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2016 June Oracle Official: 1Z0-051: Oracle Database 11g: SQL Fundamentals I Exam Questions New Updated Today! Braindump2go.com Offers 1Z0-051 PDF and VCE Dumps 303q for Free Downloading!

NEW QUESTION 11 – NEW QUESTION 10:

QUESTION 11
Examine the structure of the PROGRAMS table:
Which two SQL statements would execute successfully? (Choose two.)
 

A.    SELECT NVL(ADD_MONTHS(END_DATE,1),SYSDATE)
FROM programs;
B.    SELECT TO_DATE(NVL(SYSDATE-END_DATE,SYSDATE))
FROM programs;
C.    SELECT NVL(MONTHS_BETWEEN(start_date,end_date),’Ongoing’)
FROM programs;
D.    SELECT NVL(TO_CHAR(MONTHS_BETWEEN(start_date,end_date)),’Ongoing’) FROM programs;

Answer: AD
Explanation:
NVL Function
Converts a null value to an actual value:
Data types that can be used are date, character, and number.
Data types must match:
– NVL(commission_pct,0)
– NVL(hire_date,’01-JAN-97′)
– NVL(job_id,’No Job Yet’)
MONTHS_BETWEEN(date1, date2): Finds the number of months between date1 and date2 The result can be positive or negative. If date1 is later than date2, the result is positive; if date1 is earlier than date2, the result is negative. The noninteger part of the result represents a portion of the month.
MONTHS_BETWEEN returns a numeric value. – answer C NVL has different datatypes – numeric and strings, which is not possible!
The data types of the original and if null parameters must always be compatible. They must either be of the same type, or it must be possible to implicitly convert if null to the type of the original parameter. The NVL function returns a value with the same data type as the original parameter.

QUESTION 12
The PRODUCTS table has the following structure:
 
Evaluate the following two SQL statements:
 
Which statement is true regarding the outcome?

A.    Both the statements execute and give different results.
B.    Both the statements execute and give the same result.
C.    Only the first SQL statement executes successfully.
D.    Only the second SQL statement executes successfully.

Answer: A
Explanation:
Using the NVL2 Function
The NVL2 function examines the first expression. If the first expression is not null, the NVL2 function returns the second expression. If the first expression is null, the third expression is returned.
Syntax
NVL2(expr1, expr2, expr3)
In the syntax:
expr1 is the source value or expression that may contain a null expr2 is the value that is returned if expr1 is not null expr3 is the value that is returned if expr1 is null

QUESTION 13
Examine the structure of the INVOICE table.
Which two SQL statements would execute successfully? (Choose two.)
 

A.    SELECT inv_no,NVL2(inv_date,’Pending’,’Incomplete’)
FROM invoice;
B.    SELECT inv_no,NVL2(inv_amt,inv_date,’Not Available’)
FROM invoice;
C.    SELECT inv_no,NVL2(inv_date,sysdate-inv_date,sysdate)
FROM invoice;
D.    SELECT inv_no,NVL2(inv_amt,inv_amt*.25,’Not Available’)
FROM invoice;

Answer: AC
Explanation:
The NVL2 Function
The NVL2 function provides an enhancement to NVL but serves a very similar purpose. It evaluates whether a column or expression of any data type is null or not.
5-6 The NVL function\
If the first term is not null, the second parameter is returned, else the third parameter is returned. Recall that the NVL function is different since it returns the original term if it is not null. The NVL2 function takes three mandatory parameters. Its syntax is NVL2(original, ifnotnull, ifnull), where original represents the term being tested. Ifnotnull is returned if original is not null, and ifnull is returned if original is null. The data types of the ifnotnull and ifnull parameters must be compatible, and they cannot be of type LONG.
They must either be of the same type, or it must be possible to convert ifnull to the type of the ifnotnull parameter. The data type returned by the NVL2 function is the same as that of the ifnotnull parameter.

QUESTION 14
View the Exhibit and evaluate the structure and data in the CUST_STATUS table.
You issue the following SQL statement:
Which statement is true regarding the execution of the above query?
 

