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| Section | Objectives |
|---|
| Subqueries and Advanced SQL | - Single-row and multi-row subqueries - Correlated subqueries and nested queries
|
| Relational Database Concepts | - Data types and database objects overview - Tables, rows, columns, and relational model basics
|
| Data Manipulation and Schema Objects | - DDL operations
- 1. CREATE, ALTER, DROP tables and constraints
- DML operations
- 1. INSERT, UPDATE, DELETE, MERGE
|
| SQL Functions | - Single-row functions
- 1. Character, numeric, date functions
- Conversion and conditional functions
- 1. TO_CHAR, TO_NUMBER, CASE expressions
|
| 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 (Q206-Q211):
NEW QUESTION # 206
Which two queries execute successfully?
- A. SELECT prod_id, exp_date FROM products
UNION ALL
SELECT prod_id, NULL FROM new_products; - B. SELECT prod_id FROM products
UNION ALL
SELECT prod_id, prod_name FROM new_products; - C. SELECT * FROM products
UNION
SELECT * FROM new_products; - D. SELECT k FROM products
MINUS
SELECT prod_id FROM new_products; - E. SELECT prod_id, prod_name FROM products
INTERSECT
SELECT 100, prod_name FROM newproducts;
Answer: A,C
Explanation:
For SQL queries involving set operations like UNION, UNION ALL, INTERSECT, and MINUS, the data types and number of columns in the SELECT statements being combined must be the same. Here is a breakdown of each query option:
* Option A: This query successfully combines results from two tables using UNION ALL, which combines all rows from both SELECT statements, including duplicates. Both queries select two columns (prod_id and exp_date, with exp_date being NULL in the second query). This is valid as the number of columns and their data types match.
* Option B: This query attempts to use INTERSECT, which returns only the rows common to both SELECT statements. However, the first SELECT statement chooses prod_id and prod_name, while the second only substitutes a constant value for prod_id without specifying from which table it selects prod_name. If "newproducts" (presumably a typo and should be "new_products") does not have the exact columns as "products", or if the data types do not match, it will fail.
* Option C: This query uses UNION, which removes duplicates from the combined dataset. The use of SELECT * assumes that both "products" and "new_products" tables have the same columns in the same order with matching data types. This query will succeed if those conditions are met.
* Option D: This query uses MINUS to subtract results of the second SELECT from the first. However, it fails because 'k' does not explicitly match a column name in "products", suggesting a potential error unless 'k' is indeed a valid column.
* Option E: This query fails because the first SELECT statement fetches only one column (prod_id) while the second fetches two columns (prod_id and prod_name). UNION ALL requires the same number of columns with compatible types in all SELECT statements involved.
NEW QUESTION # 207
You want to return the current date and time from the user session, with a data type of TIMESTAMP WITH TIME ZONE.
Which function will do this?
- A. CURRENT DATE
- B. LOCALTIMESTAMP
- C. SYSDATE
- D. CURRENT_ TIMESTAMP
Answer: D
NEW QUESTION # 208
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, 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 not get created because the DEFAULT value cannot be specified in the column definition.
- D. 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.
Answer: B
NEW QUESTION # 209
Which two statements are true? (Choose two.)
- A. The USER OBJECTS view can provide information about the tables and views created by the user only.
- B. All the dynamic performance views prefixed with V$ are accessible to all the database users.
- C. The user SYSTEM owns all the base tables and user-accessible views of the data dictionary.
- D. DICTIONARY is a view that contains the names of all the data dictionary views that the user can access.
- E. The USER SYNONYMS view can provide information about private synonyms.
Answer: D,E
Explanation:
A . The USER SYNONYMS view can provide information about private synonyms: This view contains information about private synonyms that have been created by the user. It lists all synonyms that are accessible to the user in their schema.
E . DICTIONARY is a view that contains the names of all the data dictionary views that the user can access: This view lists the names of all data dictionary views available to the user, providing a directory of useful database metadata.
Incorrect options:
B: The SYSTEM user does not own all base tables of the data dictionary; some are owned by SYS.
C: Not all users have access to dynamic performance views prefixed with V$; access is restricted based on privileges.
D: The USER_OBJECTS view lists all objects owned by the user, not just tables and views but also other types of schema objects like procedures and functions.
NEW QUESTION # 210
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 'MEMBER ID', due_date '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 AS "MEMBER ID", due_date AS "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: C
NEW QUESTION # 211
......
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