Pass 1z0-071 Exam Latest Practice Questions Updated on Mar 05, 2024
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Oracle 1z1-071 Certification Exam is a popular certification exam for database professionals who want to demonstrate their skills and knowledge in Oracle Database SQL. 1z0-071 exam validates an individual's ability to write SQL queries to retrieve, insert, update, and delete data from a database. 1z0-071 exam is designed for individuals who have experience working with Oracle Database 12c or later versions.
Oracle Database SQL certification is a globally recognized credential that demonstrates a candidate's proficiency in SQL programming language. Oracle Database SQL certification is highly valued by employers as it is evidence of a candidate's ability to work with Oracle databases and use SQL to manage data. Oracle Database SQL certification also provides a foundation for advanced Oracle certifications such as the Oracle Certified Professional (OCP) and Oracle Certified Master (OCM) certifications.
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Recertification is required every three years. In general, this course takes 9 weeks to complete, but it depends on the person who is taking this course. Oracle 1Z0-071 Dumps cover all questions of 1Z0-071. In order to tune your 1Z0-071 certification after you have completed the course, you must pass Oracle's Exam Retake Retrieval Tool twice.
The Exam Retake Retrieval Tool was originally designed for people who have already passed a 1Z0-102 exam and want to retake the exam for a better score. You will need a separate username and password in order to access this tool from any computer. The process of using the tool consists of signing into your Oracle account then asking for your name in order to get your username and password. After you get your username and password, you connect to the Retake Retrieval Tool. When you are connected to the tool, it will show all of your 1Z0-071 exam histories. You will then see a link where you can retrieve your exam history.
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NEW QUESTION # 13
View the Exhibit and examine the structure of CUSTOMERS table.
Using the CUSTOMERS table, you need to generate a report that shows an increase in the credit limit by 15% for all customers. Customers whose credit limit has not been entered should have the message "Not Available" displayed.
Which SQL statement would produce the required result?
- A. SELECT NVL (TO CHAR(cust_credit_limit * .15), 'Not Available') "NEW CREDIT"FROM customers;
- B. SELECT NVL(cust_credit_limit), 'Not Available') "NEW CREDIT"FROM customers;
- C. SELECT TO_CHAR (NVL(cust_credit_limit * .15), 'Not Available') "NEW CREDIT"FROM customers;
- D. SELECT NVL(cust_credit_limit * .15), 'Not Available') "NEW CREDIT"FROM customers;
Answer: A
NEW QUESTION # 14
Which two statements are true regarding roles? (Choose two.)
- A. A role can be granted to PUBLIC.
- B. A user can be granted only one role at any point of time.
- C. Roles are named groups of related privileges that can be granted to users or other roles.
- D. A role can be granted to itself.
- E. The REVOKE command can be used to remove privileges but not roles from other users.
Answer: A,C
Explanation:
http://docs.oracle.com/cd/E25054_01/network.1111/e16543/authorization.htm#autoId28
NEW QUESTION # 15
Which statement is true regarding external tables?
- A. The default REJECT LIMIT for external tables is UNLIMITED.
- B. The data and metadata for an external table are stored outside the database.
- C. The CREATE TABLE AS SELECT statement can be used to upload data into a normal table in the database from an external table.
- D. ORACLE_LOADER and ORACLE_DATAPUMP have exactly the same functionality when used with an external table.
Answer: C
Explanation:
References:
https://docs.oracle.com/cd/B28359_01/server.111/b28310/tables013.htm
NEW QUESTION # 16
Evaluate the following query:
SQL> SELECT TRUNC (ROUND (156.00, -1),-1)
FROM DUAL;
What would be the outcome?
- A. 0
- B. 1
- C. 2
- D. 3
- E. 4
Answer: C
Explanation:
Explanation
References:
https://docs.oracle.com/cd/B19306_01/server.102/b14200/functions135.htm
https://docs.oracle.com/cd/B28359_01/olap.111/b28126/dml_functions_2127.htm
NEW QUESTION # 17
Examine the structure of the EMPLOYEEStable.
There is a parent/child relationship between EMPLOYEE_IDand MANAGER_ID.
You want to display the last names and manager IDs of employees who work for the same manager as the employee whose EMPLOYEE_IDis 123.
Which query provides the correct output?
