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NEW QUESTION 1
View the Exhibit and examine the description of the ORDERS table. (Choose two.)
1Z0-071 dumps exhibit
Which two WHERE clause conditions demonstrate the correct usage of conversion functions?

  • A. WHERE Order_date_IN ( TO_DATE('OCT 21 2003', 'MON DD YYYY'), TO_CHAR('NOV 21 2003', 'MON DD YYYY') )
  • B. WHERE Order_date > TO_CHAR(ADD_MONTHS(SYSDATE, 6), 'MON DD YYYY')
  • C. WHERE TO_CHAR(Order_date, 'MON DD YYYY') = 'JAN 20 2003'
  • D. WHERE Order_date > ( TO_DATE('JUL 10 2006', 'MON DD YYYY')

Answer: CD

NEW QUESTION 2
You issued the following command: SQL> DROP TABLE employees; Which three statements are true?

  • A. All uncommitted transactions are committed.
  • B. All indexes and constraints defined on the table being dropped are also dropped.
  • C. Sequences used in the employees table become invalid.
  • D. The space used by the employees table is reclaimed immediately.
  • E. The employees table can be recovered using the rollback command.
  • F. The employees table is moved to the recycle bin

Answer: ABF

NEW QUESTION 3
View the exhibit and examine the data in ORDERS_MASTER and MONTHLY_ORDERS tables.
ORDERS_MASTER ORDER_ID ORDER_TOTAL
1
1000
2
2000
3
3000
4
MONTHLY_ORDERS ORDER_ID ORDER_TOTAL
2
2500
3
Evaluate the following MERGE statement: MERGE_INTO orders_master o
USING monthly_orders m ON (o.order_id = m.order_id) WHEN MATCHED THEN
UPDATE SET o.order_total = m.order_total DELETE WHERE (m.order_total IS NULL) WHEN NOT MATCHED THEN
INSERT VALUES (m.order_id, m.order_total)
What would be the outcome of the above statement?

  • A. The ORDERS_MASTER table would contain the ORDER_IDs 1, 2, 3 and 4.
  • B. The ORDERS_MASTER table would contain the ORDER_IDs 1, 2 and 4.
  • C. The ORDERS_MASTER table would contain the ORDER_IDs 1, 2 and 3.
  • D. The ORDERS_MASTER table would contain the ORDER_IDs 1 and 2.

Answer: B

Explanation: References:
https://docs.oracle.com/cd/B28359_01/server.111/b28286/statements_9016.htm

NEW QUESTION 4
Which three statements are true regarding group functions? (Choose three.)

  • A. They can be used on columns or expressions.
  • B. They can be passed as an argument to another group function.
  • C. They can be used only with a SQL statement that has the GROUP BY clause.
  • D. They can be used on only one column in the SELECT clause of a SQL statement.
  • E. They can be used along with the single-row function in the SELECT clause of a SQL statement.

Answer: ABE

Explanation: References:
https://www.safaribooksonline.com/library/view/mastering-oracle-sql/0596006322/ch04.html

NEW QUESTION 5
Which three tasks can be performed using SQL functions built into Oracle Database?

  • A. displaying a date in a nondefault format
  • B. finding the number of characters in an expression
  • C. substituting a character string in a text expression with a specified string
  • D. combining more than two columns or expressions into a single column in the output

Answer: ABC

NEW QUESTION 6
Which three statements are true regarding the WHERE and HAVING clauses in a SQL statement? (Choose three.)

  • A. WHERE and HAVING clauses cannot be used together in a SQL statement.
  • B. The HAVING clause conditions can have aggregate functions.
  • C. The HAVING clause conditions can use aliases for the columns.
  • D. The WHERE clause is used to exclude rows before the grouping of data.
  • E. The HAVING clause is used to exclude one or more aggregated results after grouping data.

Answer: ABD

NEW QUESTION 7
Evaluate the following SQL statements that are issued in the given order:
CREATE TABLE emp
(emp_no NUMBER(2) CONSTRAINT emp_emp_no_pk PRIMARY KEY, ename VARCHAR2(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 would be the status of the foreign key EMP_MGR_PK?

  • A. It would remain disabled and can be enabled only by dropping the foreign key constraint and recreating it.
  • B. It would remain disabled and has to be enabled manually using the ALTER TABLE command.
  • C. It would be automatically enabled and immediate.
  • D. It would be automatically enabled and deferred.

Answer: B

NEW QUESTION 8
The first DROP operation is performed on PRODUCTS table using the following command: DROP TABLE products PURGE;
Then you performed the FLASHBACK operation by using the following command: FLASHBACK TABLE products TO BEFORE DROP;
Which statement describes the outcome of the FLASHBACK command?

