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Oracle 1z0-071 Exam Syllabus Topics:

Section	Objectives
Topic 1: SQL Retrieval	- Joins and set operations 1. INNER, OUTER, CROSS joins 2. UNION, INTERSECT, MINUS - SELECT statements and filtering data 1. WHERE clause and conditions 2. ORDER BY and row limiting
Topic 2: Relational Database Concepts	- Tables, rows, columns, and relational model basics - Data types and database objects overview
Topic 3: Data Manipulation and Schema Objects	- DDL operations 1. CREATE, ALTER, DROP tables and constraints - DML operations 1. INSERT, UPDATE, DELETE, MERGE
Topic 4: Subqueries and Advanced SQL	- Single-row and multi-row subqueries - Correlated subqueries and nested queries
Topic 5: SQL Functions	- Single-row functions 1. Character, numeric, date functions - Conversion and conditional functions 1. TO_CHAR, TO_NUMBER, CASE expressions

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Oracle Database SQL Sample Questions (Q62-Q67):

NEW QUESTION # 62
Which two are true about constraints?

A. Constraints are enforced only during insert operations.
B. All constraints can be defined at the table or column level.
C. A column with a FOREIGN key constraint can never contain a null value.
D. A constraint can be disabled even if the constrained column contains data.
E. A column with a unique constraint can contain a null value.

Answer: D,E

NEW QUESTION # 63
Which statement is true regarding the INTERSECT operator?

A. The names of columns in all SELECT statements must be identical.
B. Reversing the order of the intersected tables alters the result.
C. The number of columns and data types must be identical for all SELECT statements in the query.
D. It ignores NULL values.

Answer: C

Explanation:
INTERSECT Returns only the rows that occur in both queries' result sets, sorting them and removing duplicates.
The columns in the queries that make up a compound query can have different names, but the output result set will use the names of the columns in the first query.
References:
http://oraclexpert.com/using-the-set-operators/

NEW QUESTION # 64
Examine the description of the EMPLOYEES table:

Which two queries return rows for employees whose manager works in a different department?

A. SELECT emp.*
FROM employees emp
WHERE NOT EXISTS (
SELECT NULL
FROM employees mgr
WHERE emp.manager id = mgr.employee_ id
AND emp.department_id<>mgr.department_id
);
B. SELECT emp. *
FROM employees emp
JOIN employees mgr
ON emp. manager_ id = mgr. employee_ id
AND emp. department_ id<> mgr.department_ id;
C. SELECT emp. *
FROM employees emp
RIGHT JOIN employees mgr
ON emp.manager_ id = mgr. employee id
AND emp. department id <> mgr.department_ id
WHERE emp. employee_ id IS NOT NULL;
D. SELECT emp.*
FROM employees emp
LEFT JOIN employees mgr
ON emp.manager_ id = mgr.employee_ id
AND emp. department id < > mgr. department_ id;
E. SELECT emp. *
FROM employees emp
WHERE manager_ id NOT IN (
SELECT mgr.employee_ id
FROM employees mgr
WHERE emp. department_ id < > mgr.department_ id
);

Answer: A,B

Explanation:
To find employees whose manager works in a different department, you can use a subquery or a join that compares the DEPARTMENT_ID of the employee with the DEPARTMENT_ID of their manager.
A). This query is incorrect because the NOT IN subquery incorrectly attempts to compare EMPLOYEE_ID with MANAGER_ID, and the correlation condition inside the subquery is incorrect.
B). This query is correct. The NOT EXISTS clause correctly identifies employees whose MANAGER_ID matches the EMPLOYEE_ID of another employee (mgr) and where the DEPARTMENT_ID differs from that manager's DEPARTMENT_ID.
C). This query is incorrect because the LEFT JOIN will return all employees, and there is no WHERE clause to filter out those employees whose managers are in the same department.
D). This query is incorrect. The RIGHT JOIN does not ensure that the resulting rows are for employees whose manager works in a different department. It also returns all managers, which is not the requirement.
E). This query is correct. The JOIN ensures that the returned rows are for employees (emp) whose MANAGER_ID matches the EMPLOYEE_ID of managers (mgr), and it correctly filters to include only those employees whose DEPARTMENT_ID is different from their manager's DEPARTMENT_ID.

NEW QUESTION # 65
View the Exhibit and examine PRODUCTS and ORDER_ITEMS tables.

You executed the following query to display PRODUCT_NAME and the number of times the product has been ordered:
SQL>SELECT p.product_name, i.item_cnt
FROM (SELECT product_id, COUNT (*) item_cnt
FROM order_items
GROUP BY product_id) i RIGHT OUTER JOIN products p
ON i.product_id = p.product_id;
What would happen when the above statement is executed?

A. The statement would execute successfully to produce the required output.
B. The statement would not execute because inline views and outer joins cannot be used together.
C. The statement would not execute because the GROUP BY clause cannot be used in the inline.
D. The statement would not execute because the ITEM_CNT alias cannot be displayed in the outer query.

Answer: A

NEW QUESTION # 66
Examine the data in the ENPLOYEES table:

Which statement will compute the total annual compensation tor each employee

A. SELCECT last_namo, (monthly_salary * 12) + (menthy_salary * 12 * NVL (monthly_commission_pct, 0)) AS annual_comp FROM employees
B. SELCECT last_namo, (monthly_salary * 12) + (monthly_commission_pct * 12) AS annual_comp FROM employees
C. SECECT last_namo, (menthy_salary + monthly_commission_pct) * 12 AS annual_comp FROM employees;
D. SELCECT last_namo, (monthly_salary * 12) + (menthy_salary * 12 * monthly_commission_pct) AS annual_comp FROM employees

Answer: A

Explanation:
The correct statement for computing the total annual compensation for each employee is option C. This is because the monthly commission is a percentage of the monthly salary (indicated by the column name MONTHLY_COMMISSION_PCT). To calculate the annual compensation, we need to calculate the annual salary (which is monthly_salary * 12) and add the total annual commission to it.
Here's the breakdown of the correct statement, option C:
* (monthly_salary * 12) computes the total salary for the year.
* NVL(monthly_commission_pct, 0) replaces NULL values in the monthly_commission_pct column with 0, ensuring that the commission is only added if it exists.
* (monthly_salary * 12 * NVL(monthly_commission_pct, 0)) computes the annual commission by first determining the monthly commission (which is a percentage of the monthly salary), and then multiplying it by 12 to get the annual commission.
* Finally, (monthly_salary * 12) + (monthly_salary * 12 * NVL(monthly_commission_pct, 0)) adds the annual salary and the annual commission to get the total annual compensation.
The other options are incorrect:
* Option A is incorrect because it adds the monthly_commission_pct directly to the monthly_salary, which does not consider that monthly_commission_pct is a percentage.
* Option B is incorrect because it adds the commission percentage directly without first applying it to the monthly salary.
* Option D is incorrect because it does not handle the NULL values in the commission column, which would result in a NULL total annual compensation whenever the monthly_comission_pct is NULL.
References:
* Oracle Documentation on NVL function: NVL
* Oracle Documentation on Numeric Literals: Numeric Literals

NEW QUESTION # 67
......

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