SELECT e.last_name, m.manager_id
- A. FROM employees e LEFT OUTER JOIN employees m
on (e.employee_id = m.manager_id)
WHERE e.employee_id = 123;
SELECT e.last_name, e.manager_id - B. FROM employees e RIGHT OUTER JOIN employees m
on (e.employee_id = m.employee_id)
WHERE e.employee_id = 123;
SELECT m.last_name, e.manager_id - C. FROM employees e LEFT OUTER JOIN employees m
on (e.manager_id = m.manager_id)
WHERE e.employee_id = 123; - D. FROM employees e RIGHT OUTER JOIN employees m
on (e.manager_id = m.employee_id)
AND e.employee_id = 123;
SELECT e.last_name, m.manager_id
Answer: C
NEW QUESTION # 18
View the exhibit and examine the structures of the EMPLOYEESand DEPARTMENTStables.
EMPLOYEES
Name Null? Type
- ---------------- ----- -------------
EMPLOYEE_ID NOT NULL NUMBER(6)
FIRST_NAME VARCHAR2(20)
LAST_NAME NOT NULL VARCHAR2(25)
HIRE_DATE NOT NULL DATE
JOB_ID NOT NULL VARCHAR2(10)
SALARY NUMBER(10,2)
COMMISSION NUMBER(6,2)
MANAGER_ID NUMBER(6)
DEPARTMENT_ID NUMBER(4)
DEPARTMENTS
Name Null? Type
----------------- ----- -------------
DEPARTMENT_ID NOT NULL NUMBER(4)
DEPARTMENT_NAME NOT NULL VARCHAR2(30)
MANAGER_ID NUMBER(6)
LOCATION_ID NUMBER(4)
You want to update EMPLOYEEStable as follows:
Update only those employees who work in Boston or Seattle (locations 2900 and 2700).
Set department_idfor these employees to the department_idcorresponding to London
(location_id 2100).
Set the employees' salary in location_id2100 to 1.1 times the average salary of their department.
Set the employees' commission in location_id2100 to 1.5 times the average commission of their
department.
You issue the following command:
SQL> UPDATE employees
SET department_id
( SELECT department_id
FROM departments
WHERE location_id = 2100),
( salary, commission)
( SELECT 1.1*AVG(salary), 1.5*AVG(commission)
FROM employees, departments
WHERE departments.location_id IN(2900, 2700, 2100))
WHERE department_id IN
( SELECT department_id
FROM departments
WHERE location_id = 2900
OR location_id = 2700;
What is outcome?
- A. It generates an error because a subquery cannot have a join condition in a UPDATEstatement.
- B. It executes successfully but does not give the desired update
- C. It executes successfully and gives the desired update
- D. It generates an error because multiple columns (SALARY, COMMISSION)cannot be specified together in an UPDATEstatement.
Answer: B
NEW QUESTION # 19
Examine the structure of the EMPLOYEEStable.
There is a parent/child relationship between EMPLOYEE_IDand MANAGER_ID.
You want to display the last names and manager IDs of employees who work for the same manager as the employee whose EMPLOYEE_IDis 123.
Which query provides the correct output?
- A. SELECT m.last_name, e.manager_id
FROM employees e LEFT OUTER JOIN employees m
on (e.manager_id = m.manager_id)
WHERE e.employee_id = 123; - B. SELECT e.last_name, m.manager_id
FROM employees e LEFT OUTER JOIN employees m
on (e.employee_id = m.manager_id)
WHERE e.employee_id = 123; - C. SELECT e.last_name, m.manager_id
FROM employees e RIGHT OUTER JOIN employees m
on (e.manager_id = m.employee_id)
AND e.employee_id = 123; - D. SELECT e.last_name, e.manager_id
FROM employees e RIGHT OUTER JOIN employees m
on (e.employee_id = m.employee_id)
WHERE e.employee_id = 123;
Answer: A
NEW QUESTION # 20
Which statement adds a column called SALARY to the EMPLOYEES table having 100 rows, which cannot contain null?
- A. ALTER TABLE EMPLOYEESADD SALARY NUMBER(8,2) DEFAULT NOT NULL;
- B. ALTER TABLE EMPLOYEESADD SALARY NUMBER(8,2) DEFAULT 0 NOT NULL;
- C. ALTER TABLE EMPLOYEESADD SALARY NUMBER(8,2) DEFAULT CONSTRAINT p_nn NOT NULL;
- D. ALTER TABLE EMPLOYEESADD SALARY NUMBER(8,2) NOT NULL;
Answer: B
NEW QUESTION # 21
Which three statements are true about performing Data Manipulation Language (DML) operations on a view in an Oracle Database? (Choose three.)
- A. Views cannot be used to add or modify rows in an underlying table if the defining query of the view contains the DISTINCTkeyword.
- B. Views cannot be used to add or modify rows in an underlying table if the defining query of the view contains aggregating functions.
- C. The WITH CHECKclause has no effect when deleting rows from the underlying table through the view.
- D. Views cannot be used to add rows to an underlying table if the table has columns with NOT NULL constraints lacking default values which are not referenced in the defining query of the view.
