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| Section | Weight | Objectives |
|---|---|---|
| Topic 1: Using Subqueries to Solve Queries | - Correlated subqueries for update/delete - Single-row and multi-row subqueries | |
| Topic 2: Working with Time Zones and Date/Time Data | - Time zone conversion functions - TIMESTAMP, INTERVAL data types | |
| Topic 3: Managing Tables and Constraints | - Data types, creating/altering/dropping tables - Constraints: primary key, foreign key, unique, check, not null - Temporary and external tables | |
| Topic 4: Aggregating Data Using Group Functions | 10% | - GROUP BY and HAVING clauses - SUM, AVG, COUNT, MAX, MIN functions - Handling NULLs in group functions |
| Topic 5: Manipulating Data and Managing Transactions | - INSERT, UPDATE, DELETE, MERGE statements - Transaction control: COMMIT, ROLLBACK, SAVEPOINT - Multi-table inserts | |
| Topic 6: Using Single-Row Functions to Customize Output | 20% | - Character, number, and date functions - Nested functions usage |
| Topic 7: Using SET Operators | - Ordering results with set operations - UNION, UNION ALL, INTERSECT, MINUS operators | |
| Topic 8: Using Data Dictionary Views | - Retrieve metadata and object information | |
| Topic 9: Conversion Functions and Conditional Expressions | 10% | - Implicit and explicit data type conversion - NVL, NULLIF, COALESCE and CASE expressions - TO_CHAR, TO_NUMBER, TO_DATE functions |
| Topic 10: Managing Indexes, Synonyms and Sequences | - Create and use synonyms - Create and maintain indexes - Create and use sequences | |
| Topic 11: Managing Views | - Create, modify and drop views - Updatable and read-only views | |
| Topic 12: Relational Database Concepts | 10% | - Relationship between database and SQL - ERD and SQL statement mapping - Theoretical and physical aspects of relational databases |
| Topic 13: Displaying Data from Multiple Tables | 15% | - Cartesian products - Self-joins and outer joins - Equijoins and non-equijoins |
| Topic 14: Controlling User Access | - Roles, granting and revoking privileges - System and object privileges | |
| Topic 15: Restricting and Sorting Data | 15% | - ORDER BY clause and sorting rules - WHERE clause and filtering conditions - Operator precedence - Substitution variables, DEFINE and VERIFY commands |
| Topic 16: Retrieving Data Using SQL SELECT Statement | 20% | - SELECT statement syntax and usage - Column aliases, concatenation, literals and DISTINCT - Arithmetic operations and NULL handling |
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NEW QUESTION # 144
Which three statements are true about views in an Oracle database?
Answer: A,D,F
NEW QUESTION # 145
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?
Answer: A
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 # 146
Which three tasks can be performed using SQL functions built into Oracle Database? (Choose three.)
Answer: B,C,D
NEW QUESTION # 147
Which three actions can you perform by using the ORACLE DATAPUMP access driver?
Answer: D,E,F
NEW QUESTION # 148
View the exhibit and examine the structures of the EMPLOYEESand DEPARTMENTStables.
EMPLOYEES
Name Null? Type
- ---------------- ----- -------------
EMPLOYEE_ID NOT NULL NUMBER(6)
FIRST_NAME VARCHAR2(20)
LAST_NAME NOT NULL VARCHAR2(25)
HIRE_DATE NOT NULL DATE
JOB_ID NOT NULL VARCHAR2(10)
SALARY NUMBER(10,2)
COMMISSION NUMBER(6,2)
MANAGER_ID NUMBER(6)
DEPARTMENT_ID NUMBER(4)
DEPARTMENTS
Name Null? Type
----------------- ----- -------------
DEPARTMENT_ID NOT NULL NUMBER(4)
DEPARTMENT_NAME NOT NULL VARCHAR2(30)
MANAGER_ID NUMBER(6)
LOCATION_ID NUMBER(4)
You want to update EMPLOYEEStable as follows:
Update only those employees who work in Boston or Seattle (locations 2900 and 2700).
Set department_idfor these employees to the department_idcorresponding to London
(location_id 2100).
Set the employees' salary in location_id2100 to 1.1 times the average salary of their department.
Set the employees' commission in location_id2100 to 1.5 times the average commission of their
department.
You issue the following command:
SQL> UPDATE employees
SET department_id
( SELECT department_id
FROM departments
WHERE location_id = 2100),
( salary, commission)
( SELECT 1.1*AVG(salary), 1.5*AVG(commission)
FROM employees, departments
WHERE departments.location_id IN(2900, 2700, 2100))
WHERE department_id IN
( SELECT department_id
FROM departments
WHERE location_id = 2900
OR location_id = 2700;
What is outcome?
Answer: B
NEW QUESTION # 149
......
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