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Oracle Database SQL Sample Questions (Q168-Q173):

NEW QUESTION # 168
Examine the description of the CUSTONERS table

CUSTON is the PRIMARY KEY.
You must derermine if any customers'derails have entered more than once using a different costno,by listing duplicate name Which two methode can you use to get the requlred resuit?

A. LEFT OUTER JOIN with seif join
B. SUBQUERY
C. seif join
D. FULL OUTER JOIN with seif join
E. RIGHT OUTER JOIN with seif join

Answer: B,C

Explanation:
To determine if customer details have been entered more than once using a different custno, the following methods can be used:
* SUBQUERY: A subquery can be used to find duplicate names by grouping the names and having a count greater than one.
SELECT custname FROM customers GROUP BY custname HAVING COUNT(custname) > 1;
* Self Join: A self join can compare rows within the same table to find duplicates in the custname column, excluding matches with the same custno.
SELECT c1.custname FROM customers c1 JOIN customers c2 ON c1.custname = c2.custname AND c1.custno != c2.custno; These methods allow us to find duplicates in the custname column regardless of the custno. Options A, B, and E are not applicable methods for finding duplicates in this context.
References:
* Oracle Documentation on Group By: Group By
* Oracle Documentation on Joins: Joins

NEW QUESTION # 169
Examine the description of the PRODUCT_INFORMATION table:

A. SELECT count(nvl( list_price,0)) FROM product_information WHERE list_price is null;
B. SELECT COUNT(DISTINCT list_price) FROM product_information WHERE list_price is null.
C. BELECT COUNT(list_price) FROM product_information where list_price is NULL;
D. SELECT (COUNT(list_price) FROM Product_intormation WHERE list_price=NULL;

Answer: A

NEW QUESTION # 170
You must display details of all users whose username contains the string 'ch_'. (Choose the best answer.) Which query generates the required output?

A. SELECT * FROM users
Where user_name LIKE '%ch_';
B. SELECT * FROM users
Where user_name LIKE 'ch\_%' ESCAPE '_';
C. SELECT * FROM users
Where user_name LIKE '%ch\_%' ESCAPE '\';
D. SELECT * FROM users
Where user_name LIKE '%ch_%'ESCAPE'%';

Answer: D

NEW QUESTION # 171
Examine these SQL statements that are executed in the given order:

What will be the status of the foreign key EMP_MGR_FK?

A. It will remain disabled and can be enabled only by dropping the foreign key constraint and re-creating it.
B. It will be enabled and deferred.
C. It will be enabled and immediate.
D. It will remain disabled and can be re-enabled manually.

Answer: D

NEW QUESTION # 172
Which two are true about virtual columns?

A. They cannot have a data type explicitly specified.
B. They can be referenced In the where clause of an update or debete statement.
C. They can be referenced in the set clause of an update statement as the name of the column To be updated.
D. They can be indexed.
E. They can be referenced in the column expression of another virtxial column.

Answer: A,D

Explanation:
Regarding the properties and capabilities of virtual columns in Oracle Database:
* C. They can be indexed: In Oracle, virtual columns can indeed be indexed. This allows for performance optimization on queries that utilize these columns, despite the fact that their values are derived and not stored physically.
* D. They cannot have a data type explicitly specified: The data type of a virtual column is derived from the expression used to define it. Therefore, it is not specified explicitly but is inferred from the expression itself.
Incorrect options:
* A: Virtual columns cannot be referenced in the WHERE clause of UPDATE or DELETE statements because they are not stored and cannot be individually modified.
* B: Virtual columns cannot be referenced in the SET clause of an UPDATE statement since they are derived from other column values and cannot be updated directly.
* E: Virtual columns cannot reference other virtual columns in their expressions due to the potential for recursive and complex dependencies.

NEW QUESTION # 173
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