2013年12月25日星期三

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Exam Code: 1Z0-047
Exam Name: Oracle (Oracle Database SQL Expert)
Guaranteed success with practice guides, No help, Full refund!
260 Questions and Answers
Updated: 2013-12-24

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NO.1 View the Exhibit and examine the structure of the EMPLOYEES table. You want to retrieve hierarchical
data of the employees using the top-down hierarchy. Which SQL clause would let you choose the
direction to walk through the hierarchy tree?
A.WHERE
B.HAVING
C.GROUP BY
D.START WITH
E.CONNECT BY PRIOR
Answer:E

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NO.2 Evaluate the following SQL statement: ALTER TABLE hr.emp SET UNUSED (mgr_id); Which
statement is true regarding the effect of the above SQL statement?
A.Any synonym existing on the EMP table would have to be re-created.
B.Any constraints defined on the MGR_ID column would be removed by the above command.
C.Any views created on the EMP table that include the MGR_ID column would have to be dropped and
re-created.
D.Any index created on the MGR_ID column would continue to exist until the DROP UNUSED
COLUMNS command is executed.
Answer:B

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NO.3 View the Exhibit and examine the data in ORDERS and ORDER_ITEMS tables. You need to create a
view that displays the ORDER ID, ORDER_DATE, and the total number of items in each order. Which
CREATE VIEW statement would create the view successfully?
A.CREATE OR REPLACE VIEW ord_vu (order_id,order_date) AS SELECT o.order_id, o.order_date,
COUNT(i.line_item_id) "NO OF ITEMS" FROM orders o JOIN order_items i ON (o.order_id = i.order_id)
GROUP BY o.order_id,o.order_date;
B.CREATE OR REPLACE VIEW ord_vu AS SELECT o.order_id, o.order_date, COUNT(i.line_item_id)
"NO OF ITEMS" FROM orders o JOIN order_items i ON (o.order_id = i.order_id) GROUP BY
o.order_id,o.order_date;
C.CREATE OR REPLACE VIEW ord_vu AS SELECT o.order_id, o.order_date, COUNT(i.line_item_id)
FROM orders o JOIN order_items i ON (o.order_id = i.order_id) GROUP BY o.order_id,o.order_date;
D.CREATE OR REPLACE VIEW ord_vu AS SELECT o.order_id, o.order_date, COUNT(i.line_item_id)||'
NO OF ITEMS' FROM orders o JOIN order_items i ON (o.order_id = i.order_id) GROUP BY
o.order_id,o.order_date WITH CHECK OPTION;
Answer:B

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NO.4 Evaluate the following CREATE SEQUENCE statement: CREATE SEQUENCE seq1 START WITH 100
INCREMENT BY 10 MAXVALUE 200 CYCLE NOCACHE; The sequence SEQ1 has generated numbers
up to the maximum limit of 200. You issue the following SQL statement: SELECT seq1.nextval FROM
dual; What is displayed by the SELECT statement?
A.1
B.10
C.100
D.an error
Answer:A

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NO.5 View the Exhibit and examine the structure of the EMP table which is not partitioned and not an
index-organized table. Evaluate the following SQL statement: ALTER TABLE emp DROP COLUMN
first_name; Which two statements are true regarding the above command? (Choose two.)
A.The FIRST_NAME column would be dropped provided it does not contain any data.
B.The FIRST_NAME column would be dropped provided at least one or more columns remain in the table.
C.The FIRST_NAME column can be rolled back provided the SET UNUSED option is added to the above
SQL statement.
D.The FIRST_NAME column can be dropped even if it is part of a composite PRIMARY KEY provided the
CASCADE option is used.
Answer:B D

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NO.6 OE and SCOTT are the users in the database. The ORDERS table is owned by OE. Evaluate the
statements issued by the DBA in the following sequence: CREATE ROLE r1; GRANT SELECT, INSERT
ON oe.orders TO r1; GRANT r1 TO scott; GRANT SELECT ON oe.orders TO scott; REVOKE SELECT
ON oe.orders FROM scott; What would be the outcome after executing the statements?
A.SCOTT would be able to query the OE.ORDERS table.
B.SCOTT would not be able to query the OE.ORDERS table.
C.The REVOKE statement would remove the SELECT privilege from SCOTT as well as from the role R1.
D.The REVOKE statement would give an error because the SELECT privilege has been granted to the
role R1.
Answer:A

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NO.7 Which three tasks can be performed using regular expression support in Oracle Database 10g?
(Choose three.)
A.It can be used to concatenate two strings.
B.It can be used to find out the total length of the string.
C.It can be used for string manipulation and searching operations.
D.It can be used to format the output for a column or expression having string data.
E.It can be used to find and replace operations for a column or expression having string data.
Answer:C D E

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NO.8 Which two statements are true regarding the execution of the correlated subqueries? (Choose two.)
A.The nested query executes after the outer query returns the row.
B.The nested query executes first and then the outer query executes.
C.The outer query executes only once for the result returned by the inner query.
D.Each row returned by the outer query is evaluated for the results returned by the inner query.
Answer:A D

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NO.9 View the Exhibit and examine the description of the EMPLOYEES table. You want to display the
EMPLOYEE_ID, FIRST_NAME, and DEPARTMENT_ID for all the employees who work in the same
department and have the same manager as that of the employee having EMPLOYEE_ID 104. To
accomplish the task, you execute the following SQL statement: SELECT employee_id, first_name,
department_id FROM employees WHERE (manager_id, department_id) =(SELECT department_id,
manager_id FROM employees WHERE employee_id = 104) AND employee_id <> 104; When you
execute the statement it does not produce the desired output. What is the reason for this?
A.The WHERE clause condition in the main query is using the = comparison operator, instead of EXISTS.
B.The WHERE clause condition in the main query is using the = comparison operator, instead of the IN
operator.
C.The WHERE clause condition in the main query is using the = comparison operator, instead of the =
ANY operator.
D.The columns in the WHERE clause condition of the main query and the columns selected in the
subquery should be in the same order.
Answer:D

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NO.10 View the Exhibit and examine the description of the CUSTOMERS table. You want to add a constraint
on the CUST_FIRST_NAME column of the CUSTOMERS table so that the value inserted in the column
does not have numbers. Which SQL statement would you use to accomplish the task?
A.ALTER TABLE CUSTOMERS ADD CONSTRAINT cust_f_name
CHECK(REGEXP_LIKE(cust_first_name,'

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