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Download Oracle Database SQL Exam Dumps
NEW QUESTION 53
View the Exhibits and examine PRODUCTSand SALES tables.
Exhibit 1
Exhibit 2
You issue the following query to display product name the number of times the product has been sold:
What happens when the above statement is executed?
- A. The statement produces an error because ITEM_CNTcannot be displayed in the outer query.
- B. The statement executes successfully and produces the required output.
- C. The statement produces an error because the GROUP BYclause cannot be used in a subquery in the FROMclause.
- D. The statement produces an error because a subquery in the FROMclause and outer-joins cannot be used together.
Answer: B
NEW QUESTION 54
Examine these statements:
CREATE TABLE alter_test (c1 VARCHAR2(10), c2 NUMBER(10));
INSERT INTO alter_test VALUES ('123', 123);
COMMIT;
Which is true ahout modifyIng the columns in AITER_TEST?
- A. c1 can be changed to NUMBER(10) and c2 can be changed to VARCHAN2 (10).
- B. c1 can be changed to NUMBER(10) but c2 cannot be changed to VARCHAN2 (10).
- C. c2 can be changed to VARCHAR2(10) but c1 cannot be changed to NUMBER (10).
- D. c2 can be changed to NUMBER(5) but c1 cannot be changed to VARCHAN2 (5).
- E. c1 can be changed to VARCHAR2(5) and c2 can be changed to NUMBER (12,2).
Answer: E
NEW QUESTION 55
Evaluate the following statement.
Which statement is true regarding the evaluation of rows returned by the subquery in the INSERT statement?
- A. Each row is evaluated by the first WHEN clause and if the condition is true, then the row would be evaluated by the subsequent WHEN clauses.
- B. All rows are evaluated by all the three WHEN clauses.
- C. The INSERT statement will return an error because the ELSE clause is missing.
- D. Each row is evaluated by the first WHEN clause and if the condition is false then the row would be evaluated by the subsequent WHEN clauses.
Answer: B
NEW QUESTION 56
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*.15, 'Not Available') "NEW CREDIT" FROM customers
- B. SELECT NVL (TO_CHAR(cust_credit_limit*.15), 'Not Available') "NEW CREDIT" FROM customers
- C. SELECT NVL (cust_credit_limit, 'Not Available')*.15 "NEW CREDIT" FROM customers
- D. SELECT TO_CHAR(NVL(cust_credit_limit*.15), 'Not Available')) "NEW CREDIT" FROM customers
Answer: A
NEW QUESTION 57
View the exhibit and examine the structure of the STOREStable.
You want to display the NAME of the store along with the ADDRESS, START_DATE, PROPERTY_PRICE, and the projected property price, which is 115% of property price.
The stores displayed must have START_DATEin the range of 36 months starting from 01-Jan-2000 and above.
Which SQL statement would get the desired output?
SELECT name, concat (address| | ','| |city| |', ', country) AS full_address,
- A. start_date,
property_price, property_price*115/100
FROM stores
WHERE MONTHS_BETWEEN (start_date, TO_DATE('01-JAN-2000','DD-MON-RRRR'))
<=36; - B. start_date,
property_price, property_price*115/100
FROM stores
WHERE MONTHS_BETWEEN (start_date, TO_DATE('01-JAN-2000','DD-MON-RRRR'))
<=36;
SELECT name, concat (address||','| |city| |', ', country) AS full_address, - C. start_date,
property_price, property_price*115/100
FROM stores
WHERE MONTHS_BETWEEN (start_date, '01-JAN-2000') <=36;
SELECT name, concat (address| | ','| |city| |', ', country) AS full_address, - D. start_date,
property_price, property_price*115/100
FROM stores
WHERE TO_NUMBER(start_date-TO_DATE('01-JAN-2000','DD-MON-RRRR')) <=36;
SELECT name, address||','||city||','||country AS full_address,
Answer: A
NEW QUESTION 58
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