Oracle 1Z0-071 Dumps 2019

We offers 1z0 071 dumps. "Oracle Database 12c SQL", also known as 1Z0-071 exam, is a Oracle Certification. This set of posts, Passing the 1Z0-071 exam with 1z0 071 dumps, will help you answer those questions. The 1z0 071 dumps covers all the knowledge points of the real exam. 100% real 1z0 071 dumps and revised by experts!

Online 1Z0-071 free questions and answers of New Version:

NEW QUESTION 1
View the exhibit and examine the structure of the STORES table. STORES table
NameNull?Type
---------------------- ------------- STORE_IDNUMBER NAMEVARCHAR2(100)
ADDRESSVARCHAR2(200) CITYVARCHAR2(100) COUNTRYVARCHAR2(100) START_DATEDATE END_DATEDATE PROPERTY_PRICENUMBER
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_DATE in the range of 36 months starting from 01-Jan-2000 and above.
Which SQL statement would get the desired output?

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

Answer: D

NEW QUESTION 2
Examine the commands used to create DEPARTMENT_DETAILS and COURSE_DETAILS:
1Z0-071 dumps exhibit
You want to generate a report that shows all course IDs irrespective of whether they have corresponding department IDs or not but no department IDs if they do not have any courses.
Which SQL statement must you use?

  • A. SELECT course_id, department_id, FROM department_details d RIGHT OUTER JOIN course_details c USING (department_id)
  • B. SELECT c.course_id, d.department_id FROM course_details c RIGHT OUTER JOIN.department_details d ON (c.depatrment_id=d.department_id)
  • C. SELECT c.course_id, d.department_id FROM course_details c FULL OUTER JOIN department_details d ON (c.department_id=
  • D. department_id)
  • E. SELECT c.course_id, d.department_id FROM course_details c FULL OUTER JOIN department_details d ON (c.department_id<>
  • F. department_id)

Answer: C

NEW QUESTION 3
Examine the commands used to create the DEPARTMENT_DETAILS and the COURSE-DETAILS tables: SQL> CREATE TABLE DEPARTMfiNT_D£TAILS
DEPARTMENT_ID NUMBER PRIMARY KEY , DEPARTMEHT_NAME VARCHAR2(50) ,
HOD VARCHAP2(50));
SQL> CREATE TABLE COURSE-DETAILS (COURSE ID NUMBER PRIMARY KEY , COURS_NAME VARCHAR2 (50) ,
DEPARTMEHT_ID NUMBER REFERENCES DEPARTMENT_DETAIL
You want to generate a list of all department IDs along with any course IDs that may have been assigned to them.
Which SQL statement must you use?

  • A. SELECT d.departranc_id, c.cours«_id FROM cource_deatils c LEFT OUTER JOIN departmnt_details d ON (c.dapartmsnt_id=d.departtnent_id);
  • B. SELECT d.department_id,
  • C. course_id FROM dapartment_details d RIGHT OUTER JOIN course_dotails c ON (c.depattnient_id=d.department_id) ;
  • D. SELECT d.department i
  • E. ccours_id FROM department_details d RIGHT OUTER JOIN course_details c ON (d.department_id);
  • F. SELECT d.department_id, c.course_id FROM department_details d LEFT OUTER JOIN course_details c ON (d.department id).- (DEPARTMENT_ID) ;

Answer: D

NEW QUESTION 4
View the Exhibit and examine the structure of the ORDERS table. The ORDER_ID column is the PRIMARY KEY in the ORDERS table.
1Z0-071 dumps exhibit
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 not get created because the DEFAULT value cannot be specified in the column definition.
  • B. 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.
  • 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 get created and all the constraints defined on the specified columns in the ORDERS table would be passed to the new table.

Answer: B

NEW QUESTION 5
Examine the structure of the BOOKS_TRANSACTIONS table:
1Z0-071 dumps exhibit
Examine the SQL statement:
1Z0-071 dumps exhibit
Which statement is true about the outcome?

  • A. It displays details only for members who have borrowed before today with RM as TRANSACTION_TYPE.
  • B. It displays details for members who have borrowed before today’s date with either RM as TRANSACTION_TYPE or MEMBER_ID as A101 and A102.
  • C. It displays details for only members A101 and A102 who have borrowed before today with RM TRANSACTION_TYPE.
  • D. It displays details for members who have borrowed before today with RM as TRANSACTION_TYPE and the details for members A101 or A102.

