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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: Data Manipulation and Schema Objects	- DDL operations 1. CREATE, ALTER, DROP tables and constraints - DML operations 1. INSERT, UPDATE, DELETE, MERGE
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: SQL Retrieval	- Joins and set operations 1. INNER, OUTER, CROSS joins 2. UNION, INTERSECT, MINUS - SELECT statements and filtering data 1. ORDER BY and row limiting 2. WHERE clause and conditions

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Oracle Database SQL Sample Questions (Q234-Q239):

NEW QUESTION # 234
Examine this business rule:
Each student can work on multiple projects and earth project can have multiple students.
You must decide an Entity Relationship (ER) model for optional data storage and allow generating reports in this format:
STUDENT_ID FIRST_NAME LAST_NAME PROJECT_ID PROJECT_NAME PROJECT_TASK Which two statements are true?

A. PROJECT_ID must be the primary key in the PROJECTS entity and foreign key in the STUDENTS entity.
B. The ER must have a 1-to-many relationship between the STUDENTS and PROJECTS entities.
C. The ER must have a many-to-many relationship between the STUDENTS and PROJECTS entities that must be resolved into 1-to-many relationships.
D. STUDENT_ID must be the primary key in the STUDENTS entity and foreign key in the PROJECTS entity.
E. An associative table must be created with a composite key of STUDENT_ID and PROJECT_ID, which is the foreign key linked to the STUDENTS and PROJECTS entities.

Answer: C,E

Explanation:
For the described business rule, the relationship between the students and projects entities is a many-to-many relationship, meaning that each student can be involved in multiple projects and each project can have multiple students.
* Statement A is true because to implement a many-to-many relationship in a relational database, an associative (junction) table is typically used. This table will contain the primary keys from both related entities as foreign keys in the associative table, making a composite primary key consisting of STUDENT_ID and PROJECT_ID. This setup allows the database to effectively manage the relationships between students and projects.
* Statement B is true as it accurately describes the solution to managing a many-to-many relationship. A direct many-to-many relationship cannot be physically implemented in a relational database. It is instead resolved into two 1-to-many relationships using an associative table as described in A. This is a fundamental relational database design principle aimed at normalizing the database and avoiding redundancy.
* Statements C, D, and E are incorrect as they misrepresent the nature of the relationships and key constraints necessary for this scenario. Specifically, C and E incorrectly suggest that one entity's primary key should serve as a foreign key in another, which does not align with the many-to-many relationship requirement described. Statement D incorrectly suggests a 1-to-many relationship directly between students and projects, which does not meet the business rule requirement.

NEW QUESTION # 235
Examine the structure of the BOOKS_TRANSACTIONS table:
NameNull?Type
------------------------------------------------------------
TRANSACTION_IDNOT NULLVARCHAR2 (6)
BORROWED_DATEVARCHAR2 (50)
DUE_DATEDATE
BOOK_IDDATE
MEMBER_IDVARCHAR2 (6)
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 # 236
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 tables, which references the PRODUCTS table.
Similarly, the CUST_ID and TIME_ID columns are also foreign keys in the SALES table referencing the CUSTOMERS and TIMES tables, respectively.
Evaluate the following CREATE TABLE 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 regarding the above command?

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 NOT NULL constraints defined on the specified columns would be passed to the new table.
C. The NEW_SALES table would get created and all the FOREIGN KEY constraints defined on the specified columns would be passed to the new table.
D. The NEW_SALES table would not get created because the DEFAULT value cannot be specified in the column definition.

Answer: B

NEW QUESTION # 237
View the exhibit and examine the structure of the CUSTOMERS table.

Which two tasks would require subqueries or joins to be executed in a single statement?

A. listing of those customers, whose credit limit is the same as the credit limit of customers residing in the city 'Tokyo'.
B. listing of customers who do not have a credit limit and were born before 1980
C. finding the average credit limit of male customers residing in 'Tokyo' or 'Sydney'
D. finding the number of customers, in each city, whose credit limit is more than the average credit limit of all the customers
E. finding the number of customers, in each city, who's marital status is 'married'.

Answer: A,D

NEW QUESTION # 238
Which two are true about the NVL, NVL2, and COALESCE functions? (Choose two.)

A. NVL2 can have any number of expressions in the list.
B. The first expression in NVL2 is never returned.
C. COALESCE stops evaluating the list of expressions when it finds the first null value.
D. NVL must have expressions of the same data type.
E. COALESCE stops evaluating the list of expressions when it finds the first non-null value.
F. NVL can have any number of expressions in the list.

Answer: B,E

Explanation:
https://www.interviewsansar.com/difference-between-nvl-nvl2-nullif-and-coalesce-functions/

NEW QUESTION # 239
......

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