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Q1. - (Topic 1)
Which view should a user query to display the columns associated with the constraints on a table owned by the user?
A. USER_CONSTRAINTS
B. USER_OBJECTS
C. ALL_CONSTRAINTS
D. USER_CONS_COLUMNS
E. USER_COLUMNS
Answer: D
Explanation: view the columns associated with the constraint names in the USER_CONS_COLUMNS view. Incorrect Answer: Atable to view all constraints definition and names Bshow all object name belong to user Cdoes not display column associated Eno such view
Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 10-25
Q2. - (Topic 2)
Which two statements are true regarding indexes? (Choose two.)
A. They can be created on tables and clusters.
B. They can be created on tables and simple views.
C. You can create only one index by using the same columns.
D. You can create more than one index by using the same columns if you specify distinctly different combinations of the columns.
Answer: A,D
Q3. - (Topic 2)
The PRODUCTS table has these columns:
PRODUCT_ID NUMBER(4)
PRODUCT_NAME VARCHAR2(45)
PRICE NUMBER(8,2)
Evaluate this SQL statement:
SELECT *
FROM PRODUCTS
ORDER BY price, product_name;
What is true about the SQL statement?
A. The results are not sorted.
B. The results are sorted numerically.
C. The results are sorted alphabetically.
D. The results are sorted numerically and then alphabetically.
Answer: D
Explanation:
the result is sort by price which is numeric and follow by product_name which is alphabetically.
Incorrect Answer: Athe results are sorted Bthe results are sorted with alphabetically as well Cthe results are sorted with numerically as well
Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 2-2
Q4. - (Topic 2)
Which constraint can be defined only at the column level?
A. UNIQUE
B. NOT NULL
C. CHECK
D. PRIMARY KEY
E. FOREIGN KEY
Answer: B
Explanation:
The NOT NULL constraint can be defined only at the column level. It enforces that a value must be defined for this column such that the column may not be NULL for any row.
Incorrect Answers A:The UNIQUE constraint enforces uniqueness on values in the constrained column. It can be defined not only at the column level. C:The CHECK constraint enforces that values added to the constrained column must be present in a static list of values permitted for the column.
D:The PRIMARY KEY constraint stipulates that values in the constrained column(s) must be unique and not NULL. If the primary key applies to multiple columns, then the combination of values in the columns must be unique and not NULL. E:The FOREIGN KEY constraint enforces that only values in the primary key of a parent table may be included as values in the constrained column(s) of the child table.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 227-232 Chapter 5: Creating Oracle Database Objects
Q5. - (Topic 2)
Evaluate the following CREATE SEQUENCE statement:
CREATE SEQUENCE seq1
START WITH 100
INCREMENT BY 10
MAXVALUE 200
CYCLE
NOCACHE;
The SEQ1 sequence 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
Explanation:
But why the answer is not "C" ? Because you didn't specify the MINVALUE for the sequence. If you check the sequence definition that you created it will have the default value of 1, which it reverts to when cycling. If you wanted to keep the minimum value you would need to specify it in the sequence creation. sequence Is the name of the sequence generator INCREMENT BY n Specifies the interval between sequence numbers, where n is an integer (If this clause is omitted, the sequence increments by 1.) START WITH n Specifies the first sequence number to be generated (If this clause is omitted, the sequence starts with 1.) MAXVALUE n Specifies the maximum value the sequence can generate NOMAXVALUE Specifies a maximum value of 10^27 for an ascending sequence and –1 for a descending sequence (This is the default option.) MINVALUE n Specifies the minimum sequence value NOMINVALUE Specifies a minimum value of 1 for an ascending sequence and –(10^26) for a descending sequence (This is the default option.)
CYCLE | NOCYCLE Specifies whether the sequence continues to generate values after reaching its maximum or minimum value (NOCYCLE is the default option.) CACHE n | NOCACHE Specifies how many values the Oracle server preallocates and keeps in memory (By default, the Oracle server caches 20 values.)
