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Oracle 1z1-071 Exam Overview:

Certification Vendor:	Oracle
Exam Name:	Oracle Database SQL
Exam Number:	1Z0-071
Available Languages:	Japanese, Portuguese, Spanish, English, Chinese (Simplified)
Certificate Validity Period:	Non-expiring (Oracle certification policy subject to change)
Related Certifications:	Oracle Database Administrator Certified Associate Oracle Database Development certifications
Exam Duration:	120 minutes
Exam Format:	Proctored exam, Multiple Choice, Computer-based exam, Multiple Response
Exam Price:	USD 245 (varies by country/region)
Recommended Training:	Oracle Database SQL Training
Exam Registration:	Oracle Certification Registration
Sample Questions:	Oracle 1z1-071 Sample Questions
Exam Way:	Online proctored or test center delivery depending on region
Pre Condition:	No formal prerequisites required; basic knowledge of relational databases and SQL recommended.
Official Syllabus URL:	https://education.oracle.com/oracle-database-sql/pexam_1Z0-071

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Oracle 1z1-071 certification exam covers a wide range of topics related to Oracle Database SQL. These topics include SQL fundamentals, database design, data manipulation, and data control. 1z1-071 exam also covers topics such as subqueries, joins, set operators, and data aggregation. By passing 1z1-071 exam, individuals will have demonstrated their ability to work with Oracle Database SQL in a variety of contexts.

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Oracle Database SQL Sample Questions (Q115-Q120):

NEW QUESTION # 115
View the exhibit and examine the description of the PRODUCT_INFORMATION table.
Which SQL statement would retrieve from the table the number of products having LIST_PRICE as NULL?

A. SELECT COUNT (DISTINCT list_price)FROM product_informationWHERE list_price is NULL
B. SELECT COUNT (list_price)FROM product_informationWHERE list_price i= NULL
C. SELECT COUNT (list_price)FROM product_informationWHERE list_price is NULL
D. SELECT COUNT (NVL(list_price, 0))FROM product_informationWHERE list_price is NULL

Answer: D

NEW QUESTION # 116
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 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.
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 not get created because the column names in the CREATE TABLE command and the SELECT clause do not match.

Answer: A

NEW QUESTION # 117
View the Exhibit and examine the structure of ORDERS and ORDER_ITEMS tables.
ORDER_ID is the primary key in the ORDERS table. It is also the foreign key in the ORDER_ITEMS table wherein it is created with the ON DELETE CASCADE option.
Which DELETE statement would execute successfully?

A. DELETE ordersWHERE order_total < 1000;
B. DELETE order_idFROM ordersWHERE order_total < 1000;
C. DELETEFROM ordersWHERE (SELECT order_idFROM order_items);
D. DELETE orders o, order_items IWHERE o.order_id = i.order_id;

Answer: C

NEW QUESTION # 118
See the Exhibit and Examine the structure of the 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*.15,'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,'Not Available')*.15 "NEW CREDIT" FROM customers;

Answer: A

Explanation:
NVL Function
Converts a null value to an actual value:
Data types that can be used are date, character, and number.
Data types must match:
-
NVL(commission_pct,0)
-
NVL(hire_date,'01-JAN-97')
- NVL(job_id,'No Job Yet')

NEW QUESTION # 119
Examine the data in the ORDERS table:

Examine the data in the INVOICES table:

Examine this query:
SELECT order_ id, order_ date FROM orders
INTERSECT
SELECT order_ 1d, order_ date FROM invoices;
Which two rows will it return?

A. 2 <null>
B. 3 <null>
C. 3 01-JAN-2019
D. 4 01-FEB-2019
E. 5 01-MAR-2019
F. 1 <null>

Answer: D,F

NEW QUESTION # 120
......

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