PASS GUARANTEED ORACLE - 1Z0-071 - ORACLE DATABASE SQL–PROFESSIONAL VALID EXAM COST

Pass Guaranteed Oracle - 1z0-071 - Oracle Database SQL–Professional Valid Exam Cost

Pass Guaranteed Oracle - 1z0-071 - Oracle Database SQL–Professional Valid Exam Cost

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Oracle Database SQL Sample Questions (Q287-Q292):

NEW QUESTION # 287
Examine the description of EMPLOYEES table:
Which three queries return all rows for which SALARY+COMMISSION is greate than 20000?

  • A. SELECT * FROM employees WHERE NVL(salary+commission,0)>==20000;
  • B. SELECT * FROM employees WHERE salary+NVL2(commission,commission,0)>=20000;
  • C. SELECT * FROM employees WHERE salary+NVL(commission,0)>=20000;
  • D. SELECT * FROM employees WHERE salary+NULLF(commission,0)>=20000;
  • E. SELECT * FROM employees WHERE NVL(salary+commission,0)>=20000;
  • F. SELECT * FROM employees WHERE NVL2(salary)+commission,salary+commission,

Answer: B,C,F


NEW QUESTION # 288
View the Exhibit and examine the structure of CUSTOMERStable.
Using the CUSTOMERStable, you need to generate a report that shows an increase in the credit limit by
15% for all customers. Customers whose credit limit has not been entered should have the message "Not Available" displayed.
Which SQL statement would produce the required result?

SELECT NVL (TO CHAR(cust_credit_limit * .15), 'Not Available') "NEW CREDIT"

  • A. FROM customers;
  • B. FROM customers;
    SELECT NVL(cust_credit_limit * .15), 'Not Available') "NEW CREDIT"
  • C. FROM customers;
    SELECT TO_CHAR (NVL(cust_credit_limit * .15), 'Not Available') "NEW CREDIT"
  • D. FROM customers;
    SELECT NVL(cust_credit_limit), 'Not Available') "NEW CREDIT"

Answer: C


NEW QUESTION # 289
Which two statements are true about the ORDER BY clause?

  • A. Ina character sort, the values are case-sensitive.
  • B. NULLS are not included in the sort operation.
  • C. Numeric values are displayed in descending order if they have decimal positions.
  • D. Only columns that are specified in the SELECT list can be used in the ORDER BY cause.
  • E. Column aliases can be used In the ORDER BY cause.

Answer: A,E


NEW QUESTION # 290
Examine the commands used to create DEPARTMENT_DETAILS and
COURSE_DETAILS:
SQL>CREATE TABLE DEPARTMENT_DETAILS
(DEPARTMENT_ID NUMBER PRIMARY KEY,
DEPARTMENT_NAMEVARCHAR2(50),
HODVARCHAR2(50));
SQL>CREATE TABLE COURSE_DETAILS
(COURSE_IDNUMBER PRIMARY KEY,
COURSE_NAMEVARCHAR2(50),
DEPARTMENT_IDVARCHAR2(50));
You want to generate a list of all department IDs along with any course IDs that may have been assigned to them.
Which SQL statement must you use?

  • A. SELECT d.department_id, c.course_id FROM department_details d RIGHT OUTER JOIN course_details c ON (c.department_id=d. department_id);
  • B. SELECT d.department_id, c.course_id FROM department_details d RIGHT OUTER JOIN course_details c ON (d.department_id=c. department_id);
  • C. SELECT d.department_id, c.course_id FROM course_details c LEFT OUTER JOIN department_details d ON (c.department_id=d. department_id);
  • D. SELECT d.department_id, c.course_id FROM department_details d LEFT OUTER JOIN course_details c ON (d.department_id=c. department_id);

Answer: D


NEW QUESTION # 291
Examine the structure of the EMPLOYEES table.

You must display the details of employees who have manager with MANAGER_ID 100, who were hired in the past 6 months and who have salaries greater than 10000.
Which query would retrieve the required result?

  • A. SELECT last_name, hire_date, salary
    FROM employees
    WHERE manager_id = (SELECT employee_id FROM employees WHERE employee_id = 100) UNION ALL (SELECT last_name, hire_date, salary FROM employees WHERE hire_date > SYSDATE -180 INTERSECT SELECT last_name, hire_date, salary FROM employees WHERE salary > 10000);
  • B. SELECT last_name, hire_date, salary
    FROM employees
    WHERE salary > 10000
    UNION ALL
    SELECT last_name, hire_date, salary
    FROM employees
    WHERE manager_ID = (SELECT employee_id FROM employees WHERE employee_id = 100) INTERSECT SELECT last_name, hire_date, salary FROM employees WHERE hire_date > SYSDATE- 180;
  • C. SELECT last_name, hire_date, salary
    FROM employees
    WHERE manager_id = (SELECT employee_id FROM employees WHERE employee_id = '100') UNION SELECT last_name, hire_date, salary FROM employees WHERE hire_date > SYSDATE -180 INTERSECT SELECT last_name, hire_date, salary FROM employees WHERE salary > 10000;
  • D. (SELECT last_name, hire_date, salary
    FROM employees
    WHERE salary > 10000
    UNION ALL
    SELECT last_name, hire_date, salary
    FROM employees
    WHERE manager_ID = (SELECT employee_id FROM employees WHERE employee_id = 100)) UNION SELECT last_name, hire_date, salary FROM employees WHERE hire_date > SYSDATE -180;

Answer: C


NEW QUESTION # 292
......

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