Q6. - (Topic 1)
Which two statements about sub queries are true? (Choose two.)
A. A sub query should retrieve only one row.
B. A sub query can retrieve zero or more rows.
C. A sub query can be used only in SQL query statements.
D. Sub queries CANNOT be nested by more than two levels.
E. A sub query CANNOT be used in an SQL query statement that uses group functions.
F. When a sub query is used with an inequality comparison operator in the outer SQL statement, the column list in the SELECT clause of the sub query should contain only one column.
Answer: B,F
Explanation: Explanation: sub query can retrieve zero or more rows, sub query is used with an inequality comparison operator in the outer SQL statement, and the column list in the SELECT clause of the sub query should contain only one column.
Incorrect Answer: Asub query can retrieve zero or more rows Csub query is not SQL query statement Dsub query can be nested Egroup function can be use with sub query
Q7. - (Topic 2)
Which SQL statements would display the value 1890.55 as $1,890.55? (Choose three.)
A.
SELECT TO_CHAR(1890.55,'$0G000D00') FROM DUAL;
B.
SELECT TO_CHAR(1890.55,'$9,999V99') FROM DUAL;
C.
SELECT TO_CHAR(1890.55,'$99,999D99') FROM DUAL;
D.
SELECT TO_CHAR(1890.55,'$99G999D00') FROM DUAL;
E. SELECT TO_CHAR(1890.55,'$99G999D99') FROM DUAL;
Answer: A,D,E
Q8. - (Topic 2)
What is true about updates through a view?
A. You cannot update a view with group functions.
B. When you update a view group functions are automatically computed.
C. When you update a view only the constraints on the underlying table will be in effect.
D. When you update a view the constraints on the views always override the constraints on the underlying tables.
Answer: A
Q9. - (Topic 2)
Examine the structure of the PROMOS table:
You want to generate a report showing promo names and their duration (number of days).
If the PROMO_END_DATE has not been entered, the message 'ONGOING' should be displayed. Which queries give the correct output? (Choose all that apply.)
A. SELECT promo_name, TO_CHAR(NVL(promo_end_date -promo_start_date,'ONGOING')) FROM promos;
B. SELECT promo_name,COALESCE(TO_CHAR(promo_end_date -promo_start_date),'ONGOING') FROM promos;
C. SELECT promo_name, NVL(TO_CHAR(promo_end_date -promo_start_date),'ONGOING') FROM promos;
D. SELECT promo_name, DECODE(promo_end_date
-promo_start_date,NULL,'ONGOING',promo_end_date - promo_start_date) FROM
promos;
E. SELECT
promo_name,ecode(coalesce(promo_end_date,promo_start_date),null,'ONGOING',
promo_end_date - promo_start_date)
FROM promos;
Answer: B,C,D
Q10. - (Topic 1)
The CUSTOMERS table has the following structure: Exhibit:
You need to write a query that does the following task:
*
Display the first name and tax amount of the customers. Tax is 5% of their credit limit
*
Only those customers whose income level has a value should be considered
*
Customers whose tax amount is null should not be considered
Which statement accomplishes all the required tasks?
A.
SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers WHERE cust_income_level IS NOT NULL AND tax_amount IS NOT NULL;
B.
SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers WHERE cust_income_level IS NOT NULL AND cust_credit_limit IS NOT NULL;
C.
SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers WHERE cust_income_level <> NULL AND tax_amount <> NULL;
D.
SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers WHERE (cust_income_level,tax_amount) IS NOT NULL;
Answer: B
Q11. - (Topic 1)
You need to extract details of those products in the SALES table where the PROD_ID column contains the string '_D123'. Which WHERE clause could be used in the
SELECT statement to get the required output?
