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Oracle 1z1-071 Exam Syllabus Topics:

Section	Objectives
Topic 1: Subqueries and Advanced SQL	- Correlated subqueries and nested queries - Single-row and multi-row subqueries
Topic 2: Relational Database Concepts	- Data types and database objects overview - Tables, rows, columns, and relational model basics
Topic 3: SQL Functions	- Conversion and conditional functions 1. TO_CHAR, TO_NUMBER, CASE expressions - Single-row functions 1. Character, numeric, date functions
Topic 4: Data Manipulation and Schema Objects	- DDL operations 1. CREATE, ALTER, DROP tables and constraints - DML operations 1. INSERT, UPDATE, DELETE, MERGE
Topic 5: SQL Retrieval	- SELECT statements and filtering data 1. WHERE clause and conditions 2. ORDER BY and row limiting - Joins and set operations 1. UNION, INTERSECT, MINUS 2. INNER, OUTER, CROSS joins

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Oracle Database SQL Sample Questions (Q190-Q195):

NEW QUESTION # 190
Examine the description of the PRODUCT_INFORMATION table:

Which query retrieves the number of products with a null list price?

A. SELECT COUNT (list_price) FROM product_information WHERE list_price = NULL;
B. SELECT COUNT (DISTINCT list_price) FROM product_information WHERE list_price IS NULL;
C. SELECT COUNT(NVL(list_price, 0)) FROM product_information WHERE list_price IS NULL;
D. SELECT COUNT (list_price) FROM product_information WHERE list_price IS NULL;

Answer: C

NEW QUESTION # 191
Viev the Exhibit and examine the structure of the PRODUCT INFORMATION and INVENTORIEStables.
You have a requirement from the supplies department to give a list containing PRODUCT _ID,SUPPLIER ID, and QUANTITY_ON HAND for all the products where in QUANTITY ON HAND is lessthan five.
Which two SQL statements can accomplish the task? (Choose two)

A. SELECT i. product id, i. quantity .on hand, pi. supplier_id
FROM product_information pi JOIN inventories i
ON (pi. product. id=i. product id) AND quantity on hand < 5;
B. SELECT i. product_id, i. quantity_on hand, pi. supplier id
FROM product information pi JOIN inventories i USING (product id) AND quantity .on hand < 5;
C. SELECT i.product id, i. quantity on hand, pi. supplier id
FROM product information pi JOIN inventories i
ON (pi.product id=i. product id)WHERE quantity on hand < 5;
D. SELECT product id, quantity on hand, supplier id
FROM product information
NATURAL JOIN inventories AND quantity .on hand < 5;

Answer: A,C

Explanation:
Given the requirement to list PRODUCT_ID, SUPPLIER_ID, and QUANTITY_ON_HAND for products with QUANTITY_ON_HAND less than five:
* B. This query correctly joins the two tables on PRODUCT_ID and filters rows where QUANTITY_ON_HAND is less than five. However, the syntax presented here might be incorrect due to improper placement of the condition within the ON clause; it should be in a WHERE clause.
* D. This query correctly joins the two tables on PRODUCT_ID and applies the condition in a WHERE clause, which is the proper way to filter rows after performing the join.
Incorrect options:
* A: The syntax uses NATURAL JOIN and AND incorrectly; also, conditions should be in a WHERE clause, not combined with the JOIN clause.
* C: Similar to A, this option incorrectly places a condition directly in the JOIN clause without using a WHERE clause.

NEW QUESTION # 192
The ORDERS table has a column ORDER_DATE of date type DATE The default display format for a date is DD-MON-RR
Which two WHERE conditions demonstrate the correct usage of conversion functions?

A. WHERE order_date IN (TO_DATE ('Oct 21 2018','MON DD YYYY'), TO_CHAR('Nov 21 2018','MON DD YYYY'));
B. WHERE order_date> TO_DATE('JUL 10 2018','MON DD YYYY');
C. WHERE TO_CHAR(order_date,'MON DD YYYY') ='JAN 20 2019';
D. WHERE order_date> TO_DATE(ADD_MONTHS(SYSDATE,6),'MON DD YYYY');
E. WHERE ordet_date> TO_CHAR(ADD_MONTHS(SYSDATE, 6),'MON DD YYYY')

Answer: B,C

NEW QUESTION # 193
Examine the structure of the BOOKS_TRANSACTIONS table:

You want to display the member IDs, due date, and late fee as $2 for all transactions.
Which SQL statement must you execute?

A. SELECT member_id AS "MEMBER ID", due_date AS "DUE DATE", '$2' AS "LATE FEE" FROM BOOKS_TRANSACTIONS
B. SELECT member_id AS MEMBER_ID, due_date AS DUE_DATE, $2 AS LATE_FEE FROM BOOKS_TRANSACTIONS
C. SELECT member_id 'MEMBER ID', due_date 'DUE DATE', '$2 AS LATE FEE' FROM BOOKS_TRANSACTIONS;
D. SELECT member_id AS "MEMBER ID", due_date AS "DUE DATE", $2 AS "LATE FEE" FROM BOOKS_TRANSACTIONS

Answer: A

NEW QUESTION # 194
View the Exhibit and examine the structure of the CUSTOMERS and CUST_HISTORY tables.

The CUSTOMERS table contains the current location of all currently active customers.
The CUST_HISTORY table stores historical details relating to any changes in the location of all current as well as previous customers who are no longer active with the company.
You need to find those customers who have never changed their address.
Which SET operator would you use to get the required output?

A. MINUS
B. UNION ALL
C. INTERSECT
D. UNION

Answer: A

NEW QUESTION # 195
......

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