A.    It produces an error because the AMT_SPENT column contains a null value.
B.    It displays a bonus of 1000 for all customers whose AMT_SPENT is less than CREDIT_LIMIT.
C.    It displays a bonus of 1000 for all customers whose AMT_SPENT equals CREDIT_LIMIT, or
AMT_SPENT is null .
D.    It produces an error because the TO_NUMBER function must be used to convert the result of
the NULLIF function before it can be used by the NVL2 function.

Answer: C
Explanation:
The NULLIF Function
The NULLIF function tests two terms for equality. If they are equal the function returns a null, else it returns the first of the two terms tested.
The NULLIF function takes two mandatory parameters of any data type. The syntax is NULLIF(ifunequal, comparison_term), where the parameters ifunequal and comparison_term are compared. If they are identical, then NULL is returned. If they differ, the ifunequal parameter is returned.

QUESTION 15
Which statement is true regarding the COALESCE function?

A.    It can have a maximum of five expressions in a list.
B.    It returns the highest NOT NULL value in the list for all rows.
C.    It requires that all expressions in the list must be of the same data type.
D.    It requires that at least one of the expressions in the list must have a NOT NULL value.

Answer: C
Explanation:
The COALESCE Function
The COALESCE function returns the first nonnull value from its parameter list. If all its parameters are null, then null is returned.
The COALESCE function takes two mandatory parameters and any number of optional parameters. The syntax is COALESCE(expr1, expr2,…,exprn), where expr1 is returned if it is not null, else expr2 if it is not null, and so on. COALESCE is a general form of the NVL function, as the following two equations illustrate:
COALESCE(expr1,expr2) = NVL(expr1,expr2)
COALESCE(expr1,expr2,expr3) = NVL(expr1,NVL(expr2,expr3))
The data type COALESCE returns if a not null value is found is the same as that of the first not null parameter.
To avoid an “ORA-00932: inconsistent data types” error, all not null parameters must have data types compatible with the first not null parameter.

QUESTION 16
View the Exhibit and examine the structure of the PROMOTIONS table.
Using the PROMOTIONS table, you need to find out the average cost for all promos in the ranges $0-2000 and $2000-5000 in category A
You issue the following SQL statement:
 
What would be the outcome?
 

A.    It executes successfully and gives the required result.
B.    It generates an error because NULL cannot be specified as a return value.
C.    It generates an error because CASE cannot be used with group functions.
D.    It generates an error because multiple conditions cannot be specified for the WHEN clause.

Answer: A
Explanation:
CASE Expression
Facilitates conditional inquiries by doing the work of an IF-THEN-ELSE statement:
CASE expr WHEN comparison_expr1 THEN return_expr1
[WHEN comparison_expr2 THEN return_expr2
WHEN comparison_exprn THEN return_exprn
ELSE else_expr]
END

QUESTION 17
View the Exhibit and examine the structure of the PROMOTIONS table.
Which SQL statements are valid? (Choose all that apply.)
 

A.    SELECT promo_id, DECODE(NVL(promo_cost,0), promo_cost,
promo_cost * 0.25, 100) “Discount”
FROM promotions;
B.    SELECT promo_id, DECODE(promo_cost, 10000,
DECODE(promo_category, ‘G1’, promo_cost *.25, NULL),
NULL) “Catcost”
FROM promotions;
C.    SELECT promo_id, DECODE(NULLIF(promo_cost, 10000),
NULL, promo_cost*.25, ‘N/A’) “Catcost”
FROM promotions;
D.    SELECT promo_id, DECODE(promo_cost, >10000, ‘High’,
<10000, ‘Low’) “Range”
FROM promotions;

Answer: AB
Explanation:
The DECODE Function
Although its name sounds mysterious, this function is straightforward. The DECODE function implements ifthen-else conditional logic by testing its first two terms for equality and returns the third if they are equal and optionally returns another term if they are not. The DECODE function takes at least three mandatory parameters, but can take many more. The syntax of the function is DECODE(expr1,comp1, iftrue1, [comp2,iftrue2…[ compN,iftrueN]], [iffalse]).

QUESTION 18
Examine the data in the PROMO_BEGIN_DATE column of the PROMOTIONS table:
PROMO_BEGIN _DATE
——————-
04-jan-00
10-jan-00
15-dec-99
18-oct-98
22-aug-99
You want to display the number of promotions started in 1999 and 2000.
Which query gives the correct output?

