EXAM ORACLE 1Z0-071 DURATION | 1Z0-071 EXAM QUESTIONS ANSWERS

Exam Oracle 1z0-071 Duration | 1z0-071 Exam Questions Answers

Exam Oracle 1z0-071 Duration | 1z0-071 Exam Questions Answers

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Oracle Database SQL Sample Questions (Q263-Q268):

NEW QUESTION # 263
See the Exhibit and examine the structure of the PROMOTIONS table:

Using the PROMOTIONS table,
you need to find out the average cost for all promos in the range $0-2000 and $2000-5000 in category A.
You issue the following SQL statements:

What would be the outcome?

  • A. It generates an error because multiple conditions cannot be specified for the WHEN clause.
  • B. It generates an error because NULL cannot be specified as a return value.
  • C. It generates an error because CASE cannot be used with group functions.
  • D. It executes successfully and gives the required result.

Answer: D

Explanation:
CASE Expression
Facilitates conditional inquiries by doing the work of an IF-THEN-ELSE statement:
CASE expr WHEN comparison_expr1 THEN return_expr1
[WHEN comparison_expr2 THEN return_expr2
WHEN comparison_exprn THEN return_exprn
ELSE else_expr]
END


NEW QUESTION # 264
Examine the data in the CUST_NAME column of the CUSTOMERS table.
CUST_NAME
-------------------
Renske Ladwig
Jason Mallin
Samuel McCain
Allan MCEwen
Irene Mikilineni
Julia Nayer
You need to display customers' second names where the second name starts with "Mc" or
"MC".
Which query gives the required output?

  • A. SELECT SUBSTR (cust_name, INSTR (cust_name, ' ')+1)FROM customersWHERE SUBSTR (cust_name, INSTR (cust_name, ' ')+1)LIKE INITCAP ('MC%');
  • B. SELECT SUBSTR (cust_name, INSTR (cust_name, ' ')+1)FROM customersWHERE INITCAP (SUBSTR(cust_name, INSTR (cust_name, ' ')+1))LIKE 'Mc%';
  • C. SELECT SUBSTR (cust_name, INSTR (cust_name, ' ')+1)FROM customersWHERE INITCAP (SUBSTR(cust_name, INSTR (cust_name, ' ')+1)) ='Mc';
  • D. SELECT SUBSTR (cust_name, INSTR (cust_name, ' ')+1)FROM customersWHERE INITCAP (SUBSTR(cust_name, INSTR (cust_name, ' ')+1)) =INITCAP 'MC%';

Answer: B


NEW QUESTION # 265
View the Exhibit and examine the structure of the ORDERS table.

Which UPDATE statement is valid?

  • A. UPDATE ordersSET order_date = TO_DATE('12-mar-2007','dd-mon-yyyy'),SET order_total = TO_NUMBER (NULL)WHERE order_id = 2455;
  • B. UPDATE ordersSET order_date = '12-mar-2007',AND order_total
    TO_NUMBER(NULL)WHERE order_id = 2455;
  • C. UPDATE ordersSET order_date = '12-mar-2007',order_total = NULLWHERE order_id
    2455;
  • D. UPDATE ordersSET order_date = '12-mar-2007',order_total IS NULLWHERE order_id
    = 2455;

Answer: C


NEW QUESTION # 266
Examine the description of the ENPLYEES table:

Which two queries return all rows for employees whose salary is greater than the average salary in their department?

  • A. SELECT
    FROM employees
    WHERE salary > AVG (salary) OVER (PARTITION BY department _ id);
  • B. SELECT"
    FROM employees
    WHERE salary >
    ( SELECT AVG
    (salary) FROM
    employees
    GROUP BY department _ id
  • C. SELECT"
    FROM employees e1
    WHERE salary >!
    SELECT AVG (salary)
    FROM employees e2
    WHERE e1. Department _id = e2, department_ id
  • D. SELECT.
    FROM
    SELECT e.", AVG (salary) OVER (PARTITION BY department id) avg_ sal
    FROM employees e
    WHERE salary > avg_ sal;
  • E. SELECT "
    FROM employees
    WHERE salary > ANY
    SELECT AVG (salary)
    EROM employees
    GROUP BY department_ id);

Answer: A,C

Explanation:
To return all rows for employees whose salary is greater than the average salary in their department, you would use either a subquery or an analytic function:
* Option B:
* SELECT ... FROM employees WHERE salary > AVG(salary) OVER (PARTITION BY department_id);
* This uses the window function AVG with PARTITION BY to calculate the average salary per department, and it compares each employee's salary to this average.
* Option C:
* SELECT ... FROM employees e1 WHERE salary > (SELECT AVG(salary) FROM
employees e2 WHERE e1.department_id = e2.department_id);
* This correlated subquery compares each employee's salary to the average salary in their department using a subquery to calculate the average salary for that department.
Options A, D, and E are incorrect because:
* Option A: The use of ANY with the subquery does not ensure comparison with the average salary of their respective department.
* Option D: This is syntactically incorrect; the subquery alias avg_sal is not accessible outside the subquery.
* Option E: The subquery does not correlate with the outer query to ensure that each employee's salary is compared to the average salary of their respective department.


NEW QUESTION # 267
Examine the following query:
SQL> SELECT prod_id, amount_sold
FROM sales
ORDER BY amount_sold
FETCH FIRST 5 PERCENT ROWS ONLY;
What is the output of this query?

  • A. It results in an error because the ORDER BY clause should be the last clause.
  • B. It displays the first 5 percent of the rows from the SALES table.
  • C. It displays 5 percent of the products with the lowest amount sold.
  • D. It displays 5 percent of the products with the highest amount sold.

Answer: C

Explanation:
Explanation
References:
https://oracle-base.com/articles/12c/row-limiting-clause-for-top-n-queries-12cr1


NEW QUESTION # 268
......

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