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Oracle 1z0-071 exam covers a broad range of topics related to SQL development and database management, including SQL queries, data manipulation, data control language, data definition language, transactions, and database security. Candidates must demonstrate their ability to write complex SQL queries, manage database objects, and perform database administration tasks. 1z1-071 Exam is challenging, but passing it can open up many opportunities for IT professionals in various industries.

Oracle Database SQL Sample Questions (Q390-Q395):

NEW QUESTION # 390
View the Exhibit and examine the structure of the CUSTOMERS table.

Using the CUSTOMERS table, you must generate a report that displays a credit limit increase of 15% for all customers.
Customers with no credit limit should have "Not Available" displayed.
Which SQL statement would produce the required result?

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

Answer: C

NEW QUESTION # 391
Examine the description of the CUSTOMERS table:

You want to display details of all customers who reside in cities starting with the letter D followed by at least two character.
Which query can be used?

A. SELECT * FROM customers WHERE city ='%D_';
B. SELECT * FROM customers WHERE city LIKE'D %';
C. SELECT * FROM customers WHERE city ='D_%';
D. SELECT * FROM customers WHERE city LIKE'D_';

Answer: C

Explanation:
The LIKE operator is used in SQL to search for a specified pattern in a column. To find all customers residing in cities starting with 'D' followed by at least two characters, you need to use the LIKE operator with the appropriate wildcard pattern:
* A. SELECT * FROM customers WHERE city LIKE 'D_%'; The underscore () wildcard character in SQL represents a single character. The percent (%) wildcard character represents zero or more characters. So 'D%' will match any city starting with 'D' followed by at least one character (as there is one underscore).
References:
* Oracle Database SQL Language Reference 12c, specifically the section on pattern matching with the
* LIKE condition.

NEW QUESTION # 392

Which two queries only return CUBE?

A. SELECT shape FROM bricks JOIN boxes ON weight >= min_weight AND weight < max_weight;
B. SELECT shape FROM bricks JOIN boxes ON NOT (weight > max_weight);
C. SELECT shape FROM bricks JOIN boxes ON weight > min_weight;
D. SELECT shape FROM bricks JOIN boxes ON weight < max_weight;
E. SELECT shape FROM bricks JOIN boxes ON weight BETWEEN min_weight AND max_weight;

Answer: A,B

Explanation:
Based on the table structure given in the image, to return the value 'CUBE' from the 'bricks' table when joined with 'boxes', the condition must ensure that the weight of the bricks is within the allowed weight range specified in the 'boxes' table for a 'SMALL' box size.
A). True. Since MAX_WEIGHT is 0, a comparison using >= min_weight AND weight < max_weight will only return rows where the weight is less than 0, which is impossible for actual weight values, suggesting there might be a mistake in the data provided or the comparison logic.
E). True. NOT (weight > max_weight) effectively translates to 'where weight is less than or equal to max_weight'. However, since MAX_WEIGHT is 0, this condition would only be true if the weight is not greater than 0, which can only happen if the weight is 0 or less. This seems to indicate an anomaly where either the data is incorrect, or the condition is meant to handle a case where the weight is zero or possibly a negative placeholder value.
Both B and D will potentially return more than just 'CUBE' if there are bricks with weights greater than MIN_WEIGHT. C is incorrect because BETWEEN is inclusive, and there are no weights that are both greater than or equal to MIN_WEIGHT and less than or equal to MAX_WEIGHT when MAX_WEIGHT is
0.

NEW QUESTION # 393
View the exhibit and examine the description of the PRODUCT_INFORMATION table.

Which SQL statement would retrieve from the table the number of products having LIST_PRICE as NULL?

A. SELECT COUNT (NVL(list_price, 0))FROM product_informationWHERE list_price is NULL
B. SELECT COUNT (list_price)FROM product_informationWHERE list_price is NULL
C. SELECT COUNT (list_price)FROM product_informationWHERE list_price i= NULL
D. SELECT COUNT (DISTINCT list_price)FROM product_informationWHERE list_price is NULL

Answer: A

NEW QUESTION # 394
Which two statements are true regarding the COUNTfunction?

A. The COUNTfunction can be used only for CHAR, VARCHAR2, and NUMBERdata types.
B. A SELECTstatement using the COUNTfunction with a DISTINCTkeyword cannot have a WHERE clause.
C. COUNT(cust_id)returns the number of rows including rows with duplicate customer IDs and NULL value in the CUST_IDcolumn.
D. COUNT(DISTINCT inv_amt)returns the number of rows excluding rows containing duplicates and NULLvalues in the INV_AMTcolumn.
E. COUNT (*)returns the number of rows including duplicate rows and rows containing NULLvalue in any of the columns.

Answer: D,E

NEW QUESTION # 395
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