A.    SELECT SUM(DECODE(SUBSTR(promo_begin_date,8),’00’,1,0)) “2000”, SUM(DECODE(SUBSTR
(promo_begin_date,8),’99’,1,0)) “1999”
FROM promotions;
B.    SELECT SUM(CASE TO_CHAR(promo_begin_date,’yyyy’) WHEN ’99’ THEN 1 ELSE 0 END) “1999”,
SUM(CASE TO_CHAR(promo_begin_date,’yyyy’) WHEN ’00’ THEN 1 ELSE 0 END) “2000”
FROM promotions;
C.    SELECT COUNT(CASE TO_CHAR(promo_begin_date,’yyyy’) WHEN ’99’ THEN 1 ELSE 0 END) “1999”,
COUNT(CASE TO_CHAR(promo_begin_date,’yyyy’) WHEN ’00’ THEN 1 ELSE 0 END) “2000”
FROM promotions;
D.    SELECT COUNT(DECODE(SUBSTR(TO_CHAR(promo_begin_date,’yyyy’), 8), ‘1999’, 1, 0)) “1999”, COUNT(DECODE(SUBSTR(TO_CHAR(promo_begin_date,’yyyy’), 8),’2000′, 1,
0)) “2000”
FROM promotions;

Answer: A

QUESTION 19
Examine the structure of the TRANSACTIONS table:
name             Null            Type
————-  ———–    ——————–
TRANS_ID        NOT NULL       NUMBER(3)
CUST_NAME                       VARCHAR2(30)
TRANS_DATE                      TIMESTAMP
TRANS_AMT                       NUMBER(10,2)
You want to display the date, time, and transaction amount of transactions that where done before 12 noon.
The value zero should be displayed for transactions where the transaction amount has not been entered.
Which query gives the required result?

A.    SELECT TO_CHAR(trans_date,’dd-mon-yyyy hh24:mi:ss’),
TO_CHAR(trans_amt,’$99999999D99′)
FROM transactions
WHERE TO_NUMBER(TO_DATE(trans_date,’hh24′)) < 12 AND COALESCE(trans_amt,NULL)<>NULL;
B.    SELECT TO_CHAR(trans_date,’dd-mon-yyyy hh24:mi:ss’),
NVL(TO_CHAR(trans_amt,’$99999999D99′),0)
FROM transactions
WHERE TO_CHAR(trans_date,’hh24′) < 12;
C.    SELECT TO_CHAR(trans_date,’dd-mon-yyyy hh24:mi:ss’),
COALESCE(TO_NUMBER(trans_amt,’$99999999.99′),0)
FROM transactions
WHERE TO_DATE(trans_date,’hh24′) < 12;
D.    SELECT TO_DATE (trans_date,’dd-mon-yyyy hh24:mi:ss’),
NVL2(trans_amt,TO_NUMBER(trans_amt,’$99999999.99′), 0)
FROM transactions
WHERE TO_DATE(trans_date,’hh24′) < 12;

Answer: B

QUESTION 20
Examine the structure of the TRANSACTIONS table:
name            Null            Type
———–  ———–    ——————–
TRANS_ID       NOT NULL       NUMBER(3)
CUST_NAME                      VARCHAR2(30)
TRANS_DATE                     DATE
TRANS_AMT                      NUMBER(10,2)
You want to display the transaction date and specify whether it is a weekday or weekend.
Evaluate the following two queries:
Which statement is true regarding the above queries?
 

A.    Both give wrong results.
B.    Both give the correct result.
C.    Only the first query gives the correct result.
D.    Only the second query gives the correct result.

Answer: C
Explanation:
Range Conditions Using the BETWEEN Operator
Use the BETWEEN operator to display rows based on a range of values:
SELECT last_name, salary
FROM employees
WHERE salary BETWEEN 2500 AND 3500;
Range Conditions Using the BETWEEN Operator
You can display rows based on a range of values using the BETWEEN operator. The range that you specify contains a lower limit and an upper limit. The SELECT statement in the slide returns rows from the EMPLOYEES table for any employee whose salary is between $2,500 and $3,500. Values that are specified with the BETWEEN operator are inclusive. However, you must specify the lower limit first.
You can also use the BETWEEN operator on character values:
SELECT last_name
FROM employees
WHERE last_name BETWEEN ‘King’ AND ‘Smith’;


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