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| Section | Weight | Objectives |
|---|---|---|
| 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 |
| Retrieving Data Using the SQL SELECT Statement | 5% | - Differentiate between SQL statements and SQL*Plus commands - Execute a basic SELECT statement - List the capabilities of SQL SELECT statements |
| Working with Oracle Database SQL | 8% | - Explain the installation of Oracle Database software - Start up and shut down a database - Create a container database - Describe the Oracle Database architecture |
| Using Single-Row Functions to Customize Output | 10% | - Manipulate dates with the date functions - Manipulate strings with character function in the SELECT and WHERE clauses - Describe the differences between single-row and multiple-row functions - Manipulate numbers with the ROUND, TRUNC, and MOD functions - Perform arithmetic with date data |
| Displaying Data from Multiple Tables Using Joins | 15% | - 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 - Join a table to itself by using a self-join - Cross join tables |
| Manipulating Data | 14% | - Delete rows from a table - Insert rows into a table - Describe each data manipulation language (DML) statement - Update rows in a table - Control transactions |
| Creating Other Schema Objects | 10% | - Create, modify, and retrieve data from a view - Create and maintain indexes - Create, maintain, and use sequences - Create synonyms for database objects |
| Controlling User Access | 5% | - Differentiate system privileges from object privileges - Distinguish between granting privileges and roles - Grant privileges on tables and on a view |
| 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 - Apply the NVL, NULLIF, and COALESCE functions to data - Use conditional IF THEN ELSE logic in a SELECT statement |
| Using Subqueries to Solve Queries | 15% | - Define subqueries - Write single-row and multi-row subqueries - List the types of subqueries - Describe the types of problems that subqueries can solve |
| Reporting Aggregated Data Using Group Functions | 15% | - Group data by using the GROUP BY clause - Include or exclude grouped rows by using the HAVING clause - Describe the use of group functions - Identify the available group functions |
| Using Set Operators | 5% | - Describe the order of precedence when the set operators are used - Use a set operator to combine multiple queries into a single query - Describe set operators |
| Using DDL Statements to Create and Manage Tables | 14% | - Alter table definitions - Create a simple table - Drop, rename, and truncate tables - Categorize the main database objects - Review the table structure - Describe the data types that are available for columns |
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NEW QUESTION # 224
An Oracle database server session has an uncommitted transaction in progress which updated 5000 rows in a table.
In which three situations does the transactions complete thereby committing the updates? (Choose three.)
Answer: B,C,F
NEW QUESTION # 225
Examine the structure of the employees table.
There is a parent/child relationship between EMPLOYEE_Id and MANAGER_ID.
You want to display the last names and manager IDs of employees who work for the same manager asthe employee whose EMPLOYEE_ID is 123.
Which query provides the correct output?
Answer: C
NEW QUESTION # 226
Examine this Statement which returns the name of each employee and their manager, SELECT e.last name AS emp,,m.last_name AS mgr FROM employees e JOIN managers m ON e.manager_ id = m. employee_ id ORDER BY emp; You want to extend the query to include employees with no manager. What must you add before JOIN to do this?
Answer: A
NEW QUESTION # 227
Examine the data in the PRODUCTS table:
Examine these queries:
1. SELECT prod name, prod list
FROM products
WHERE prod 1ist NOT IN(1020) AND category _id=1;
2. SELECT prod name, | prod _ list
FROM products
WHERE prod list < > ANY (1020) AND category _id= 1;
SELECT prod name, prod _ list
FROM products
WHERE prod_ list <> ALL (10 20) AND category _ id= 1;
Which queries generate the same output?
Answer: A
Explanation:
Based on the given PRODUCTS table and the SQL queries provided:
* Query 1: Excludes rows where prod_list is 10 or 20 and category_id is 1.
* Query 2: Includes rows where prod_list is neither 10 nor 20 and category_id is 1.
* Query 3: Excludes rows where prod_list is both 10 and 20 (which is not possible for a single value) and category_id is 1.
The correct answer is A, queries 1 and 3 will produce the same result. Both queries exclude rows where prod_list is 10 or 20 and include only rows from category_id 1. The NOT IN operator excludes the values within the list, and <> ALL operator ensures that prod_list is not equal to any of the values in the list, which effectively excludes the same set of rows.
Query 2, using the <> ANY operator, is incorrect because this operator will return true if prod_list is different from any of the values in the list, which is not the logic represented by the other two queries.
NEW QUESTION # 228
Examine the contents of the EMP table:
Examine this query that executes successfully:
What is the result?
Answer: D
Explanation:
CREATE TABLE EMP
(
ID NUMBER(10),
NAME VARCHAR2(10),
SALARY NUMBER(10)
)
INSERT INTO EMP VALUES (101, 'JOHN', 26000);
INSERT INTO EMP VALUES (102, 'NEENA', 24000);
INSERT INTO EMP VALUES (103, 'DEHAAN', 12000);
INSERT INTO EMP VALUES (104, 'LEX', 17000);
INSERT INTO EMP VALUES (105, 'BILL', 18000);
INSERT INTO EMP VALUES (106, 'DANIEL', 26000);
INSERT INTO EMP VALUES (107, 'BEN', 12000);
INSERT INTO EMP VALUES (108, 'GEORGE', 25000);
SELECT * FROM EMP
ORDER BY SALARY
FETCH FIRST 5 ROWS WITH TIES;
NEW QUESTION # 229
......
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