Answer: D

NEW QUESTION 6
The following are the steps for a correlated subquery, listed in random order:
The WHERE clause of the outer query is evaluated.
The candidate row is fetched from the table specified in the outer query.
This is repeated for the subsequent rows of the table, till all the rows are processed.
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. 4, 2, 1, 3
  • D. 2, 4, 1, 3

Answer: D

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

NEW QUESTION 7
Evaluate the following SQL statement:
SELECT product_name || 'it's not available for order' FROM product_information
WHERE product_status = 'obsolete';
You received the following error while executing the above query: ERROR
ORA-01756: quoted string not properly terminated What would you do to execute the query successfully?

  • A. Use Quote (q) operator and delimiter to allow the use of single quotation mark in the literal character string.
  • B. Enclose the literal character string in the SELECT clause within the double quotation marks.
  • C. Do not enclose the character literal string in the SELECT clause within the single quotation marks.
  • D. Use escape character to negate the single quotation mark inside the literal character string in the SELECT clause.

Answer: A

Explanation: References:
http://docs.oracle.com/cd/B19306_01/server.102/b14200/sql_elements003.htm

NEW QUESTION 8
Which statements are true? (Choose all that apply.)

  • A. The data dictionary is created and maintained by the database administrator.
  • B. The data dictionary views consists of joins of dictionary base tables and user-defined tables.
  • C. The usernames of all the users including the database administrators are stored in the data dictionary.
  • D. The USER_CONS_COLUMNS view should be queried to find the names of the columns to which a constraint applies.
  • E. Both USER_OBJECTS and CAT views provide the same information about all the objects that are owned by the user.
  • F. Views with the same name but different prefixes, such as DBA, ALL and USER, use the same base tables from the data dictionary.

Answer: CDF

Explanation: References:
https://docs.oracle.com/cd/B10501_01/server.920/a96524/c05dicti.htm

NEW QUESTION 9
Which two statements are true regarding savepoints? (Choose two.)

  • A. Savepoints may be used to ROLLBACK.
  • B. Savepoints can be used for only DML statements.
  • C. Savepoints are effective only for COMMIT.
  • D. Savepoints are effective for both COMMIT and ROLLBACK.
  • E. Savepoints can be used for both DML and DDL statements.

Answer: AB

NEW QUESTION 10
View the Exhibit and examine the data in the PRODUCT_INFORMATION table.
1Z0-071 dumps exhibit
Which two tasks would require subqueries? (Choose two.)

  • A. displaying all the products whose minimum list prices are more than average list price of products having the status orderable
  • B. displaying the total number of products supplied by supplier 102071 and having product status OBSOLETE
  • C. displaying the number of products whose list prices are more than the average list price
  • D. displaying all supplier IDs whose average list price is more than 500
  • E. displaying the minimum list price for each product status

Answer: AC

NEW QUESTION 11
Which statement is true regarding the USING clause in table joins? (Choose two.)

  • A. It can be used to join a maximum of three tables.
  • B. It can be used to access data from tables through equijoins as well as nonequijoins.
  • C. It can be used to join tables that have columns with the same name and compatible data types.
  • D. It can be used to restrict the number of columns used in a NATURAL join.

Answer: CD

NEW QUESTION 12
View the Exhibit and examine the structure of the CUSTOMERS table.
1Z0-071 dumps exhibit
Using the CUSTOMERS table, you must generate a report that displays a credit limit increase of 15% for all customers.
Customers with no credit limit should have “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 TO_CHAR(NVL(cust_credit_limit*.15), ‘Not Available’)) “NEW CREDIT” FROMcustomers
  • C. SELECT NVL (cust_credit_limit*.15, ‘Not Available’) “NEW CREDIT” FROM customers
  • D. SELECT NVL (cust_credit_limit, ‘Not Available’)*.15 “NEW CREDIT” FROM customers

Answer: C

NEW QUESTION 13
Which statement correctly grants a system privilege?

