIT column does not contain any null values. 

D. It produces an error because the AND operator cannot be used to combine multiple BETWEEN clauses. 

Answer:


Q12. - (Topic 1) 

Evaluate the following query: 

What would be the outcome of the above query? 

A. It produces an error because flower braces have been used. 

B. It produces an error because the data types are not matching. 

C. It executes successfully and introduces an 's at the end of each promo_name in the output. 

D. It executes successfully and displays the literal" {'s start date was} " for each row in the output. 

Answer:

Explanation: 

So, how are words that contain single quotation marks dealt with? There are essentially 

two mechanisms available. The most popular of these is to add an additional single 

quotation mark next to each naturally occurring single quotation mark in the character 

string 

Oracle offers a neat way to deal with this type of character literal in the form of the 

alternative quote (q) operator. Notice that the problem is that Oracle chose the single quote 

characters as the special pair of symbols that enclose or wrap any other character literal. 

These character-enclosing symbols could have been anything other than single quotation 

marks. 

Bearing this in mind, consider the alternative quote (q) operator. The q operator enables 

you to choose from a set of possible pairs of wrapping symbols for character literals as 

alternatives to the single quote symbols. The options are any single-byte or multibyte 

character or the four brackets: (round brackets), {curly braces}, [squarebrackets], or <angle 

brackets>. Using the q operator, the character delimiter can effectively be changed from a 

single quotation mark to any other character 

The syntax of the alternative quote operator is as follows: 

q'delimiter'character literal which may include the single quotes delimiter' where delimiter 

can be any character or bracket. 

Alternative Quote (q) Operator 

Specify your own quotation mark delimiter. 

Select any delimiter. 

Increase readability and usability. 

SELECT department_name || q'[ Department's Manager Id: ]' 

|| manager_id 

AS "Department and Manager" 

FROM departments; 

Alternative Quote (q) Operator 

Many SQL statements use character literals in expressions or conditions. If the literal itself contains a single quotation mark, you can use the quote (q) operator and select your own quotation mark delimiter. You can choose any convenient delimiter, single-byte or multibyte, or any of the following character pairs: [ ], { }, ( ), or < >. In the example shown, the string contains a single quotation mark, which is normally interpreted as a delimiter of a character string. By using the q operator, however, brackets [] are used as the quotation mark delimiters. The string between the brackets delimiters is interpreted as a literal character string. 


Q13. - (Topic 2) 

Examine the structure of the PRODUCTS table: 

You want to display the names of the products that have the highest total value for UNIT_PRICE *QTY_IN_HAND. 

Which SQL statement gives the required output? 

A. 

SELECT prod_name FROM products WHERE (unit_price * qty_in_hand) = (SELECT MAX(unit_price * qty_in_hand) FROM products); 

B. 

SELECT prod_name FROM products WHERE (unit_price * qty_in_hand) = (SELECT MAX(unit_price * qty_in_hand) FROM products GROUP BY prod_name); 

C. 

SELECT prod_name FROM products GROUP BY prod_name HAVING MAX(unit_price * qty_in_hand) = (SELECT MAX(unit_price * qty_in_hand) FROM products GROUP BY prod_name); 

D. 

SELECT prod_name 

FROM products 

WHERE (unit_price * qty_in_hand) = (SELECT MAX(SUM(unit_price * qty_in_hand)) 

FROM products) 

GROUP BY prod_name; 

Answer:


Q14. - (Topic 2) 

In which two cases would you use an outer join? (Choose two.) 

A. The tables being joined have NOT NULL columns. 

B. The tables being joined have only matched data. 

C. The columns being joined have NULL values. 

D. The tables being joined have only unmatched data. 

E. The tables being joined have both matched and unmatched data. 

F. Only when the tables have a primary key/foreign key relationship. 

Answer: C,E 

Explanation: 

You use an outer join to also see rows that do not meet the join condition. 

Incorrect Answer: Ameet a join condition Bmeet a join condition Dmeet non join condition only Fdoes not take into consideration of primary key and foreign key relationship 

Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 4-17 


Q15. - (Topic 2) 

Examine the data in the LIST_PRICE and MIN_PRICE columns of the PRODUCTS table: 

Which two expressions give the same output? (Choose two.) 

A. NVL(NULLIF(list_price, min_price), 0) 

B. NVL(COALESCE(list_price, min_price), 0) 

C. NVL2(COALESCE(list_price, min_price), min_price, 0) 

D. COALESCE(NVL2(list_price, list_price, min_price), 0) 

Answer: B,D 

Explanation: 

Using the COALESCE Function 

The advantage of the COALESCE function over the NVL function is that the COALESCE 

function can take multiple alternate values. 

If the first expression is not null, the COALESCE function returns that expression; 

otherwise, it does a COALESCE of the remaining expressions. 

Using the COALESCE Function 

The COALESCE function returns the first non-null expression in the list. 

Syntax 

COALESCE (expr1, expr2, ... exprn) In the syntax: 

expr1 returns this expression if it is not null 

expr2 returns this expression if the first expression is null and this expression is not null 

exprn returns this expression if the preceding expressions are null Note that all expressions must be of the same data type. 


Q16. - (Topic 2) 

Examine the description of the CUSTOMERS table: 

The CUSTOMER_ID column is the primary key for the table. 

Which statement returns the city address and the number of customers in the cities Los Angeles or San Francisco? 

