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Oracle 1z1-071 exam consists of 73 multiple-choice questions that need to be completed within two hours. 1z1-071 exam covers a wide range of topics, including SQL fundamentals, data retrieval, data manipulation, table creation, schema objects, and much more. 1z1-071 Exam can be taken online or in a test center, and the passing score is 63%. By passing the exam, a candidate can earn the Oracle Database SQL certification, which is valid for three years.

Oracle Database SQL Sample Questions (Q380-Q385):

NEW QUESTION # 380
Which three statements are true about sequences in a single instance Oracle database?

A. A sequence can issue duplicate values.
B. A sequence can only be dropped by a DBA.
C. Sequences can always have gaps.
D. Two or more tables cannot have keys generated from the same sequence.
E. A sequence's unallocated cached values are lost if the instance shuts down.
F. A sequence number that was allocated can be rolled back if a transaction fails.

Answer: C,E,F

Explanation:
Sequences are database objects used to generate unique numeric identifiers. Here's the correct understanding of sequences in Oracle:
* A: Correct. Cached sequence values that are not yet used are lost if the database instance shuts down, as the cache is held in memory.
* B: Incorrect. The same sequence can be used to generate keys for more than one table. There is no such limitation.
* C: Correct. If a transaction using a sequence number is rolled back, the sequence number that was used or allocated is not reused.
* D: Incorrect. By their nature and configuration, sequences are designed to avoid issuing duplicate values unless explicitly designed to cycle.
* E: Correct. Sequences may have gaps, which can occur due to caching, sequence increment settings, or if a transaction using a sequence number is rolled back.
* F: Incorrect. A sequence can be dropped by any user with adequate privileges, not just a DBA.

NEW QUESTION # 381
Examine the data in the CUST_NAME column of the CUSTOMERS table.

You want to extract only those customer names that have three names and display the * symbol in place of the first name as follows:

Which two queries give the required output?

A. SELECT LPAD(SUBSTR(cust_name, INSTR(cust_name, ' ')),LENGTH(cust_name),'*') "CUST NAME"FROM customersWHERE INSTR(cust_name, ' ',1,2)<>0;
B. SELECT LPAD(SUBSTR(cust_name ' ')),LENGTH(cust_name) - INSTR(cust_name, ' '), '*') "CUST NAME"FROM customersWHERE INSTR(cust_name, ' ',1,2)<>0;
C. SELECT LPAD(SUBSTR(cust_name ' ')),LENGTH(cust_name) - INSTR(cust_name, ' '), '*') "CUST NAME"FROM customersWHERE INSTR(cust_name, ' ',1,-2)<>0;
D. SELECT LPAD(SUBSTR(cust_name, INSTR(cust_name, ' ')),LENGTH(cust_name),'*') "CUST NAME"FROM customersWHERE INSTR(cust_name, ' ',-1,2)<>0;

Answer: A,D

NEW QUESTION # 382
Evaluate these commands which execute successfully:

Which two statements are true about the ORD_ITEMStable and the ORD_SEQsequence? (Choose two.)

A. If sequence ORD_SEQis dropped then the default value for column ORD_NOwill be NULL for rows inserted into ORD_ITEMS.
B. Sequence ORD_SEQcycles back to 1 after every 5000 numbers and can cycle 20 times.
C. Sequence ORD_SEQis guaranteed not to generate duplicate numbers.
D. Any user inserting rows into table ORD_ITEMSmust have been granted access to sequence ORD_SEQ.
E. Column ORD_NOgets the next number from sequence ORD_SEQwhenever a row is inserted into ORD_ITEMSand no explicit value is given for ORD_NO.

Answer: C,D

NEW QUESTION # 383
Examine the data in the EMPLOYEES table:

Which statement will compute the total annual compensation for each employee?

A. SELECT last _ NAME (monthly_ salary + monthly _commission _ pct) * 12 AS annual_ comp FROM employees;
B. SELECT last _ name, (monthly_ salary * 12) + (monthly_ salary * 12 * NVL (monthly_ commission_pct, 0)) AS annual _comp
C. select last _ name, (monthly_ salary * 12) + (monthly_ salary * 12 *monthly_ commission_ pct) AS annual_ camp FROM employees
D. SELECT last _ name, (monthly _ salary * 12) + (monthly_ commission _ pct * 12) AS FROM employees:

Answer: B

Explanation:
The correct way to compute the total annual compensation, which includes the monthly salary and the monthly commission (if any), is:
* Option C: SELECT last_name, (monthly_salary * 12) + (monthly_salary * 12 * NVL(monthly_commission_pct, 0)) AS annual_comp FROM employees;
* This statement takes the monthly salary and multiplies it by 12 to get the annual salary, and then adds the annual commission which is the monthly salary multiplied by the commission percentage (if any, else 0) and then by 12.
Options A, B, and D are incorrect because:
* Option A: Does not handle the case where the commission percentage is NULL which would result in NULL for the entire expression when added to the monthly salary.
* Option B: Does not consider that the commission percentage might be NULL which could lead to incorrect calculations (or NULL values if commission is NULL).
* Option D: Incorrectly adds the monthly commission percentage directly to the annual salary without considering that the percentage needs to be applied to the salary.

NEW QUESTION # 384
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_idFROM product_informationWHERE UPPER(product_name) IN ('LASERPRO',
'CABLE');
B. SELECT UPPER(product_name)FROM product_informationWHERE product_id = 2254;
C. SELECT UPPER(product_name)FROM product_information;
D. SELECT product_id, UPPER(product_name)FROM product_informationWHERE
UPPER(product_name)='LASERPRO' OR list_price > 1000;

Answer: A

NEW QUESTION # 385
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