A. WHERE prod_id LIKE '%_D123%' ESCAPE '_'
B. WHERE prod_id LIKE '%\_D123%' ESCAPE '\'
C. WHERE prod_id LIKE '%_D123%' ESCAPE '%_'
D. WHERE prod_id LIKE '%\_D123%' ESCAPE '\_'
Answer: B
Explanation:
A naturally occurring underscore character may be escaped (or treated as a regular nonspecial symbol) using the ESCAPE identifier in conjunction with an ESCAPE character. The second example in Figure 3-12 shows the SQL statement that retrieves the JOBS table records with JOB_ID values equal to SA_MAN and SA_REP and which conforms to the original requirement: select job_id from jobs where job_id like 'SA\_%' escape '\'
Q12. - (Topic 2)
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
Q13. - (Topic 1)
Examine the structure of the EMPLOYEES and NEW_EMPLOYEES tables:
Which MERGE statement is valid?
A.
MERGE INTO new_employees c USING employees e ON (c.employee_id = e.employee_id) WHEN MATCHED THEN UPDATE SET
B. name = e.first_name ||','|| e.last_name WHEN NOT MATCHED THEN INSERT value S(e.employee_id, e.first_name ||', '||e.last_name);
C.
MERGE new_employees c USING employees e ON (c.employee_id = e.employee_id) WHEN EXISTS THEN UPDATE SET
D. name = e.first_name ||','|| e.last_name WHEN NOT MATCHED THEN INSERT valueS(e.employee_id, e.first_name ||', '||e.last_name);
E.
MERGE INTO new_employees cUSING employees e ON (c.employee_id = e.employee_id) WHEN EXISTS THEN UPDATE SET
F. name = e.first_name ||','|| e.last_name WHEN NOT MATCHED THEN INSERT value S(e.employee_id, e.first_name ||', '||e.last_name);
G.
MERGE new_employees c FROM employees e ON (c.employee_id = e.employee_id) WHEN MATCHED THEN UPDATE SET
H. name = e.first_name ||','|| e.last_name WHEN NOT MATCHED THEN INSERT INTO new_employees valueS(e.employee_id, e.first_name ||', '||e.last_name);
Answer: A
Explanation: Explanation: this is the correct MERGE statement syntax
Incorrect Answer: Bit should MERGE INTO table_name Cit should be WHEN MATCHED THEN Dit should MERGE INTO table_name Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 8-29
Q14. - (Topic 2)
You need to produce a report for mailing labels for all customers. The mailing label must have only the customer name and address. The CUSTOMERS table has these columns:
CUST_IDNUMBER(4)NOT NULL CUST_NAMEVARCHAR2(100)NOT NULL CUST_ADDRESSVARCHAR2(150)
CUST_PHONEVARCHAR2(20)
Which SELECT statement accomplishes this task?
A. SELECT * FROM customers
B. SELECT name, address FROM customers;
C. SELECT id, name, address, phone FROM customers;
D. SELECT cust_name, cust_address FROM customers;
E. SELECT cust_id, cust_name, cust_address, cust_phone FROM customers;
Answer: D
Explanation:
This answer provides correct list of columns for the output.
Incorrect Answers
A:This answer does not provide correct list of columns for the output. It is not required to
show all columns of the table. Symbol “*” is used in the SELECT command to substitute a
list of all columns of the table.
B:This answer does not provide correct list of columns for the output. There are not NAME
and ADDRESS columns in the CUSTOMERS table.
C:This answer does not provide correct list of columns for the output. There are not ID,
NAME, ADDRESS or PHONE columns in the CUSTOMERS table.
E:This answer does not provide correct list of columns for the output. It is not required to
show all columns of the table.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 20-24
Chapter 1: Overview of Oracle Databases
Q15. - (Topic 1)
You created an ORDERS table with the following description: Exhibit:
You inserted some rows in the table. After some time, you want to alter the table by creating the PRIMARY KEY constraint on the ORD_ID column.
Which statement is true in this scenario?
A. You cannot add a primary key constraint if data exists in the column
B. You can add the primary key constraint even if data exists, provided that there are no duplicate values
C. The primary key constraint can be created only a the time of table creation
D. You cannot have two constraints on one column
Answer: B