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Oracle 1z1-071 Exam Syllabus Topics:

Section	Weight	Objectives
Topic 1: Using Conversion Functions and Conditional Expressions	10%	- Use the TO_CHAR, TO_NUMBER, and TO_DATE conversion functions - Describe implicit and explicit data type conversion - Use conditional IF THEN ELSE logic in a SELECT statement - Apply the NVL, NULLIF, and COALESCE functions to data
Topic 2: Using DDL Statements to Create and Manage Tables	14%	- Categorize the main database objects - Review the table structure - Drop, rename, and truncate tables - Create a simple table - Describe the data types that are available for columns - Alter table definitions
Topic 3: Using Set Operators	5%	- Describe set operators - Describe the order of precedence when the set operators are used - Use a set operator to combine multiple queries into a single query
Topic 4: Working with Oracle Database SQL	8%	- Describe the Oracle Database architecture - Start up and shut down a database - Create a container database - Explain the installation of Oracle Database software
Topic 5: Manipulating Data	14%	- Update rows in a table - Delete rows from a table - Insert rows into a table - Describe each data manipulation language (DML) statement - Control transactions
Topic 6: Using Subqueries to Solve Queries	15%	- List the types of subqueries - Define subqueries - Write single-row and multi-row subqueries - Describe the types of problems that subqueries can solve
Topic 7: Controlling User Access	5%	- Differentiate system privileges from object privileges - Grant privileges on tables and on a view - Distinguish between granting privileges and roles
Topic 8: Restricting and Sorting Data	6%	- Sort the rows that are retrieved by a query - Limit the rows that are retrieved by a query - Use ampersand substitution to restrict and sort output at runtime - Use DEFINE and UNDEFINE commands
Topic 9: Using Single-Row Functions to Customize Output	10%	- Manipulate strings with character function in the SELECT and WHERE clauses - Manipulate dates with the date functions - Describe the differences between single-row and multiple-row functions - Perform arithmetic with date data - Manipulate numbers with the ROUND, TRUNC, and MOD functions
Topic 10: Retrieving Data Using the SQL SELECT Statement	5%	- List the capabilities of SQL SELECT statements - Execute a basic SELECT statement - Differentiate between SQL statements and SQL*Plus commands
Topic 11: Displaying Data from Multiple Tables Using Joins	15%	- View data that does not meet a join condition by using outer joins - Cross join tables - Write SELECT statements to access data from more than one table using equijoins and non-equijoins - Join a table to itself by using a self-join
Topic 12: Reporting Aggregated Data Using Group Functions	15%	- Include or exclude grouped rows by using the HAVING clause - Describe the use of group functions - Identify the available group functions - Group data by using the GROUP BY clause
Topic 13: Creating Other Schema Objects	10%	- Create synonyms for database objects - Create and maintain indexes - Create, modify, and retrieve data from a view - Create, maintain, and use sequences

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Oracle Database SQL Sample Questions (Q338-Q343):

NEW QUESTION # 338
Examine the description of the PRODUCTS table:

Which two statements execute without errors?

A. MERGE INTO new_prices n
USING products p
ON (p.prod_id =n.prod_id)
WHEN NOT MATCHED THEN
INSERT (n.prod _id, n.price) VALUES (p.prod_id, cost*.01)
WHERE (p.cost<200);
B. MERGE INTO new_prices n
USING (SELECT * FROM products) p
WHEN MATCHED THEN
UPDATE SET n.price= p.cost* 01
WHEN NOT MATCHED THEN
INSERT(n.prod_id, n.price) VALUES(p.prod_id, cost*.01)
WHERE(p.cost<200);
C. MERGE INTO new_prices n
USING (SELECT * FROM products WHERE cost>150) p
ON (n.prod_id= p.prod_id)
WHEN MATCHED THEN
DELETE WHERE (p.cost<200)
D. MERGE INTO new_prices n
USING (SELECT * FROM products WHERE cost>150) p
ON (n.prod_id= p.prod_id)
WHEN MATCHED THEN
UPDATE SET n.price= p.cost*.01
DELETE WHERE (p.cost<200);

Answer: A,D

NEW QUESTION # 339
View the Exhibits and examine the structure of the COSTS and PROMOTIONS tables.
You want to display PROD IDS whose promotion cost is less than the highest cost PROD ID in a pro motion time interval.
Examine this SQL statement:
SELECT prod id
FROM costs
WHERE promo id IN
(SELECT promo id
FROM promotions
WHERE promo_cost < ALL
(SELECT MAX (promo cost)
FROM promotions
GROUP BY (promo_end date-promo_begin_date)) );
What will be the result?

A. It gives an error because the ALL keyword is not valid.
B. It gives an error because the GROUP BY clause is not valid
C. It executes successfully but does not give the required result.
D. It executes successfully and gives the required result.

Answer: D

Explanation:
This SQL query checks for product IDs (prod_id) from the costs table where the promotion ID (promo_id) is associated with promotions whose costs are less than the maximum promotional cost of all promotions, grouped by the duration of each promotion (calculated as promo_end_date - promo_begin_date). The query is correctly written to fulfill these conditions.
* The subquery SELECT MAX(promo_cost) FROM promotions GROUP BY (promo_end_date - promo_begin_date) computes the maximum cost of promotions grouped by their duration, returning the highest cost for each distinct promotion length.
* The WHERE promo_cost < ALL (...) clause then filters out the promotions where the promo_cost is less than all of the maximum costs returned by the subquery, ensuring that only promo_ids associated with costs less than the highest cost for any promotion length are considered.
* Since this logical structure correctly implements the given requirement, the statement executes successfully and yields the correct results.

NEW QUESTION # 340
See the Exhibit and examine the structure of the PROMOTIONS table: Exhibit:

Using the PROMOTIONS table, you need to find out the average cost for all promos in the
range $0-2000 and $2000-5000 in category A.
You issue the following SQL statements:
Exhibit:

What would be the outcome?

A. It generates an error because multiple conditions cannot be specified for the WHEN clause
B. It executes successfully and gives the required result
C. It generates an error because CASE cannot be used with group functions
D. It generates an error because NULL cannot be specified as a return value

Answer: B

Explanation:
CASE Expression Facilitates conditional inquiries by doing the work of an IF-THEN-ELSE statement: CASE expr WHEN comparison_expr1 THEN return_expr1 [WHEN comparison_expr2 THEN return_expr2 WHEN comparison_exprn THEN return_exprn ELSE else_expr] END

NEW QUESTION # 341
Examine this SQL statement:

Which two are true? (Choose two.)

A. The subquery is not a correlated subquery.
B. All existing rows in the EMPLOYEES table are deleted.
C. The subquery is executed for every row in the EMPLOYEES table.
D. The subquery is executed before the DELETE statement is executed.
E. The DELETE statement executes successfully even if the subquery selects multiple rows.

Answer: C,D

NEW QUESTION # 342
Which statement adds a column called SALARYto the EMPLOYEEStable having 100 rows, which cannot contain null?

A. ALTER TABLE EMPLOYEES
ADD SALARY NUMBER(8,2) DEFAULT NOT NULL;
B. ALTER TABLE EMPLOYEES
ADD SALARY NUMBER(8,2) DEFAULT 0 NOT NULL;
C. ALTER TABLE EMPLOYEES
ADD SALARY NUMBER(8,2) DEFAULT CONSTRAINT p_nn NOT NULL;
D. ALTER TABLE EMPLOYEES
ADD SALARY NUMBER(8,2) NOT NULL;

Answer: B

NEW QUESTION # 343
......

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