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| Section | Weight | Objectives |
|---|---|---|
| Displaying Data from Multiple Tables Using Joins | 15% | - Join a table to itself by using a self-join - View data that does not meet a join condition by using outer joins - Write SELECT statements to access data from more than one table using equijoins and non-equijoins - Cross join tables |
| Using Subqueries to Solve Queries | 15% | - List the types of subqueries - Describe the types of problems that subqueries can solve - Define subqueries - Write single-row and multi-row subqueries |
| Controlling User Access | 5% | - Distinguish between granting privileges and roles - Grant privileges on tables and on a view - Differentiate system privileges from object privileges |
| Using Single-Row Functions to Customize Output | 10% | - Describe the differences between single-row and multiple-row functions - Manipulate strings with character function in the SELECT and WHERE clauses - Manipulate numbers with the ROUND, TRUNC, and MOD functions - Perform arithmetic with date data - Manipulate dates with the date functions |
| Restricting and Sorting Data | 6% | - Use ampersand substitution to restrict and sort output at runtime - Sort the rows that are retrieved by a query - Use DEFINE and UNDEFINE commands - Limit the rows that are retrieved by a query |
| Manipulating Data | 14% | - Control transactions - Delete rows from a table - Insert rows into a table - Update rows in a table - Describe each data manipulation language (DML) statement |
| Reporting Aggregated Data Using Group Functions | 15% | - Identify the available group functions - Describe the use of group functions - Group data by using the GROUP BY clause - Include or exclude grouped rows by using the HAVING clause |
| Using Conversion Functions and Conditional Expressions | 10% | - Describe implicit and explicit data type conversion - Use the TO_CHAR, TO_NUMBER, and TO_DATE conversion functions - Apply the NVL, NULLIF, and COALESCE functions to data - Use conditional IF THEN ELSE logic in a SELECT statement |
| 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 |
| Working with Oracle Database SQL | 8% | - Describe the Oracle Database architecture - Explain the installation of Oracle Database software - Start up and shut down a database - Create a container database |
| Using DDL Statements to Create and Manage Tables | 14% | - Drop, rename, and truncate tables - Alter table definitions - Create a simple table - Review the table structure - Categorize the main database objects - Describe the data types that are available for columns |
| Creating Other Schema Objects | 10% | - Create synonyms for database objects - Create, maintain, and use sequences - Create and maintain indexes - Create, modify, and retrieve data from a view |
| 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 |
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NEW QUESTION # 432
Examine the description of the EMPLOYEES table:
Examine this query:
Which line produces an error?
Answer: D
NEW QUESTION # 433
Examine the structure of the MEMBERStable:
Name Null? Type
------------------ --------------- ------------------------------
MEMBER_ID NOT NULL VARCHAR2 (6)
FIRST_NAME VARCHAR2 (50)
LAST_NAME NOT NULL VARCHAR2 (50)
ADDRESS VARCHAR2 (50)
You execute the SQL statement:
SQL > SELECT member_id, ' ' , first_name, ' ' , last_name "ID FIRSTNAME LASTNAME " FROM members; What is the outcome?
Answer: A
NEW QUESTION # 434
View the Exhibit and examine the data in ORDERS_MASTER and MONTHLYjDRDERS tables.
Evaluate the following MERGE statement:
MERGE INTO orders_master o
USING monthly_orders m
ON (o.order_id = m.order_id)
WHEN MATCHED THEN
UPDATE SET o.order_total = m.order_total
DELETE WHERE (m.order_total IS NULL)
WHEN NOT MATCHED THEN
INSERT VALUES (m.order_id, m.order_total);
What would be the outcome of the above statement?
Answer: A
NEW QUESTION # 435
Examine the description of the countries table:
Examine the description of the departments table:
Examine the description of the locations table:
Which two queries will return a list of countries with no departments?
Answer: B,D
Explanation:
The query's goal is to return a list of countries that have no departments linked to them.
Option B and Option D are the correct answers because they use set operations that will effectively return countries that do not have a corresponding entry in the departments table:
Option B uses the NOT IN subquery to exclude countries that have departments linked to them. It looks for country_id values in the countries table that are not present in the list of country_id values associated with locations that are, in turn, associated with departments. This will correctly return countries that have no departments.
Option D uses the MINUS set operator, which subtracts the results of the second SELECT statement from the results of the first. This statement will return all countries from the countries table minus those that have an associated department_id in the departments table, effectively listing countries with no departments.
Option A and Option C are incorrect because:
Option A will not execute successfully as it tries to join tables using a column (country_id) that doesn't exist in the departments table, which will lead to an error.
Option C's use of INTERSECT is incorrect for this requirement. INTERSECT returns only the rows that exist in both queries. Since we want countries with no departments, using INTERSECT would actually return the opposite of what is required.
Reference:
Oracle Documentation on NOT IN clause: SQL Language Reference - Subquery Oracle Documentation on MINUS operator: SQL Language Reference - Set Operators Therefore, the correct options are B and D, which use subquery exclusion and the MINUS set operator, respectively, to accurately identify and return countries without departments.
NEW QUESTION # 436
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?
Answer: D
NEW QUESTION # 437
......
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