A. SELECT city_address, COUNT(*) FROM customers 

WHERE city_address IN ( ‘Los Angeles’, ‘San Fransisco’); 

B. SELECT city_address, COUNT (*) 

FROMcustomers 

WHERE city address IN ( ‘Los Angeles’, ‘San Fransisco’) 

GROUP BY city_address; 

C. SELECT city_address, COUNT(customer_id) 

FROMcustomers 

WHERE city_address IN ( ‘Los Angeles’, ‘San Fransisco’) 

GROUP BYcity_address, customer_id; 

D. SELECT city_address, COUNT (customer_id) 

FROM customers 

GROUP BY city_address IN ( ‘Los Angeles’, ‘San Fransisco’); 

Answer:

Explanation: 

Not C: The customer ID in the GROUP BY clause is wrong 


Q17. - (Topic 2) 

Which describes the default behavior when you create a table? 

A. The table is accessible to all users. 

B. Tables are created in the public schema. 

C. Tables are created in your schema. 

D. Tables are created in the DBA schema. 

E. You must specify the schema when the table is created. 

Answer:

Explanation: 

sorted by highest to lowest is DESCENDING order 

Incorrect Answer: Agrant the table privilege to PUBLIC Blogin as sysoper Dlogin as DBA or sysdba Eno such option is allow. 

Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 9-9 


Q18. - (Topic 1) 

Using the CUSTOMERS table, you need to generate a report that shows 50% of each credit amount in each income level. The report should NOT show any repeated credit amounts in each income level. Which query would give the required result? 

A. SELECT cust_income_level, DISTINCT cust_credit_limit * 0.50 AS "50% Credit Limit" 

FROM customers; 

B. SELECT DISTINCT cust_income_level, DISTINCT cust_credit_limit * 0.50 AS "50% 

Credit Limit" 

FROM customers; 

C. SELECT DISTINCT cust_income_level ' ' cust_credit_limit * 0.50 AS "50% Credit Limit" 

FROM customers; 

D. SELECT cust_income_level ' ' cust_credit_limit * 0.50 AS "50% Credit Limit" FROM 

customers; 

Answer:

Explanation: Duplicate Rows Unless you indicate otherwise, SQL displays the results of a query without eliminating the duplicate rows. To eliminate duplicate rows in the result, include the DISTINCT keyword in the SELECT clause immediately after the SELECT keyword. You can specify multiple columns after the DISTINCT qualifier. The DISTINCT qualifier affects all the selected columns, and the result is every distinct combination of the columns. 


Q19. - (Topic 1) 

View the Exhibit and examine the structure of the CUSTOMERS table .Which statement would display the highest credit limit available in each income level in each city in the CUSTOMERS table? 

A. SELECT cust_city, cust_income_level, MAX(cust_credit_limit ) FROM customers GROUP BY cust_city, cust_income_level, cust_credit_limit; 

B. SELECT cust_city, cust_income_level, MAX(cust_credit_limit) FROM customers GROUP BY cust_city, cust_income_level; 

C. SELECT cust_city, cust_income_level, MAX(cust_credit_limit) FROM customers GROUP BY cust_credit_limit, cust_income_level, cust_city ; 

D. SELECT cust_city, cust_income_level, MAX(cust_credit_limit) FROM customers GROUP BY cust_city, cust_income_level, MAX(cust_credit_limit); 

Answer:


Q20. - (Topic 1) 

You need to display the first names of all customers from the CUSTOMERS table that contain the character 'e' and have the character 'a' in the second last position. 

Which query would give the required output? 

A. 

SELECT cust_first_name FROM customers WHERE INSTR(cust_first_name, 'e')<>0 AND SUBSTR(cust_first_name, -2, 1)='a' 

B. 

SELECT cust_first_name FROM customers WHERE INSTR(cust_first_name, 'e')<>'' AND SUBSTR(cust_first_name, -2, 1)='a' 

C. 

SELECT cust_first_name FROM customers WHERE INSTR(cust_first_name, 'e')IS NOT NULL AND SUBSTR(cust_first_name, 1,-2)='a' 

D. 

SELECT cust_first_name FROM customers WHERE INSTR(cust_first_name, 'e')<>0 AND SUBSTR(cust_first_name, LENGTH(cust_first_name),-2)='a' 

Answer:

Explanation: 

The SUBSTR(string, start position, number of characters) function accepts three 

parameters and returns a string consisting of the number of characters extracted from the 

source string, beginning at the specified start position: 

substr('http://www.domain.com',12,6) = domain 

The position at which the first character of the returned string begins. 

When position is 0 (zero), then it is treated as 1. 

When position is positive, then the function counts from the beginning of string to find the 

first character. 

When position is negative, then the function counts backward from the end of string. 

substring_length 

The length of the returned string. SUBSTR calculates lengths using characters as defined 

by the input character set. SUBSTRB uses bytes instead of characters. SUBSTRC uses 

Unicode complete characters. 

SUBSTR2 uses UCS2 code points. SUBSTR4 uses UCS4 code points. 

When you do not specify a value for this argument, then the function 

The INSTR(source string, search item, [start position],[nth occurrence of search item]) 

function returns a number that represents the position in the source string, beginning from 

the given start position, where the nth occurrence of the search item begins: 

instr('http://www.domain.com','.',1,2) = 18