  • A. It recovers only the table structure.
  • B. It recovers the table structure, data, and the indexes.
  • C. It recovers the table structure and data but not the related indexes.
  • D. It is not possible to recover the table structure, data, or the related indexes.

Answer: D

Explanation: References:
https://docs.oracle.com/cd/B19306_01/server.102/b14200/statements_9003.htm

NEW QUESTION 9
Evaluate the following ALTER TABLE statement:
ALTER TABLE orders
SET UNUSED (order_date); Which statement is true?

  • A. After executing the ALTER TABLE command, you can add a new column called ORDER_DATE to the ORDERS table.
  • B. The ORDER_DATE column should be empty for the ALTER TABLE command to execute succsessfully.
  • C. ROLLBACK can be used to get back the ORDER_DATE column in the ORDERS table.
  • D. The DESCRIBE command would still display the ORDER_DATE column.

Answer: A

NEW QUESTION 10
View the Exhibit and examine the details of the PRODUCT_INFORMATION table. (Choose two.)
1Z0-071 dumps exhibit
Evaluate this SQL statement:
SELECT TO_CHAR (list_price, '$9,999') From product_information;
Which two statements are true regarding the output?

  • A. A row whose LIST_PRICE column contains value 11235.90 would be displayed as #######.
  • B. A row whose LIST_PRICE column contains value 1123.90 would be displayed as $1,123.
  • C. A row whose LIST_PRICE column contains value 1123.90 would be displayed as $1,124.
  • D. A row whose LIST_PRICE column contains value 11235.90 would be displayed as $1,123.

Answer: AC

NEW QUESTION 11
View the Exhibit and examine PRODUCTS and ORDER_ITEMS tables.
1Z0-071 dumps exhibit
You executed the following query to display PRODUCT_NAME and the number of times the product has been ordered:
SQL>SELECT p.product_name, i.item_cnt
FROM (SELECT product_id, COUNT (*) item_cnt FROM order_items
GROUP BY product_id) i RIGHT OUTER JOIN products p ON i.product_id = p.product_id;
What would happen when the above statement is executed?

  • A. The statement would execute successfully to produce the required output.
  • B. The statement would not execute because inline views and outer joins cannot be used together.
  • C. The statement would not execute because the ITEM_CNT alias cannot be displayed in the outer query.
  • D. The statement would not execute because the GROUP BY clause cannot be used in the inline.

Answer: A

NEW QUESTION 12
You must create a table EMPLOYEES in which the values in the columns EMPLOYEES_ID and LOGIN_ID must be unique and not null. (Choose two.)
Which two SQL statements would create the required table?

  • A. CREATE TABLE employees(employee_id NUMBER,Login_id NUMBER,Employee_name VARCHAR2(100),Hire_date DATE,CONSTRAINT emp_id_ukUNIQUE (employee_id, login_id));
  • B. CREATE TABLE employees(employee_id NUMBER,login_id NUMBER,employee_name VARCHAR2(25),hire_date DATE,CONSTRAINT emp_id_pk PRIMARY KEY (employee_id, login_id));
  • C. CREATE TABLE employees(employee_id NUMBER CONSTRAINT emp_id_pk PRIMARY KEY, Login_id NUMBER UNIQUE, Employee_name VARCHAR2(25),Hire_date DATE);
  • D. CREATE TABLE employees(employee_id NUMBER,Login_id NUMBER,Employee_name VARCHAR2(100),Hire_date DATE,CONSTRAINT emp_id_uk UNIQUE (employee_id, login_id);CONSTRAINT emp_id_nn NOT NULL (employee_id, login_id));
  • E. CREATE TABLE employees(employee_id NUMBER CONSTRAINT emp_id_nn NOT NULL, Login_id NUMBER CONSTRAINT login_id_nn NOT NULL,Employee_name VARCHAR2(100),Hire_date DATE,CONSTRAINT emp_id_ukUNIQUE (employee_id, login_id));

Answer: BE

NEW QUESTION 13
You issue the following command to drop the PRODUCTS table: (Choose all that apply.) SQL > DROP TABLE products;
Which three statements are true about the implication of this command?

  • A. All data along with the table structure is deleted.
  • B. A pending transaction in the session is committed.
  • C. All indexes on the table remain but they are invalidated.
  • D. All views and synonyms on the table remain but they are invalidated.
  • E. All data in the table is deleted but the table structure remains.

Answer: ABD

NEW QUESTION 14
View the Exhibit and examine the structure in the DEPARTMENTS tables. (Choose two.)
1Z0-071 dumps exhibit
Examine this SQL statement:
SELECT department_id "DEPT_ID", department_name, 'b' FROM departments
WHERE departments_id=90 UNION
SELECT department_id, department_name DEPT_NAME, 'a' FROM departments
WHERE department_id=10
Which two ORDER BY clauses can be used to sort output?

