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Oracle 1z1-071 Exam Syllabus Topics:

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

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Oracle Database SQL Sample Questions (Q352-Q357):

NEW QUESTION # 352
Examine the command:

What does ON DELETE CASCADE imply?

A. When a value in the BOOKS.BOOK_IDcolumn is deleted, the corresponding value is updated in the BOOKS_TRANSACTIONS.BOOK_IDcolumn.
B. When a row in the BOOKStable is deleted, the rows in the BOOK_TRANSACTIONStable whose BOOK_IDmatches that of the deleted row in the BOOKStable are also deleted.
C. When the BOOKStable is dropped, the BOOK_TRANSACTIONStable is dropped.
D. When the BOOKStable is dropped, all the rows in the BOOK_TRANSACTIONStable are deleted but the table structure is retained.

Answer: B

NEW QUESTION # 353
View the Exhibit and examine the structure of the SALES and PRODUCTS tables. (Choose two.) In the SALES table, PROD_ID is the foreign key referencing PROD_ID in the PRODUCTS table. You must list each product ID and the number of times it has been sold.
Examine this query which is missing a JOIN operator:
SQL > SELECT p.prod_id, count(s.prod_id)
FROM products p ______________ sales s
ON p.prod_id = s.prod_id
GROUP BY p.prod_id;
Which two JOIN operations can be used to obtain the required output?

A. LEFT OUETR JOIN
B. RIGHT OUTER JOIN
C. JOIN
D. FULL OUTER JOIN

Answer: A,D

NEW QUESTION # 354
Examine the data in the CUST_NAME column of the CUSTOMERS table.
CUST_NAME
-------------------
Renske Ladwig
Jason Mallin
Samuel McCain
Allan MCEwen
Irene Mikkilineni
Julia Nayer
You need to display customers' second names where the second name starts with "Mc" or "MC".
Which query gives the required output?

A. SELECT SUBSTR(cust_name, INSTR(cust_name,' ')+1)FROM customersWHERE
INITCAP(SUBSTR(cust_name, INSTR(cust_name,' ')+1))='Mc';
B. SELECT SUBSTR(cust_name, INSTR(cust_name,' ')+1)FROM customersWHERE
SUBSTR(cust_name, INSTR(cust_name,' ')+1) LIKE INITCAP('MC%');
C. SELECT SUBSTR(cust_name, INSTR(cust_name,' ')+1)FROM customersWHERE
INITCAP(SUBSTR(cust_name, INSTR(cust_name,' ')+1)) LIKE 'Mc%';
D. SELECT SUBSTR(cust_name, INSTR(cust_name,' ')+1)FROM customersWHERE
INITCAP(SUBSTR(cust_name, INSTR(cust_name,' ')+1)) = INITCAP('MC%');

Answer: C

NEW QUESTION # 355
Examine the data in the COLORS table:

Examine the data in the BRICKS table:

Which two queries return all the rows from COLORS?

A. SELECT
FROM bricks b
FULL JOIN colors C
ON b. color rgb _ hex_ value = c. rgb _hex_ value;
B. SELECT.
FROM bricks b
RIGHT JOIN colors c
ON b. color _rgb_ hex_ value = c. rgb hex_ value;
C. SELECT
EROM colors C
LEET JOIN bricks b
ON b. color_ rgb_ hex value = c. rgb. hex.
value
WHERE b. brick_ id > 0;
D. SELECT *
EROM bricks | b
JOIN colors C
ON b. color_ rgb_ hex_ value =c. rgb _hex value;
E. SELECT
EROM colors C
LEFT JOIN bricks
USING (rgb _ hex_ value) ;

Answer: B,E

Explanation:
The queries that will return all the rows from the COLORS table are those that ensure every record from COLORS is selected, regardless of whether there's a matching record in the BRICKS table:
* Option A:
* SELECT * FROM bricks b RIGHT JOIN colors c ON b.color_rgb_hex_value = c.rgb_hex_value;
* A right join will return all the rows from the right table (COLORS), with the matching rows from the left table (BRICKS). If there is no match, NULL will be returned for columns from BRICKS.
* Option B:
* SELECT * FROM colors c LEFT JOIN bricks b USING (rgb_hex_value);
* A left join will return all the rows from the left table (COLORS), with the matching rows from the right table (BRICKS). If there is no match, NULL will be returned for columns from BRICKS. The USING clause indicates a join condition where columns with the same names are compared for equality.
Options C, D, and E will not return all the rows from the COLORS table:
* Option C: The full join will return all rows from both tables, but it is not restricted to only the rows from COLORS.
* Option D: An inner join will return only the matching rows between both tables, not all rows from COLORS.
* Option E: This is a left join, which would typically return all rows from COLORS, but the WHERE clause restricts the result set to only those rows from COLORS that have a matching BRICK_ID in BRICKS which is greater than 0, potentially excluding rows from COLORS.

NEW QUESTION # 356
The BOOKS_TRANSACTIONStable exists in your database.
SQL>SELECT * FROM books_transactions ORDER BY 3;
What is the outcome on execution?

A. Rows are displayed in the order that they are stored in the table only for the first three rows.
B. Rows are displayed sorted in ascending order of the values in the third column in the table.
C. The execution fails unless the numeral 3 in the ORDER BYclause is replaced by a column name.
D. Rows are displayed in the order that they are stored in the table only for the three rows with the lowest values in the key column.

Answer: B

NEW QUESTION # 357
......

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