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To prepare for the exam, Oracle recommends that candidates have a solid understanding of SQL fundamentals and a good understanding of Oracle Database concepts. Candidates can take advantage of training courses, study guides, and practice exams to prepare for the exam. Oracle also provides an exam preparation guide that outlines the topics covered in the exam and provides sample questions for each topic.
Oracle 1z1-071 certification exam is a computer-based test that consists of 73 multiple-choice questions. 1z0-071 exam duration is 105 minutes, and candidates must score at least 63% to pass the exam. 1z0-071 exam can be taken at any Oracle testing center or through an online proctoring system. 1z0-071 Exam Fee may vary depending on the location, but the average cost is around $245.
Oracle 1z1-071 exam is designed for SQL developers, database administrators, and anyone who works with Oracle databases. It covers a wide range of topics, including SQL queries, data manipulation, table creation, and database schema design. 1z0-071 exam is challenging, and it requires a thorough understanding of SQL concepts and principles. However, with the right preparation and study materials, passing the exam is achievable.
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Oracle Database SQL Sample Questions (Q35-Q40):
NEW QUESTION # 35
Examine the description of the EMPLOYEES table:
NLS_DATE FORMAT is DD-MON-RR.
Which two queries will execute successfully?
- A. SELECT AVG(MAX(salary)) FROM employees GROUP BY salary;
- B. SELECT dept id, MAX (SUM(salary)) FROM employees GROUP BY dept_id;
- C. SELECT dept_ id, AVG(MAX(salary)) FROM employees GROUP BY dept_id, salary;
- D. SELECT dept_ id, AVG (MAX(salary)) FROM employees GROUP By dept_id HAVING hire_date> ' O1-JAN-19';
- E. SELECT dept_ iD, sum(salary) FROM employees WHERE hire_date > '01-JAN-9' GROUP BY dept_id;
Answer: E
Explanation:
In Oracle SQL, aggregation functions such as AVG and MAX cannot be nested directly inside each other and must be used in conjunction with GROUP BY on the column(s) that are not included in the aggregate function.
Also, the HAVING clause filters groups after aggregation is applied.
A). This query will not execute successfully because it improperly nests MAX inside AVG. Oracle SQL does not allow this type of nested aggregation without a subquery.
B). This query will not execute successfully for the same reason as A, it improperly nests MAX inside AVG.
C). Similar to A and B, this query improperly nests SUM inside MAX, which is not allowed without a subquery.
D). This query will execute successfully. It filters rows based on the HIRE_DATE using a correct date format (assuming '9' refers to '09' or '1999' due to the NLS_DATE_FORMAT being 'DD-MON-RR'), then groups the remaining rows by DEPT_ID and calculates the sum of SALARY for each department.
E). This query will not execute successfully because it improperly nests MAX inside AVG without a subquery, and it incorrectly attempts to GROUP BY SALARY, which is already being aggregated.
References:
* Oracle Database SQL Language Reference, 12c Release 1 (12.1): "Aggregate Functions"
* Oracle Database SQL Language Reference, 12c Release 1 (12.1): "GROUP BY Clause"
* Oracle Database SQL Language Reference, 12c Release 1 (12.1): "HAVING Clause"
NEW QUESTION # 36
View the Exhibit and examine the structure of the PROMOTION table.
You have to generate a report that displays the promo named start data for all promos that started after that last promo in the 'INTTERNET' category.
- A. SELECT promo-name, promo-being_date FROM promotion
WHERE promo-being-date IN (SELECT promo_biing_date
FROM promotions
WHERE promo_category='INTYERNET'); - B. Select promo_name, promo_being_date FROM promoptions
WHERE promo_being_data > ANY (SELCT promo_being-date
FROM promotions
WHERE promo_category = 'INTERNET' - C. SELECT promo-name, promo-being _date FROM promotions
Where promo_being_data >ALL (SELECT MAX (promo_being-date)
FROM promotions ) AND
Promo-category ='INTERNET'; - D. SELECT promo_neme, promo_being_date FROM promotions
WHERE promo_being_date > All (SELECT promo_beinjg-date
FROM promotions
WHERE promo_category ='INTERNET' );
Answer: D
NEW QUESTION # 37
Examine this statement:
What is returned upon execution?
- A. 0 rows
- B. 1 row
- C. 2 rows
- D. an error
Answer: B
NEW QUESTION # 38
View the exhibit and examine the structure of the SALES, CUSTOMERS, PRODUCTSand TIMEStables.
The PROD_IDcolumn is the foreign key in the SALEStable, which references the PRODUCTStable.
Similarly, the CUST_IDand TIME_IDcolumns are also foreign keys in the SALEStable referencing the CUSTOMERSand TIMEStables, respectively.
Evaluate the following CREATE TABLEcommand:
CREATE TABLE new_sales (prod_id, cust_id, order_date DEFAULT SYSDATE)
AS
SELECT prod_id, cust_id, time_id
FROM sales;
Which statement is true regarding the above command?
- A. The NEW_SALEStable would not get created because the DEFAULTvalue cannot be specified in the column definition.
- B. The NEW_SALEStable would not get created because the column names in the CREATETABLE command and the SELECTclause do not match.
- C. The NEW_SALEStable would get created and all the NOTNULLconstraints defined on the specified columns would be passed to the new table.
- D. The NEW_SALEStable would get created and all the FOREIGNKEYconstraints defined on the specified columns would be passed to the new table.
Answer: C
NEW QUESTION # 39
View the Exhibit and examine the structure of the CUSTOMERS and CUST_HISTORY tables.
The CUSTOMERS table contains the current location of all currently active customers.
The CUST_HISTORY table stores historical details relating to any changes in the location of all current as well as previous customers who are no longer active with the company.
You need to find those customers who have never changed their address.
Which SET operator would you use to get the required output?
- A. INTERSECT
- B. UNION
- C. UNION ALL
- D. MINUS
Answer: D
NEW QUESTION # 40
......
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