  • A. ORDER BY DEPT_NAME;
  • B. ORDER BY DEPT_ID;
  • C. ORDER BY 'b';
  • D. ORDER BY 3;

Answer: BD

NEW QUESTION 15
Examine the structure of the EMPLOYEES table. NameNull?Type
---------------------- ------------ EMPLOYEE_IDNOT NULLNUMBER(6) FIRST_NAMEVARCHAR2(20) LAST_NAMENOT NULLVARCHAR2(25) EMAILNOT NULLVARCHAR2(25) PHONE NUMBERVARCHAR2(20) HIRE_DATENOT NULLDATE JOB_IDNOT NULLVARCHAR2(10) SALARYNUMBER(8,2) COMMISSION_PCTNUMBER(2,2) MANAGER_IDNUMBER(6) DEPARTMENT_IDNUMBER(4)
There is a parent/child relationship between EMPLOYEE_ID and 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_ID is 123.
Which query provides the correct output?

  • A. SELECT e.last_name, m.manager_idFROM employees e RIGHT OUTER JOIN employees mon (e.manager_id = m.employee_id)AND e.employee_id = 123;
  • B. SELECT e.last_name, m.manager_idFROM employees e RIGHT OUTER JOIN employees mon (e.employee_id = m.manager_id)WHERE e.employee_id = 123;
  • C. SELECT e.last_name, e.manager_idFROM employees e RIGHT OUTER JOIN employees mon (e.employee_id = m.employee_id)WHERE e.employee_id = 123;
  • D. SELECT m.last_name, e.manager_idFROM employees e LEFT OUTER JOIN employees mon (e.manager_id = m.manager_id)WHERE e.employee_id = 123;

Answer: B

NEW QUESTION 16
Examine the command:
SQL> ALTER TABLE books_transactions
ADD CONSTRAINT fk_book_id FOREIGN KEY (book_id) REFERENCES books (book_id) ON DELETE CASCADE; What does ON DELETE CASCADE imply?

  • A. When the BOOKS table is dropped, the BOOK_TRANSACTIONS table is dropped.
  • B. When the BOOKS table is dropped, all the rows in the BOOK_TRANSACTIONS table are deleted butthe table structure is retained.
  • C. When a row in the BOOKS table is deleted, the rows in the BOOK_TRANSACTIONS table whose BOOK_ID matches that of the deleted row in the BOOKS table are also deleted.
  • D. When a value in the BOOKS.BOOK_ID column is deleted, the corresponding value is updated in the BOOKS_TRANSACTIONS.BOOK_ID column.

Answer: C

NEW QUESTION 17
You need to display the date 11-oct-2007 in words as ‘Eleventh of October, Two Thousand Seven’. Which SQL statement would give the required result?

  • A. SELECT TO_CHAR (TO_DATE (’11-oct-2007’), ‘fmDdthsp “of” Month, Year’)FROM DUAL
  • B. SELECT TO_CHAR (‘11-oct-2007’, ‘fmDdspth “of” Month, Year’)FROM DUAL
  • C. SELECT TO_CHAR (TO_DATE (‘11-oct-2007’), ‘fmDdspth of month, year’)FROM DUAL
  • D. SELECT TO_DATE (TO_CHAR (’11-oct-2007’), ‘fmDdspth “of” Month, Year’))FROM DUAL

Answer: C

NEW QUESTION 18
Examine the business rule:
Each student can work on multiple projects and each project can have multiple students.
You need to design an Entity Relationship Model (ERD) for optimal data storage and allow for generating reports in this format:
STUDENT_ID FIRST_NAME LAST_NAME PROJECT_ID PROJECT_NAME PROJECT_TASK
Which two statements are true in this scenario?

  • A. The ERD must have a 1:M relationship between the STUDENTS and PROJECTS entities.
  • B. The ERD must have a M:M relationship between the STUDENTS and PROJECTS entities that must be resolved into 1:M relationships.
  • C. STUDENT_ID must be the primary key in the STUDENTS entity and foreign key in the PROJECTS entity.
  • D. PROJECT_ID must be the primary key in the PROJECTS entity and foreign key in the STUDENTS entity.
  • E. An associative table must be created with a composite key of STUDENT_ID and PROJECT_ID, which is the foreign key linked to the STUDENTS and PROJECTS entities.

Answer: BE

Explanation: References:
http://www.oracle.com/technetwork/issue-archive/2011/11-nov/o61sql-512018.html

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