RELIABLE 1Z0-071 PRACTICE QUESTIONS, CERTIFICATION 1Z0-071 DUMP

Reliable 1z0-071 Practice Questions, Certification 1z0-071 Dump

Reliable 1z0-071 Practice Questions, Certification 1z0-071 Dump

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Oracle 1z0-071 Certification Exam is designed for SQL developers, database administrators, and database analysts who want to validate their knowledge and skills in SQL. 1z0-071 exam covers the basics of SQL, including data manipulation, data definition, and data control language. It also includes advanced topics such as subqueries, set operators, and data aggregation.

Oracle 1z1-071 Exam is an important certification test for anyone interested in working with Oracle databases and SQL. Passing the exam demonstrates that the candidate has a solid understanding of SQL fundamentals and the ability to work with Oracle databases. With the right preparation and study, candidates can successfully pass the exam and earn their Oracle Database SQL Certified Associate certification.

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Certification 1z0-071 Dump | Exam 1z0-071 Quiz

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The Oracle 1z0-071 Exam covers topics such as SQL fundamentals, data manipulation, data definition, and control and management of transactions. It is a comprehensive exam that tests the candidate's ability to work with Oracle databases and write SQL queries.

Oracle Database SQL Sample Questions (Q171-Q176):

NEW QUESTION # 171
Examine the structure of the EMPLOYEES table. (Choose the best answer.)

You must display the details of employees who have manager with MANAGER_ID 100, who were hired in the past 6 months and who have salaries greater than 10000.

  • A. (SELECT last_name, hire_date, salaryFROM employeesWHERE salary > 10000UNION ALLSELECT last_name, hire_date, salaryFROM employeesWHERE manager_ID = (SELECT employee_id FROM employees WHERE employee_id = 100))UNIONSELECT last_name, hire_date, salaryFROM employeesWHERE hire_date > SYSDATE -180;
  • B. SELECT last_name, hire_date, salaryFROM employeesWHERE salary > 10000UNION ALL SELECT last_name, hire_date, salaryFROM employeesWHERE manager_ID = (SELECT employee_id FROM employees WHERE employee_id
    100)INETRSECTSELECT last_name, hire_date, salaryFROM employees WHERE
    hire_date > SYSDATE- 180;
  • C. SELECT last_name, hire_date, salaryFROM employeesWHERE manager_id
    (SELECT employee_id FROM employees WHERE employee_id = 100)UNION
    ALL(SELECT last_name, hire_date, salaryFROM employeesWHERE hire_date >
    SYSDATE -180INTERSECTSELECT last_name, hire_date, salaryFROM
    employeesWHERE salary > 10000);
  • D. SELECT last_name, hire_date, salaryFROM employeesWHERE manager_id
    (SELECT employee_id FROM employees WHERE employee_id = '100')UNIONSELECT last_name, hire_date, salaryFROM employeesWHERE hire_date > SYSDATE -
    1 80INTERSECTSELECT last_name, hire_date, salaryFROM employeesWHERE salary >
    1 0000;

Answer: D


NEW QUESTION # 172
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 remain disabled and can be enabled only by dropping the foreign key constraint and re-creating it.
  • C. It will be enabled and deferred.
  • D. It will remain disabled and can be re-enabled manually.

Answer: D


NEW QUESTION # 173
View the exhibit and examine the structure of the SALES, CUSTOMERS, PRODUCTS and TIMES tables.

The PROD_ID column is the foreign key in the SALES table referencing the PRODUCTS table.
The CUST_ID and TIME_ID columns are also foreign keys in the SALES table referencing the CUSTOMERS and TIMES tables, respectively.
Examine this command:
CREATE TABLE new_sales (prod_id, cust_id, order_date DEFAULT SYSDATE)
AS
SELECT prod_id, cust_id, time_id
FROM sales;
Which statement is true?

  • A. The NEW_SALES table would not get created because the column names in the CREATE TABLE command and the SELECT clause do not match.
  • B. The NEW_SALES table would get created and all the FOREIGN KEY constraints defined on the selected columns from the SALES table would be created on the corresponding columns in the NEW_SALES table.
  • C. The NEW_SALES table would not get created because the DEFAULT value cannot be specified in the column definition.
  • D. The NEW_SALES table would get created and all the NOT NULL constraints defined on the selected columns from the SALES table would be created on the corresponding columns in the NEW_SALES table.

Answer: D


NEW QUESTION # 174
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(cust_credit_limit), 'Not Available') "NEW CREDIT"FROM customers;
  • B. SELECT TO_CHAR (NVL(cust_credit_limit * .15), 'Not Available') "NEW CREDIT"FROM customers;
  • C. SELECT NVL (TO CHAR(cust_credit_limit * .15), 'Not Available') "NEW CREDIT"FROM customers;
  • D. SELECT NVL(cust_credit_limit * .15), 'Not Available') "NEW CREDIT"FROM customers;

Answer: C


NEW QUESTION # 175
Examine the description of the BOOKS_TRANSACTIONS table:

Examine this partial SQL statement:
SELECT * FROM books_transactions
Which two WHERE conditions give the same result?

  • A. WHERE borrowed_date = SYSDATE AND transaction_type = 'RM' OR member_id IN('A101','A102');
  • B. WHERE borrowed_date = SYSDATE AND transaction_type = 'RM' AND member_id = 'A101' OR member_id = 'A102');
  • C. WHERE borrowed_date = SYSDATE AND transaction_type = 'RM' OR member_id IN('A101','A102');
  • D. WHERE (borrowed_date = SYSDATE AND transaction_type = 'RM') OR member_id IN ('A101','A102');
  • E. WHERE borrowed_date = SYSDATE AND transaction_type = 'RM' AND (member_id = 'A101' OR member_id = 'A102'));

Answer: C,D


NEW QUESTION # 176
......

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