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Oracle 1z1-071 Exam Syllabus Topics:

Section	Weight	Objectives
Topic 1: Retrieving Data Using SQL SELECT Statement	20%	- SELECT statement syntax and usage - Arithmetic operations and NULL handling - Column aliases, concatenation, literals and DISTINCT
Topic 2: Using Single-Row Functions to Customize Output	20%	- Nested functions usage - Character, number, and date functions
Topic 3: Relational Database Concepts	10%	- ERD and SQL statement mapping - Relationship between database and SQL - Theoretical and physical aspects of relational databases
Topic 4: Aggregating Data Using Group Functions	10%	- Handling NULLs in group functions - SUM, AVG, COUNT, MAX, MIN functions - GROUP BY and HAVING clauses
Topic 5: Using SET Operators		- Ordering results with set operations - UNION, UNION ALL, INTERSECT, MINUS operators
Topic 6: Displaying Data from Multiple Tables	15%	- Self-joins and outer joins - Equijoins and non-equijoins - Cartesian products
Topic 7: Managing Views		- Create, modify and drop views - Updatable and read-only views
Topic 8: Conversion Functions and Conditional Expressions	10%	- TO_CHAR, TO_NUMBER, TO_DATE functions - NVL, NULLIF, COALESCE and CASE expressions - Implicit and explicit data type conversion
Topic 9: Using Data Dictionary Views		- Retrieve metadata and object information
Topic 10: Manipulating Data and Managing Transactions		- INSERT, UPDATE, DELETE, MERGE statements - Multi-table inserts - Transaction control: COMMIT, ROLLBACK, SAVEPOINT
Topic 11: Using Subqueries to Solve Queries		- Single-row and multi-row subqueries - Correlated subqueries for update/delete
Topic 12: Managing Tables and Constraints		- Data types, creating/altering/dropping tables - Constraints: primary key, foreign key, unique, check, not null - Temporary and external tables
Topic 13: Restricting and Sorting Data	15%	- Substitution variables, DEFINE and VERIFY commands - Operator precedence - WHERE clause and filtering conditions - ORDER BY clause and sorting rules
Topic 14: Working with Time Zones and Date/Time Data		- TIMESTAMP, INTERVAL data types - Time zone conversion functions
Topic 15: Controlling User Access		- System and object privileges - Roles, granting and revoking privileges
Topic 16: Managing Indexes, Synonyms and Sequences		- Create and maintain indexes - Create and use sequences - Create and use synonyms

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Oracle Database SQL Sample Questions (Q212-Q217):

NEW QUESTION # 212
View the exhibit and examine the structure and data in the INVOICE table.

Which two SQL statements would execute successfully? (Choose two.)

A. SELECT AVG(inv_date)
FROM invoice
B. SELECT MAX(AVG(SYSDATE -inv_date))
FROM invoice
C. SELECT AVG( inv_date -SYSDATE), AVG(inv_amt)
FROM invoice
D. SELECT MAX(inv_date), MIN(cust_id)
FROM invoice

Answer: C,D

NEW QUESTION # 213
View the Exhibit and examine the structure of the PRODUCTS table.

You must display the category with the maximum number of items.
You issue this query:

What is the result?

A. It generates an error because = is not valid and should be replaced by the IN operator.
B. It executes successfully but does not give the correct output.
C. It executes successfully and gives the correct output.
D. It generate an error because the subquery does not have a GROUP BY clause.

Answer: D

NEW QUESTION # 214
View the exhibit and examine the structure of ORDERSand CUSTOMERStables.

Which INSERT statement should be used to add a row into the ORDERStable for the customer whose CUST_LAST_NAMEis Robertsand CREDIT_LIMITis 600? Assume there exists only one row with CUST_LAST_NAME as Roberts and CREDIT_LIMIT as 600.

A. INSERT INTO orders
VALUES (1,'10-mar-2007', 'direct',
( SELECT customer_id
FROM customers
WHERE cust_last_name='Roberts' AND credit_limit=600), 1000);
B. INSERT INTO (SELECT o.order_id, o.order_date, o.order_mode, c.customer_id, o.order_total FROM orders o, customers c
WHERE o.customer_id = c.customer_id AND c.cust_last_name='Roberts' AND
c.credit_limit=600)
VALUES (1,'10-mar-2007', 'direct', (SELECT customer_id
FROM customers
WHERE cust_last_name='Roberts' AND credit_limit=600), 1000);
C. INSERT INTO orders (order_id, order_date, order_mode,
( SELECT customer_id
FROM customers
WHERE cust_last_name='Roberts' AND credit_limit=600), order_total);
VALUES (1,'10-mar-2007', 'direct', &customer_id, 1000);
D. INSERT INTO orders (order_id, order_date, order_mode,
( SELECT customer id
FROM customers
WHERE cust_last_name='Roberts' AND credit_limit=600), order_total);
VALUES (1,'10-mar-2007', 'direct', &customer_id, 1000);

Answer: A

NEW QUESTION # 215
Examine this SELECT statement and view the Exhibit to see its output:

SELECT constraints_name, constraints_type, search_condition, r_constraint_name, delete_rule, status,
FROM user_constraints
WHERE table_name = 'ORDERS';
Which two statements are true about the output? (Choose two.)

A. The DELETE_RULE column indicates the desired state of related rows in the child table when the corresponding row is deleted from the parent table.
B. The R_CONSTRAINT_NAME column contains an alternative name for the constraint.
C. In the second column, 'c' indicates a check constraint.
D. The STATUS column indicates whether the table is currently in use.

Answer: A,C

NEW QUESTION # 216
View the Exhibit and examine the structure of the CUSTOMERS table.

Using the CUSTOMERS table, you must generate a report that displays a credit limit increase of
15% for all customers.
Customers with no credit limit should have "Not Available" displayed.
Which SQL statement would produce the required result?

A. SELECT NVL(cust_credit_limit, `Not Available')*.15 "NEW CREDIT" FROM customers;
B. SELECT NVL(TO_CHAR(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*.15, `Not Available') "NEW CREDIT" FROM customers;

Answer: B

NEW QUESTION # 217
......

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