  • A. GRANT CREATE VIEWON table1 TOuser1;
  • B. GRANT ALTER TABLETO PUBLIC;
  • C. GRANT CREATE TABLETO user1, user2;
  • D. GRANT CREATE SESSIONTO ALL;

Answer: C

NEW QUESTION 14
Examine these SQL statements that are executed in the given order:
CREATE TABLE emp
(emp_no NUMBER (2) CONSTRAINT emp_emp_no_pk PRIMARY KEY, ename VARCHAR 2 (15),
salary NUMBER (8, 2),
mgr_no NUMBER(2) CONSTRAINT emp_mgr_fk REFERENCES emp (emp_no)); ALTER TABLE emp
DISABLE CONSTRAINT emp_emp_no_pk CASCADE; ALTER TABLE emp
ENABLE CONSTRAINT emp_emp_no_pk;
What will be the status of the foreign key EMP_MGR_FK?

  • A. It will be enabled and immediate.
  • B. It will be enabled and deferred.
  • C. It will remain disabled and can be re-enabled manually.
  • D. It will remain disabled and can be enabled only by dropping the foreign key constraint and re-creating it.

Answer: C

NEW QUESTION 15
View the exhibit and examine the structures of the EMPLOYEES and DEPARTMENTS tables. EMPLOYEES
NameNull?Type
---------------------- ------------- EMPLOYEE_IDNOT NULLNUMBER(6) FIRST_NAMEVARCHAR2(20) LAST_NAMENOT NULLVARCHAR2(25) HIRE_DATENOT NULLDATE JOB_IDNOT NULLVARCHAR2(10) SALARYNUMBER(10,2) COMMISSIONNUMBER(6,2) MANAGER_IDNUMBER(6) DEPARTMENT_IDNUMBER(4) DEPARTMENTS
NameNull?Type
---------------------- -------------
DEPARTMENT_IDNOT NULLNUMBER(4) DEPARTMENT_NAMENOT NULLVARCHAR2(30) MANAGER_IDNUMBER(6) LOCATION_IDNUMBER(4)
You want to update EMPLOYEES table as follows: 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 multiple columns (SALARY, COMMISSION) cannot be specified together in an UPDATE statement.
  • B. It generates an error because a subquery cannot have a join condition in a UPDATE statement.
  • C. It executes successfully and gives the desired update
  • D. It executes successfully but does not give the desired update

Answer: D

NEW QUESTION 16
View the exhibit and examine the structure of the EMPLOYEES table.
1Z0-071 dumps exhibit
You want to display all employees and their managers having 100 as the MANAGER_ID. You want the output in two columns: the first column would have the LAST_NAME of the managers and the second column would have LAST_NAME of the employees.
Which SQL statement would you execute?

  • A. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON m.employee_id = e.manager_idWHERE m.manager_id = 100;
  • B. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON m.employee_id = e.manager_idWHERE e.manager_id = 100;
  • C. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON e.employee_id = m.manager_idWHERE m.manager_id = 100;
  • D. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eWHERE m.employee_id = e.manager_id and AND e.manager_id = 100

Answer: B

NEW QUESTION 17
Which task can be performed by using a single Data Manipulation Language (DML) statement?

  • A. Removing all data only from a single column on which a primary key constraint is defined.
  • B. Removing all data from a single column on which a unique constraint is defined.
  • C. Adding a column with a default value while inserting a row into a table.
  • D. Adding a column constraint while inserting a row into a table.

Answer: A

NEW QUESTION 18
Examine the SQL statement used to create the TRANSACTION table. (Choose the best answer.)
SQL > CREATE TABLE transaction (trn_id char(2) primary key,
Start_date date DEFAULT SYSDATE, End_date date NOT NULL);
The value 'A1' does not exist for trn_id in this table.
Which SQL statement successfully inserts a row into the table with the default value for START_DATE?

  • A. INSERT INTO transaction VALUES ('A1', DEFAULT, TO_DATE(DEFAULT+10))
  • B. INSERT INTO transaction VALUES ('A1', DEFAULT, TO_DATE('SYSDATE+10'))
  • C. INSERT INTO transaction (trn_id, end_date) VALUES ('A1', '10-DEC-2014')
  • D. INSERT INTO transaction (trn_id, start_date, end_date) VALUES ('A1', , '10-DEC-2014')

Answer: C

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