Real Exam Questions 1z0-071 Dumps Exam Questions in here [Jan-2022]

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NEW QUESTION 33
View the exhibit and examine the structure in ORDERS and ORDER_ITEMS tables.

You need to create a view that displays the ORDER_ID, ORDER_DATE, and the total number of items in each order.
Which CREATE VIEW statement would create the views successfully?

• A. CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date, COUNT (i.line_item_id)"NO OF ITEMS"FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id, o.order_date;
• B. CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date, COUNT (i.line_item_id)FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id, o.order_date;
• C. CREATE OR REPLACE VIEW ord_vu (order_id, order_date)AS SELECT o.order_id, o.order_date, COUNT (i.line_item_id)"NO OF ITEMS"FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id, o.order_date;
• D. CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date, COUNT (i.line_item_id)
  ||"NO OF ITEMS"FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id, o.order_dateWHITH CHECK OPTION;

Answer: A

 

NEW QUESTION 34
View the exhibit and examine the structure of the STOREStable.

You want to display the NAME of the store along with the ADDRESS, START_DATE, PROPERTY_PRICE, and the projected property price, which is 115% of property price.
The stores displayed must have START_DATEin the range of 36 months starting from 01-Jan-2000 and above.
Which SQL statement would get the desired output?
SELECT name, concat (address| | ','| |city| |', ', country) AS full_address,

• A. start_date,
  property_price, property_price*115/100
  FROM stores
  WHERE TO_NUMBER(start_date-TO_DATE('01-JAN-2000','DD-MON-RRRR')) <=36;
  SELECT name, address||','||city||','||country AS full_address,
• B. start_date,
  property_price, property_price*115/100
  FROM stores
  WHERE MONTHS_BETWEEN (start_date, TO_DATE('01-JAN-2000','DD-MON-RRRR'))
  <=36;
• C. start_date,
  property_price, property_price*115/100
  FROM stores
  WHERE MONTHS_BETWEEN (start_date, '01-JAN-2000') <=36;
  SELECT name, concat (address| | ','| |city| |', ', country) AS full_address,
• D. start_date,
  property_price, property_price*115/100
  FROM stores
  WHERE MONTHS_BETWEEN (start_date, TO_DATE('01-JAN-2000','DD-MON-RRRR'))
  <=36;
  SELECT name, concat (address||','| |city| |', ', country) AS full_address,

Answer: B

 

NEW QUESTION 35
View the Exhibit and examine the structure of the ORDERS table.

You must select ORDER_ID and ORDER_DATE for all orders that were placed after the last order placed by CUSTOMER_ID 101.
Which query would give you the desired result?

• A. SELECT order_id, order_date FROM ordersWHERE order_date > ALL(SELECT MAX(order_date) FROM orders ) AND customer_id = 101;
• B. SELECT order_id, order_date FROM ordersWHERE order_date > ALL(SELECT order_date FROM orders WHERE customer_id = 101);
• C. SELECT order_id, order_date FROM ordersWHERE order_date >ANY(SELECT order_date FROM orders WHERE customer_id = 101);
• D. SELECT order_id, order_date FROM ordersWHERE order_date > IN(SELECT order_date FROM orders WHERE customer_id = 101);

Answer: B

 

NEW QUESTION 36
Examine this query:
SELECT employee_id,first_name,salary
FROM employees
WHERE hire_date>'&1';
Which two methods should you use to prevent prompting for a hire date value when this query is executed?

• A. Use the UNDEFINE command before executing the query.
• B. Use the DEFINE command before executing the query.
• C. Execute the SET VERIFY OFF command before executing the query.
• D. Replace'&1' with'&&1' in the query.
• E. Store the query in a script and pass the substitution value to the script when executing it.
• F. Execute the SET VERIFY ON command before executing the query.

Answer: B,E

 

NEW QUESTION 37
The CUSTOMERStable has a CUST_CREDIT_LIMITcolumn of data type NUMBER.
Which two queries execute successfully?

• A. SELECT NVL2(cust_credit_limit, TO_CHAR(cust_credit_limit * .15), 'Not
• B. Available') FROM customers;
  SELECT NVL2 (cust_credit_limit * .15, 'Not Available') FROM customers;
• C. SELECT TO_CHAR(NVL(cust_credit_limit * .15, 'Not Available')) FROM customers;
• D. SELECT NVL(TO_CHAR(cust_credit_limit * .15), 'Not Available') FROM customers;
• E. SELECT NVL (cust_credit_limit * .15, 'Not Available') FROM customers;

Answer: B,D

 

NEW QUESTION 38
Examine the data in the ORD_ITEMS table:

Evaluate this query:

Which statement is true regarding the result?

• A. It displays the item nos with their average quantity where the average quantity is more than double the overall minimum quantity of all the items in the table.
• B. It returns an error because the HAVING clause should be specified after the GROUP BY clause.
• C. It returns an error because all the aggregate functions used in the HAVING clause must be specified in the SELECT list.
• D. It displays the item nos with their average quantity where the average quantity is more than double the minimum quantity of that item in the table.

