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Oracle 1z1-071 Exam Overview:

Certification Vendor:	Oracle
Exam Name:	Oracle Database SQL
Exam Number:	1z0-071
Passing Score:	63%
Exam Format:	Multiple Choice
Available Languages:	Korean, Japanese, Portuguese, English, Simplified Chinese
Certificate Validity Period:	No expiration for this exam/certification
Exam Duration:	120 minutes
Exam Price:	$245 USD
Real Exam Qty:	70
Related Certifications:	Oracle Database Administrator Certified Associate (DBA) Oracle Database SQL Certified Associate
Sample Questions:	Oracle 1z1-071 Sample Questions
Exam Way:	Pearson VUE testing center (online proctored option available)
Pre Condition:	No formal prerequisites required, but basic knowledge of relational database concepts is recommended
Official Syllabus URL:	https://education.oracle.com/oracle-database-sql-exam

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Oracle 1z0-071 Practice Test Questions, Oracle 1z0-071 Exam Practice Test Questions

The Oracle 1Z0-071 exam is required for obtaining several certifications provided by the Oracle University, namely Oracle PL/SQL Developer Certified Associate, Oracle Database 12c Administrator Certified Associate, and Oracle Database SQL Certified Associate. This test validates the candidates’ competency in Oracle SQL as well as PL/SQL technologies.

Oracle Database SQL Sample Questions (Q296-Q301):

NEW QUESTION # 296
Which four statements are true about constraints on Oracle tables?

A. A UNIQUE constraint permits NULLS.
B. A CHECK Constraint can refer to values in other rows.
C. A FOREIGN KEY Column can contain NULLS.
D. A PRIMARY KEY Constraint can be added after a table has been created and populated.
E. A NOT NULL Constraint can be defined at the table level.
F. A Column can have only one CHECK Constraint.
G. A UNIQUE Constraint can use a pre-existing index on the constrained column or columns.

Answer: A,C,D,G

NEW QUESTION # 297
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' AND (member_id = 'A101' OR member_id = 'A102'));
E. WHERE (borrowed_date = SYSDATE AND transaction_type = 'RM') OR member_id IN ('A101','A102');

Answer: D,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 # 298
The PRODUCT_INFORMATION table has a UNIT_PRICE column of data type NUMBER(8, 2).
Evaluate this SQL statement:
SELECT TO_CHAR(unit_price,'$9,999') FROM PRODUCT_INFORMATION;
Which two statements are true about the output?

A. A row whose UNIT_PRICE column contains the value 1023.99 will be displayed as $1,023.
B. A row whose UNIT_PRICE column contains the value 10235.99 will be displayed as #####
C. A row whose UNIT_PRICE column contains the value 10235.99 will be displayed as $1,0236.
D. A row whose UNIT_PRICE column contains the value 1023.99 will be displayed as $1,024.
E. A row whose UNIT_PRICE column contains the value 10235.99 will be displayed as $1,023.

Answer: A,B

Explanation:
The TO_CHAR function is used to convert a number to a string format in Oracle SQL. In this format mask '$9,999', the dollar sign is a literal, and the 9 placeholders represent a digit in the output. The comma is a digit group separator.
A . This statement is incorrect because the format model does not have enough digit placeholders to display the full number 1023.99; it would round it to $1,023 not $1,024. B. This statement is correct. Given the format '$9,999', the number 1023.99 will be formatted as $1,023 because the format rounds the number to no decimal places. C. This is incorrect because the format '$9,999' cannot display the number 10235.99; it exceeds the format's capacity. D. This is incorrect for the same reason as C, and the format would not change the thousands to hundreds. E. This statement is correct. If the number exceeds the maximum length of the format mask, which is 4 digits in this case, Oracle SQL displays a series of hash marks (#) instead of the number.
These formatting rules are described in the Oracle Database SQL Language Reference, which covers the TO_CHAR function and its number formatting capabilities.

NEW QUESTION # 299
View the Exhibit and examine the structure in the DEPARTMENTS tables. (Choose two.)

Examine this SQL statement:
SELECT department_id "DEPT_ID", department_name, 'b' FROM
departments
WHERE departments_id=90
UNION
SELECT department_id, department_name DEPT_NAME, 'a' FROM
departments
WHERE department_id=10
Which two ORDER BY clauses can be used to sort the output?

A. ORDER BY 3;
B. ORDER BY DEPT_ID;
C. ORDER BY DEPT_NAME;
D. ORDER BY 'b';

Answer: A,B

NEW QUESTION # 300
Examine the structure of the INVOICEtable.
Name Null? Type
- ----------------- ------------------- -------------
INV_NO NOT NULL NUMBER(3)
INV_DATE DATE
INV_AMT NUMBER(10,2)
Which two SQL statements would execute successfully?

A. SELECT inv_no, NVL2(inv_amt, inv_date, 'Not Available')
FROM invoice;
B. SELECT inv_no, NVL2(inv_date, sysdate-inv_date, sysdate)
FROM invoice;
C. SELECT inv_no, NVL2(inv_amt, inv_amt*.25, 'Not Available')
FROM invoice;
D. SELECT inv_no, NVL2(inv_date, 'Pending', 'Incomplete')
FROM invoice;

Answer: B,D

NEW QUESTION # 301
......

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