- E. Views cannot be used to query rows from an underlying table if the table has a PRIMARY KEYand the PRIMARY KEYcolumns are not referenced in the defining query of the view.
- F. Insert statements can always be done on a table through a view.
Answer: B,D,E
NEW QUESTION # 22
Evaluate the following statement.
Which statement is true regarding the evaluation of rows returned by the subquery in the INSERT statement?
- A. They are evaluated by all the three WHEN clauses regardless of the results of the evaluation of any other WHEN clause.
- B. They are evaluated by the first WHEN clause. If the condition is true, then the row would be evaluated by the subsequent WHEN clauses.
- C. The insert statement would give an error because the ELSE clause is not present for support in case none of WHEN clauses are true.
- D. They are evaluated by the first WHEN clause. If the condition is false, then the row would be evaluated by the subsequent WHEN clauses.
Answer: A
Explanation:
Explanation
http://psoug.org/definition/WHEN.htm
NEW QUESTION # 23
Examine the description of the ORDER_ITEMS table:
Examine this incomplete query:
SELECT DISTINCT quantity * unit_price total_paid FROM order_items ORDER BY <clause>;
Which two can replace <clause> so the query completes successfully?
- A. total_paid
- B. quantity, unit_price
- C. product_id
- D. quantity * unit_price
- E. quantity
Answer: A,D
NEW QUESTION # 24
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;
SELECT name, concat (address||','| |city| |', ', country) AS full_address, - C. start_date,
property_price, property_price*115/100
FROM stores
WHERE MONTHS_BETWEEN (start_date, TO_DATE('01-JAN-2000','DD-MON-RRRR'))
<=36; - D. 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,
Answer: C
NEW QUESTION # 25
The EMPLOYEES table contains columns EMP_ID of data type NUMBER and HIRE_DATE of data type DATE
You want to display the date of the first Monday after the completion of six months since hiring.
The NLS_TERRITORY parameter is set to AMERICA in the session and, therefore, Sunday is the first day of the week Which query can be used?
- A. SELECT emp_id,NEXT_DAY(MONTHS_BETWEEN(hire_date,SYSDATE),6) FROM employees;
- B. SELECT emp_id,ADD_MONTHS(hire_date,6), NEXT_DAY('MONDAY') FROM employees;
- C. SELECT emp_id,NEXT_DAY(ADD_MONTHS(hire_date,6),1) FROM employees;
- D. SELECT emp_id,NEXT_DAY(ADD_MONTHS(hite_date,6),'MONDAY') FROM employees;
Answer: D
NEW QUESTION # 26
Which three statements are true regarding the data types?
- A. Only one LONGcolumn can be used per table.
- B. A TIMESTAMPdata type column stores only time values with fractional seconds.
- C. The minimum column width that can be specified for a VARCHAR2data type column is one.
- D. The BLOBdata type column is used to store binary data in an operating system file.
- E. The value for a CHARdata type column is blank-padded to the maximum defined column width.
Answer: A,C,E
NEW QUESTION # 27
View the Exhibit and examine the structure of the ORDER_ITEMS table.
You must select the ORDER_ID of the order that has the highest total value among all the orders in the ORDER_ITEMS table.
Which query would produce the desired result?
- A. SELECT order_idFROM order_itemsGROUP BY order_idHAVING SUM(unit_price*quantity) = (SELECT MAX(SUM(unit_price*quantity))FROM order_items GROUP BY order_id);
- B. SELECT order_idFROM order_itemsWHERE (unit_price*quantity) = (SELECT MAX(unit_price*quantity)FROM order_itemsGROUP BY order_id)
- C. SELECT order_idFROM order_itemsWHERE(unit_price*quantity) = MAX(unit_price*quantity)GROUP BY order_id;
- D. SELECT order_idFROM order_itemsWHERE(unit_price*quantity) = (SELECT MAX(unit_price*quantity)FROM order_items)GROUP BY order_id;
Answer: A
NEW QUESTION # 28
View the Exhibit and examine the structure of the ORDERS table. The ORDER_ID column is the PRIMARY KEY in the ORDERS table.
Evaluate the following CREATE TABLE command:
CREATE TABLE new_orders(ord_id, ord_date DEFAULT SYSDATE, cus_id)
AS
SELECT order_id.order_date,customer_id
FROM orders;
Which statement is true regarding the above command?
- A. The NEW_ODRDERS table would get created and only the NOT NULL constraint defined on the specified columns would be passed to the new table.
- B. The NEW_ODRDERS table would get created and all the constraints defined on the specified columns in the ORDERS table would be passed to the new table.