Answer: D

 

NEW QUESTION 39
The following are the steps for a correlated subquery, listed in random order:
1. The WHEREclause of the outer query is evaluated.
2. The candidate row is fetched from the table specified in the outer query.
3. This is repeated for the subsequent rows of the table, till all the rows are processed.
4. Rows are returned by the inner query, after being evaluated with the value from the candidate row in the outer query.
Which is the correct sequence in which the Oracle server evaluates a correlated subquery?

• A. 2, 1, 4, 3
• B. 4, 1, 2, 3
• C. 2, 4, 1, 3
• D. 4, 2, 1, 3

Answer: C

Explanation:
Explanation/Reference:
References:
http://rajanimohanty.blogspot.co.uk/2014/01/correlated-subquery.html

 

NEW QUESTION 40
View the Exhibit and examine the description of the PRODUCT_INFORMATION table. Which SQL statement would retrieve from the table the number of products having
LIST_PRICE as NULL?

• A. SELECT COUNT(list_price)
  FROM product_information
  WHERE list_price IS NULL;
• B. SELECT COUNT(DISTINCT list_price)
  FROM product_information
  WHERE list_price IS NULL;
• C. SELECT COUNT(NVL(list_price, 0))
  FROM product_information
  WHERE list_price IS NULL;
• D. SELECT COUNT(list_price)
  FROM product_information
  WHERE list_price = NULL;

Answer: C

 

NEW QUESTION 41
Evaluate the following query:

What would be the outcome of the above query?

• A. It executes successfully and introduces an 'sat the end of each promo_namein the output.
• B. It executes successfully and displays the literal " {'s start date was \> "for each row in the output.
• C. It produces an error because flower braces have been used.
• D. It produces an error because the data types are not matching.

Answer: A

 

NEW QUESTION 42
View the Exhibits and examine PRODUCTSand SALES tables.
Exhibit 1

Exhibit 2

You issue the following query to display product name the number of times the product has been sold:

What happens when the above statement is executed?

• A. The statement executes successfully and produces the required output.
• B. The statement produces an error because ITEM_CNTcannot be displayed in the outer query.
• C. The statement produces an error because the GROUP BYclause cannot be used in a subquery in the FROM clause.
• D. The statement produces an error because a subquery in the FROMclause and outer-joins cannot be used together.

Answer: A

 

NEW QUESTION 43
Which three are true about subqueries?

• A. A subquery cannot be used in the select list.
• B. < any returns true If the argument Is less than the lowest value returned by the subquery.
• C. A subquery can be used in a having clause.
• D. A subquery can be used in a WHERE clause.
• E. < ANY returns true If the argument is less than the highest value returned by the subquery.
• F. = ANY can only evaluate the argument against a subquery If it returns two or more values.
• G. A subquery cannot be used in a from clause.

Answer: C,D,E

 

NEW QUESTION 44
Examine the description of the PRODUCT_DETAILStable:

Which two statements are true?

• A. PRODUCT_IDcan be assigned the PRIMARY KEYconstraint.
• B. EXPIRY_DATEcontains the SYSDATEby default if no date is assigned to it.
• C. PRODUCT_NAMEcannot contain duplicate values.
• D. EXPIRY_DATEcannot be used in arithmetic expressions.
• E. PRODUCT_PRICEcan be used in an arithmetic expression even if it has no value stored in it.
• F. PRODUCT_PRICEcontains the value zero by default if no value is assigned to it.

Answer: C,D

 

NEW QUESTION 45
View the Exhibit and examine the description of the ORDERS table.

Which two WHERE clause conditions demonstrate the correct usage of conversion functions? (Choose two.) WHERE order_date_IN ( TO_DATE('OCT 21 2003','MON DD YYYY'), TO_CHAR('NOV 21

• A. WHERE order_date > TO_CHAR(ADD_MONTHS(SYSDATE,6),'MON DD YYYY')
• B. WHERE order_date > TO_DATE('JUL 10 2006','MON DD YYYY')
• C. WHERE TO_CHAR(order_date,'MON DD YYYY') = 'JAN 20 2003'
• D. 2003','MON DD YYYY') )

Answer: B,C

 

NEW QUESTION 46
Examine the structure of the BOOKS_TRANSACTIONS table.

You want to update this table such that BOOK_ID is set to 'INVALID' for all rows where no MEMBER_ID has been entered.
Examine this partial SQL statement:

Which condition must be used in the WHERE clause to perform the required update?

• A. MEMBER_ID = NULL;
• B. MEMBER_ID = '';
• C. MEMBER_ID IS NULL;
• D. MEMBER_ID = "";

Answer: C

 

NEW QUESTION 47
The customers table has the following structure:
You need to write a query that does the following tasks:
1. Display the first name and tax amount of the customers. Tax is 5% of their credit limit.
2. Only those customers whose income level has a value should be considered.
3. Customers whose tax amount is null should not be considered.
Which statement accomplishes all the required tasks?

• A. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNTFROM customersWHERE (cust_income_level, tax_amount) IS NOT NULL;
• B. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNTFROM customersWHERE cust_income_level <> NULL ANDtax_amount <> NULL;
• C. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNTFROM customersWHERE cust_income_level IS NOT NULL ANDcust_credit_limit IS NOT NULL;
• D. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNTFROM customersWHERE cust_income_level IS NOT NULL ANDtax_amount IS NOT NULL;

Answer: C

 

NEW QUESTION 48
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