ORACLE 1Z0-071 ACCURATE STUDY MATERIAL - 1Z0-071 RELIABLE STUDY GUIDE

Oracle 1z0-071 Accurate Study Material - 1z0-071 Reliable Study Guide

Oracle 1z0-071 Accurate Study Material - 1z0-071 Reliable Study Guide

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Oracle Database SQL Sample Questions (Q94-Q99):

NEW QUESTION # 94
View the Exhibit and examine the structure of the PRODUCT_INFORMATION table.

PRODUCT_ID column is the primary key.
You create an index using this command:
SQL > CREATE INDEX upper_name_idx
ON product_information(UPPER(product_name));
No other indexes exist on the PRODUCT_INFORMATION table.
Which query would use the UPPER_NAME_IDX index? (Choose the best answer.)

  • A. SELECT product_id, UPPER(product_name)FROM product_informationWHERE UPPER(product_name)='LASERPRO' OR list_price > 1000;
  • B. SELECT UPPER(product_name)FROM product_information;
  • C. SELECT UPPER(product_name)FROM product_informationWHERE product_id = 2254;
  • D. SELECT product_idFROM product_informationWHERE UPPER(product_name) IN ('LASERPRO', 'CABLE');

Answer: D


NEW QUESTION # 95
The SALES table has columns PROD_IDand QUANTITY_SOLDof data type NUMBER.
Which two queries execute successfully? (Choose two.)

  • A. BY prod_id HAVING COUNT(*) >10;
    SELECT COUNT (prod_id) FROM sales WHERE quantity_sold > 55000 GROUP BY
  • B. SELECT prod_id FROM sales WHERE quantity_sold > 55000 GROUP BY prod_id HAVING
  • C. prod_id;
    SELECT prod_id FROM sales WHERE quantity_sold > 55000 AND COUNT(*) > 10 GROUP
  • D. COUNT(*) >10;
    SELECT prod_id FROM sales WHERE quantity_sold > 55000 AND COUNT(*) > 10 GROUP
  • E. BY COUNT(*) >10;
    SELECT COUNT(prod_id) FROM sales GROUP BY prod_id WHERE quantity_sold > 55000;

Answer: A,B


NEW QUESTION # 96
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 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;
  • C. (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;
  • 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) INTERSECT SELECT last_name, hire_date, salary FROM employees WHERE hire_date > SYSDATE- 180;

Answer: B


NEW QUESTION # 97
The BOOKS_TRANSACTIONStable exists in your database.
SQL>SELECT * FROM books_transactions ORDER BY 3;
What is the outcome on execution?

  • A. The execution fails unless the numeral 3 in the ORDER BYclause is replaced by a column name.
  • B. Rows are displayed in the order that they are stored in the table only for the three rows with the lowest values in the key column.
  • C. Rows are displayed sorted in ascending order of the values in the third column in the table.
  • D. Rows are displayed in the order that they are stored in the table only for the first three rows.

Answer: C


NEW QUESTION # 98
Examine the data in the CUST NAME column of the CUSTOMERS table:
CUST_NAME
------------------------------
Renske Ladwig
Jason Mallin
Samuel McCain
Allan MCEwen
Irene Mikkilineni
Julia Nayer
You want to display the CUST_NAME values where the last name starts with Mc or MC. Which two WHERE clauses give the required result?

  • A. WHERE INITCAP (SUBSTR(cust_name, INSTR(cust_name,'') +1)) IN ('MC%','Mc%)
  • B. WHERE INITCAP(SUBSTR(cust_name, INSTR(cust_name,'') +1)) LIKE'Mc%'
  • C. WHERE SUBSTR(cust_name,INSTR(cust_name,'') +1) LIKE'Mc%' OR'MC%'
  • D. WHERE SUBSTR(cust_name, INSTR(cust_name,'') +1) LIKE'Mc%'
  • E. WHERE UPPER (SUBSTR(cust_name, INSTR(cust_name, '') +1)) LIKE UPPER('MC%')

Answer: B,E

Explanation:
To find customers whose last names start with "Mc" or "MC", we need to ensure our SQL query correctly identifies and compares these prefixes regardless of case variations. Let's analyze the given options:
Option B: WHERE UPPER(SUBSTR(cust_name, INSTR(cust_name, ' ') + 1)) LIKE UPPER('MC%') This clause uses UPPER to convert both the extracted substring (starting just after the first space, assuming it indicates the start of the last name) and the comparison string 'MC%' to uppercase. This ensures case-insensitive comparison. The LIKE operator is used to match any last names starting with "MC", which will correctly capture both "Mc" and "MC". This option is correct.
Option C: WHERE INITCAP(SUBSTR(cust_name, INSTR(cust_name, ' ') + 1)) LIKE 'Mc%' This clause applies INITCAP to the substring, which capitalizes the first letter of each word and makes other letters lowercase. The result is compared to 'Mc%', assuming only the last name follows the space. This approach will match last names starting with "Mc" (like "McEwen"), but not "MC". However, considering we're looking for "Mc" specifically, this clause works under the assumption that "Mc" is treated as proper capitalization for these last names. Thus, it can also be considered correct, albeit less inclusive than option B.
The other options either use incorrect syntax or apply case-sensitive matches without ensuring that both "Mc" and "MC" are captured:
Option A: Contains syntax errors (unmatched quotes and wrong use of IN).
Option D: Uses case-sensitive match without combining both "Mc" and "MC".
Option E: Only matches "Mc", which is too specific.


NEW QUESTION # 99
......

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