Oracle 1z0-071 Valid Test Materials - 1z0-071 Pdf Braindumps
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The Oracle 1z0-071 certification exam offers a great opportunity to advance your career. With the Oracle Database SQL certification exam beginners and experienced professionals can demonstrate their expertise and knowledge. After passing the Oracle Database SQL (1z0-071) exam you can stand out in a crowded job market. The Oracle Database SQL (1z0-071) certification exam shows that you have taken the time and effort to learn the necessary skills and have met the standards in the market.
Oracle 1z1-071 is a certification exam that is designed for individuals seeking to validate their skills in Oracle Database SQL. 1z0-071 exam is intended for database administrators, developers, and IT professionals who want to enhance their knowledge and expertise in SQL database management using Oracle technologies. 1z0-071 exam is a part of the Oracle Certified Associate (OCA) certification program, which is recognized globally and highly valued in the IT industry.
Oracle 1z0-071 certification exam is a test designed to evaluate a candidate's knowledge and skills in SQL programming for Oracle databases. 1z0-071 exam tests the candidate's ability to write queries, manipulate data, and manage database objects using SQL. Passing 1z0-071 Exam is a requirement for obtaining the Oracle Database SQL certification.
Oracle 1z0-071 exam is a valuable certification for individuals who want to demonstrate their knowledge and skills in SQL databases. It is an ideal certification for professionals who want to enhance their career prospects in the field of database management. It is a must-have certification for those who are seeking job opportunities in database development, management, or administration. 1z0-071 exam covers a wide range of topics related to SQL databases, and candidates are required to score at least 63% to pass.
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Oracle Database SQL Sample Questions (Q151-Q156):
NEW QUESTION # 151
You must create a table EMPLOYEES in which the values in the columns EMPLOYEES_ID and LOGIN_ID must be unique and not null.
Which two SQL statements would create the required table? (Choose two.)
- A. 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)); - 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_ukUNIQUE (employee_id, login_id)); - E. 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));
Answer: A,B
NEW QUESTION # 152
View the Exhibit and examine the structure of CUSTOMERStable.
Using the CUSTOMERStable, 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?
SELECT NVL (TO CHAR(cust_credit_limit * .15), 'Not Available') "NEW CREDIT"
- A. FROM customers;
SELECT NVL(cust_credit_limit * .15), 'Not Available') "NEW CREDIT" - B. FROM customers;
- C. FROM customers;
SELECT NVL(cust_credit_limit), 'Not Available') "NEW CREDIT" - D. FROM customers;
SELECT TO_CHAR (NVL(cust_credit_limit * .15), 'Not Available') "NEW CREDIT"
Answer: D
NEW QUESTION # 153
Evaluate the following query:
What would be the outcome of the above query?
- A. It produces an error because the data types are not matching.
- B. It executes successfully and introduces an 's at the end of each promo_name in the output.
- C. It produces an error because flower braces have been used.
- D. It executes successfully and displays the literal " {'s start date was > " for each row in the output.
Answer: B
NEW QUESTION # 154
Which two are true about self joins?
- A. They require the exists operator In the join condition.
- B. They require table aliases,
- C. They require the not exists operator In the join condition.
- D. They are always condition.
- E. They have no Join condition.
- F. They can use INNER join and LEFT .join.
Answer: B,F
NEW QUESTION # 155
Examine the description or the BOOKS_TRANSACTIONS table:
FOR customers whose income level has a value, you want to display the first name and due amount as 5% of their credit limit. Customers whose due amount is null should not be displayed.
Which query should be used?
- A. SELECT cust_first_name, cust_credit_limit * . 05 AS DUE AMOUNT
FROM customers
WHERE cust income_level IS NOT NULL
AND due_amount IS NOT NULL; - B. SELECT cust_first_name, cust_credit_limit * . 05 AS DUE AMOUNT
FROM customers
WHERE cust income_level !=NULL
AND due_amount !=NULL; - C. SELECT cust_first_name, cust_credit_limit * . 05 AS DUE AMOUNT
FROM customers
WHERE cust income_level !=NULL
AND cust credit_level !=NULL; - D. SELECT cust_first_name, cust_credit_limit * . 05 AS DUE AMOUNT
FROM customers
WHERE cust income_level <> NULL
AND due_amount <> NULL; - E. SELECT cust_first_name, cust_credit_limit * . 05 AS DUE AMOUNT
FROM customers
WHERE cust_income_level IS NOT NULL
AND cust_credit_limit IS NOT NULL;
Answer: E
NEW QUESTION # 156
......
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