- C. The NEW_ODRDERS table would not get created because the column names in the CREATE TABLE command and the SELECT clause do not match.
- D. The NEW_ODRDERS table would not get created because the DEFAULT value cannot be specified in the column definition.
Answer: A
NEW QUESTION # 29
Which statements is true about using functions in WHERE and HAVING?
- A. using aggregate functions in the WHERE clause requires a subquery
- B. using aggregate functions in the HAVING clause requires a subquery
- C. using single-row functions in the WHERE clause requires a subquery
- D. using single-row functions in the HAVING clause requires a subquery
Answer: B,C
NEW QUESTION # 30
The SQL statements executed in a user session are as follows:
Which two statements describe the consequences of issuing the ROLLBACK TO SAVE POINT a command in the session? (Choose two.)
- A. Only the second DELETEstatement is rolled back.
- B. The rollback generates an error.
- C. Both the DELETEstatements and the UPDATEstatement are rolled back.
- D. Only the DELETEstatements are rolled back.
- E. No SQL statements are rolled back.
Answer: B,E
NEW QUESTION # 31
Examine the description of the CUSTONERS table:
CUSTNO is the PRIMARY KEY.
You must determine if any customers' details have been entered more than once using a different CUSTNO, by listing all duplicate names.
Which two methods can you use to get the required result?
- A. subquery
- B. PULL OUTER JOIN with self join
- C. LEFT OUTER JOIN with self join
- D. RIGHT OUTER JOIN with self join
- E. self Join
Answer: A,E
NEW QUESTION # 32
View the Exhibit and examine the structure of the PRODUCTS table.
You must display the category with the maximum number of items.
You issue this query:
What is the result?
- A. It generates an error because = is not valid and should be replaced by the IN operator.
- B. It generate an error because the subquery does not have a GROUP BY clause.
- C. It executes successfully but does not give the correct output.
- D. It executes successfully and gives the correct output.
Answer: B
NEW QUESTION # 33
The CUSTOMERStable has a CUST_CREDIT_LIMITcolumn of data type NUMBER.
Which two queries execute successfully?
- A. SELECT NVL (cust_credit_limit * .15, 'Not Available') FROM customers;
- B. SELECT NVL2(cust_credit_limit, TO_CHAR(cust_credit_limit * .15), 'Not
Available') FROM customers; - C. SELECT NVL2 (cust_credit_limit * .15, 'Not Available') FROM customers;
- D. SELECT NVL(TO_CHAR(cust_credit_limit * .15), 'Not Available') FROM customers;
- E. SELECT TO_CHAR(NVL(cust_credit_limit * .15, 'Not Available')) FROM customers;
Answer: C,D
NEW QUESTION # 34
Which three statements are true regarding subqueries? (Choose three.)
- A. Subqueries can contain GROUP BY and ORDER BY clauses.
- B. Multiple columns or expressions can be compared between the main query and subquery.
- C. Subqueries can contain ORDER BY but not the GROUP BY clause.
- D. Main query and subquery can get data from different tables.
- E. Only one column or expression can be compared between the main query and subquery.
- F. Main query and subquery must get data from the same tables.
Answer: A,B,D
Explanation:
http://docs.oracle.com/javadb/10.6.2.1/ref/rrefsqlj13658.html
NEW QUESTION # 35
Which three statements are true about indexes and their administration in an Oracle database?
- A. A descending index is a type of function-based index.
- B. If a query filters on an indexed column then it will always be used during execution of the query.
- C. An INVISIBLE index is not maintained when Data Manipulation Language (DML) is performed on its underlying table.
- D. A DROP INDEX statement always prevents updates to the table during the drop operation.
- E. An index can be created as part of a CREATE TABLE statement.
- F. A UNIQUE and non-unique index can be created on the same table column.
Answer: A,C,E
NEW QUESTION # 36
Which three statements are true about multiple row subqueries?
- A. Two or more values are always returned from the subquery.
- B. They can return multiple columns.
- C. They can contain HAVING clauses.
- D. They cannot contain a subquery.
- E. They can contain GROUP BY clauses.
Answer: A,B,E
NEW QUESTION # 37
Which statement is true about transactions?
- A. A set of DDL statements executed in a sequence ending with a COMMIT forms a single transaction.
- B. A combination of DDL and DML statements executed in a sequence ending with a COMMIT forms a single transaction.
- C. Each Data Definition Language (DDL) statement executed forms a single transaction.
- D. A set of Data Manipulation Language (DML) statements executed in a sequence ending with a SAVEPOINT forms a single transaction.
Answer: C
Explanation:
Explanation
References:
https://docs.oracle.com/database/121/CNCPT/transact.htm#CNCPT038
NEW QUESTION # 38
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