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試験番号：1z0-061問題集
試験科目：Oracle 「Oracle Database 12c: SQL Fundamentals」
最近更新時間：2014-02-16
問題と解答：75
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NO.1 Examine the types and examples of relationships that follow:
1.One-to-one a) Teacher to students
2.One-to-many b) Employees to Manager
3.Many-to-one c) Person to SSN
4.Many-to-many d) Customers to products
Which option indicates the correctly matched relationships?
A. 1-a, 2-b, 3-c, and 4-d
B. 1-c, 2-d, 3-a, and 4-b
C. 1-c, 2-a, 3-b, and 4-d
D. 1-d, 2-b, 3-a, and 4-c
Answer: C
Oracle認定証 1z0-061問題集 1z0-061 1z0-061認定資格
NO.2 View the Exhibit for the structure of the student and faculty tables.
You need to display the faculty name followed by the number of students handled by the faculty at
the base location.
Examine the following two SQL statements:
Which statement is true regarding the outcome?
A. Only statement 1 executes successfully and gives the required result.
B. Only statement 2 executes successfully and gives the required result.
C. Both statements 1 and 2 execute successfully and give different results.
D. Both statements 1 and 2 execute successfully and give the same required result.
Answer: D
Oracle認証試験 1z0-061 1z0-061認証試験 1z0-061認定試験
NO.3 You need to create a table for a banking application. One of the columns in the table has the
following requirements:
1. You want a column in the table to store the duration of the credit period.
2) The data in the column should be stored in a format such that it can be easily added and
subtracted with date data type without using conversion functions.
3) The maximum period of the credit provision in the application is 30 days.
4) The interest has to be calculated for the number of days an individual has taken a credit for.
Which data type would you use for such a column in the table?
A. DATE
B. NUMBER
C. TIMESTAMP
D. INTERVAL DAY TO SECOND
E. INTERVAL YEAR TO MONTH
Answer: D
Oracle認証試験 1z0-061 1z0-061 1z0-061
NO.4 Which three tasks can be performed using SQL functions built into Oracle Database?
A. Displaying a date in a nondefault format
B. Finding the number of characters in an expression
C. Substituting a character string in a text expression with a specified string
D. Combining more than two columns or expressions into a single column in the output
Answer: A,B,C
Oracle 1z0-061 1z0-061 1z0-061問題集
NO.5 In the customers table, the CUST_CITY column contains the value 'Paris' for the
CUST_FIRST_NAME 'Abigail'.
Evaluate the following query:
What would be the outcome?
A. Abigail PA
B. Abigail Pa
C. Abigail IS
D. An error message
Answer: B
Oracle認証試験 1z0-061 1z0-061問題集 1z0-061認定資格
NO.6 Examine the structure proposed for the transactions table:
Which two statements are true regarding the creation and storage of data in the above table
structure?
A. The CUST_STATUS column would give an error.
B. The TRANS_VALIDITY column would give an error.
C. The CUST_STATUS column would store exactly one character.
D. The CUST_CREDIT_LIMIT column would not be able to store decimal values.
E. The TRANS_VALIDITY column would have a maximum size of one character.
F. The TRANS_DATE column would be able to store day, month, century, year, hour, minutes,
seconds, and fractions of seconds
Answer: B,C
Oracle参考書 1z0-061 1z0-061問題集 1z0-061過去問
Explanation:
VARCHAR2(size)Variable-length character data (A maximum size must be specified:
minimum size is 1; maximum size is 4, 000.)
CHAR [(size)] Fixed-length character data of length size bytes (Default and minimum size
is 1; maximum size is 2, 000.)
NUMBER [(p, s)] Number having precision p and scale s (Precision is the total number of
decimal digits and scale is the number of digits to the right of the decimal point; precision
can range from 1 to 38, and scale can range from -84 to 127.)
DATE Date and time values to the nearest second between January 1, 4712 B.C., and
December 31, 9999 A.D.
NO.7 Evaluate the following SQL statement:
Which statement is true regarding the outcome of the above query?
A. It executes successfully and displays rows in the descending order of PROMO_CATEGORY .
B. It produces an error because positional notation cannot be used in the order by clause with set
operators.
C. It executes successfully but ignores the order by clause because it is not located at the end of the
compound statement.
D. It produces an error because the order by clause should appear only at the end of a compound
query-that is, with the last select statement.
Answer: D
Oracle過去問 1z0-061認定資格 1z0-061参考書
NO.8 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
Oracle 1z0-061問題集 1z0-061
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.
NO.9 View the Exhibit and examine the structure of the product, component, and PDT_COMP
tables.
In product table, PDTNO is the primary key.
In component table, COMPNO is the primary key.
In PDT_COMP table, <PDTNO, COMPNO) is the primary key, PDTNO is the foreign key referencing
PDTNO in product table and COMPNO is the foreign key referencing the COMPNO in component
table.
You want to generate a report listing the product names and their corresponding component names,
if the component names and product names exist.
Evaluate the following query:
SQL>SELECT pdtno, pdtname, compno, compname
FROM product _____________ pdt_comp
USING (pdtno) ____________ component USING (compno)
WHERE compname IS NOT NULL;
Which combination of joins used in the blanks in the above query gives the correct output?
A. JOIN; JOIN
B. FULL OUTER JOIN; FULL OUTER JOIN
C. RIGHT OUTER JOIN; LEFT OUTER JOIN
D. LEFT OUTER JOIN; RIGHT OUTER JOIN
Answer: C
Oracle 1z0-061認定証 1z0-061参考書
NO.10 Which normal form is a table in if it has no multi-valued attributes and no partial
dependencies?
A. First normal form
B. Second normal form
C. Third normal form
D. Fourth normal form
Answer: B
Oracle 1z0-061 1z0-061 1z0-061
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