Free PDF Oracle - 1z0-071 - Useful Oracle Database SQL Actual Test Answers

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The Oracle Database SQL certification exam consists of 73 questions that must be answered within 100 minutes. Test-takers can register for the exam online and may take it at a Pearson VUE testing center or remotely through an online proctoring system. 1z0-071 exam evaluates candidates on various topics, including SQL schema manipulation, data retrieval, and manipulation, and database management. A passing score of 63% is required to earn the certification, which is valid for three years.

Oracle 1z1-071 (Oracle Database SQL) certification exam is designed for individuals seeking to demonstrate their expertise in SQL database management. As a comprehensive and industry-recognized exam, it measures a candidate's understanding of database design, data modeling, and query techniques, as well as their ability to troubleshoot and optimize SQL queries. It is a crucial certification for individuals working in database administration, data warehousing, and business intelligence.

Oracle 1z0-071 Exam covers a wide range of topics related to SQL programming and Oracle database management. These topics include data modeling, database design, SQL queries, data manipulation, data control, query optimization, and security management. 1z0-071 exam is designed to test your knowledge in each of these areas, and passing it requires a good understanding of the underlying concepts and principles of Oracle database management.

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Oracle Database SQL Sample Questions (Q20-Q25):

NEW QUESTION # 20
View the Exhibit and examine the data in the EMPLOYEES table.
Exhibit

You want to generate a report showing the total compensation paid to each employee to date.
You issue the following query:

What is the outcome?

A. It generates an error because the usage of the ROUND function in the expression is not valid.
B. It generates an error because the alias is not valid.
C. IT executes successfully and gives the correct output.
D. It executes successfully but does not give the correct output.
E. It generates an error because the concatenation operator can be used to combine only two items.

Answer: D

NEW QUESTION # 21
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 SUBSTR(cust_name,INSTR(cust_name,'') +1) LIKE'Mc%' OR'MC%'
B. WHERE INITCAP(SUBSTR(cust_name, INSTR(cust_name,'') +1)) LIKE'Mc%'
C. WHERE UPPER (SUBSTR(cust_name, INSTR(cust_name, '') +1)) LIKE UPPER('MC%')
D. WHERE SUBSTR(cust_name, INSTR(cust_name,'') +1) LIKE'Mc%'
E. WHERE INITCAP (SUBSTR(cust_name, INSTR(cust_name,'') +1)) IN ('MC%','Mc%)

Answer: B,C

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 # 22
Examine this SQL statement:
DELETE FROM employees e
WHERE EXISTS
(SELECT'dummy'
FROM emp_history
WHERE employee_id = e.employee_id)
Which two are true?

A. The subquery is not a correlated subquery.
B. The subquery is executed for every row in the EMPLOYEES table.
C. All existing rows in the EMPLOYEE table are deleted.
D. The DELETE statement executes successfully even if the subquery selects multiple rows.
E. The subquery is executed before the DELETE statement is executed.

Answer: B,D

Explanation:
The provided DELETE statement uses a correlated subquery to determine which rows should be deleted from the EMPLOYEES table.
* A. The subquery is indeed executed for every row in the EMPLOYEES table. This is because it references e.employee_id, which is a column from the outer query, making it a correlated subquery.
* B. The subquery is a correlated subquery, as explained above.
* C. The subquery is executed for each row, not before the DELETE statement.
* D. Not all existing rows in the EMPLOYEES table are deleted, only those that have a corresponding employee_id in the EMP_HISTORY table.
* E. The DELETE statement executes successfully even if the subquery selects multiple rows because the EXISTS condition only checks for the presence of rows, not their count.
References:
* Oracle Database SQL Language Reference 12c Release 1 (12.1), DELETE Statement
* Oracle Database SQL Language Reference 12c Release 1 (12.1), Subquery Factoring
* Oracle Database SQL Language Reference 12c Release 1 (12.1), Correlated Subqueries

NEW QUESTION # 23
View the Exhibit and examine the structure of CUSTOMERS table.
Using the CUSTOMERS table, 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?

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

Answer: D

NEW QUESTION # 24
Examine the description of the BOOKS_TRANSACTIONS table:

Examine this partial SQL statement:
SELECT * FROM books_transactions
Which two WHERE conditions give the same result?

A. WHERE borrowed_date = SYSDATE AND transaction_type = 'RM' AND member_id = 'A101' OR member_id = 'A102');
B. WHERE (borrowed_date = SYSDATE AND transaction_type = 'RM') OR member_id IN ('A101','A102');
C. WHERE borrowed_date = SYSDATE AND transaction_type = 'RM' OR member_id IN('A101','A102');
D. WHERE borrowed_date = SYSDATE AND transaction_type = 'RM' OR member_id IN('A101','A102');
E. WHERE borrowed_date = SYSDATE AND transaction_type = 'RM' AND (member_id = 'A101' OR member_id = 'A102'));

Answer: B,E

Explanation:
When writing SQL statements with multiple conditions in the WHERE clause, it's important to understand how logical operators like AND and OR work. These operators are used to filter records based on more than one condition:
The AND operator displays a record if all the conditions separated by AND are TRUE.
The OR operator displays a record if any of the conditions separated by OR is TRUE.
The precedence of these operators is also important: AND operations are always evaluated before OR operations unless parentheses are used to explicitly define the order of operations. Therefore, conditions enclosed in parentheses are evaluated first as per the standard SQL operator precedence.
Given this understanding, let's evaluate the options:
Option A uses parentheses to group the borrowed_date and transaction_type conditions together, ensuring these are evaluated first before the OR operator. This means the query will return records that meet both of these conditions or records with member_id 'A101' or 'A102', regardless of other conditions.
Option D does not use parentheses around the borrowed_date and transaction_type conditions but does use them to group the member_id conditions. Since the AND operator has higher precedence than the OR, this query will first evaluate the borrowed_date and transaction_type conditions, then evaluate the grouped member_id conditions. The outcome is records that have a borrowed_date of SYSDATE, a transaction_type of 'RM', and a member_id of either 'A101' or 'A102'.
The results of Options A and D are effectively the same, even though the use of parentheses is different. This is because in both cases, the evaluation of the borrowed_date and transaction_type conditions will be performed first due to their grouping by parentheses in Option A and the precedence of AND over OR in Option D. Therefore, they will both return all records where borrowed_date equals SYSDATE and transaction_type equals 'RM', plus any records where member_id is either 'A101' or 'A102'.
For further details on SQL operator precedence and logical operators, you can refer to Oracle Database SQL Language Reference 12c documentation, specifically the sections on conditional expressions.

NEW QUESTION # 25
......

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