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Question 1:
Which task can be performed by using a single Data Manipulation Language (DML) statement?
A. Removing all data only from a single column on which a primary key constraint is defined.
B. Removing all data from a single column on which a unique constraint is defined.
C. Adding a column with a default value while inserting a row into a table.
D. Adding a column constraint while inserting a row into a table.
Correct Answer: A
Question 2:
View the Exhibit and examine the details of the PRODUCT_INFORMATION table.
You have the requirement to display PRODUCT_NAME and LIST_PRICE from the table where the CATEGORYJD column has values 12 or 13, and the SUPPLIER_ID column has the value 102088. You executed the following SQL
statement:
SELECT product_name, list_price
FROM product_information
WHERE (category_id = 12 AND category_id = 13) AND supplier_id = 102088;
Which statement is true regarding the execution of the query?
A. It would execute but the output would return no rows.
B. It would execute and the output would display the desired result.
C. It would not execute because the entire WHERE clause condition is not enclosed within the parentheses.
D. It would not execute because the same column has been used in both sides of the AND logical operator to form the condition.
Correct Answer: A
Question 3:
Which statement is true regarding the default behavior of the ORDER BY clause?
A. In a character sort, the values are case-sensitive.
B. NULL values are not considered at all by the sort operation.
C. Only those columns that are specified in the SELECT list can be used in the ORDER BY clause.
D. Numeric values are displayed from the maximum to the minimum value if they have decimal positions.
Correct Answer: A
Question 4:
View the Exhibit and examine the structure of the EMP table which is not partitioned and not an index-organized table. (Choose two.)
Evaluate this SQL statement:
ALTER TABLE emp
DROP COLUMN first_name;
Which two statements are true?
A. The FIRST_NAME column can be dropped even if it is part of a composite PRIMARY KEY provided the CASCADE option is added to the SQL statement.
B. The FIRST_NAME column would be dropped provided at least one column remains in the table.
C. The FIRST_NAME column would be dropped provided it does not contain any data.
D. The drop of the FIRST_NAME column can be rolled back provided the SET UNUSED option is added to the SQL statement.
Correct Answer: B
Question 5:
Which statement is true regarding the default behaviour of the ORDER by clause?
A. Numeric values are displayed in descending order if they have decimal positions.
B. Only columns that are specified in the SELECT list can be used in the ORDER by clause.
C. In a character sort, the values are case-sensitive.
D. NULLs are not including in the sort operation
Correct Answer: C
Latest 1Z0-071 Dumps1Z0-071 VCE Dumps1Z0-071 Exam Questions
Question 6:
View the exhibit and examine the structure of the CUSTOMERS table.
Which two tasks would require subqueries or joins to be executed in a single statement?
A. finding the number of customers, in each city, whose credit limit is more than the average credit limit of all the customers
B. finding the average credit limit of male customers residing in \’Tokyo\’ or \’Sydney\’
C. listing of customers who do not have a credit limit and were born before 1980
D. finding the number of customers, in each city, who\’s marital status is \’married\’.
E. listing of those customers, whose credit limit is the same as the credit limit of customers residing in the city \’Tokyo\’.
Correct Answer: AE
Question 7:
Examine the data in the CUST_NAME column of the CUSTOMERS table. CUST_NAME
Renske Ladwig Jason Mallin
Samuel McCain Allan MCEwen Irene Mikilineni Julia Nayer You need to display customers\’ second names where the second name starts with “Mc” or “MC”. Which query gives the required output?
A. SELECT SUBSTR (cust_name, INSTR (cust_name, \’ \’) 1)FROM customersWHERE SUBSTR (cust_name, INSTR (cust_name, \’ \’) 1)LIKE INITCAP (\’MC%\’);
B. SELECT SUBSTR (cust_name, INSTR (cust_name, \’ \’) 1)FROM customersWHERE INITCAP (SUBSTR(cust_name, INSTR (cust_name, \’ \’) 1)) =\’Mc\’;
C. SELECT SUBSTR (cust_name, INSTR (cust_name, \’ \’) 1)FROM customersWHERE INITCAP (SUBSTR(cust_name, INSTR (cust_name, \’ \’) 1))LIKE \’Mc%\’;
D. SELECT SUBSTR (cust_name, INSTR (cust_name, \’ \’) 1)FROM customersWHERE INITCAP (SUBSTR(cust_name, INSTR (cust_name, \’ \’) 1)) =INITCAP \’MC%\’;
Correct Answer: C
Question 8:
Examine the structure of the EMPLOYEES table. NameNull-Type
EMPLOYEE_IDNOT NULLNUMBER(6)
FIRST_NAMEVARCHAR2(20)
LAST_NAMENOT NULLVARCHAR2(25)
EMAILNOT NULLVARCHAR2(25)
PHONE NUMBERVARCHAR2(20)
HIRE_DATENOT NULLDATE
JOB_IDNOT NULLVARCHAR2(10)
SALARYNUMBER(8,2)
COMMISSION_PCTNUMBER(2,2)
MANAGER_IDNUMBER(6)
DEPARTMENT_IDNUMBER(4)
There is a parent/child relationship between EMPLOYEE_ID and MANAGER_ID.
You want to display the last names and manager IDs of employees who work for the same manager as the employee whose EMPLOYEE_ID is 123.
Which query provides the correct output?
A. SELECT e.last_name, m.manager_idFROM employees e RIGHT OUTER JOIN employees mon (e.manager_id = m.employee_id)AND e.employee_id = 123;
B. SELECT e.last_name, m.manager_idFROM employees e RIGHT OUTER JOIN employees mon (e.employee_id = m.manager_id)WHERE e.employee_id = 123;
C. SELECT e.last_name, e.manager_idFROM employees e RIGHT OUTER JOIN employees mon (e.employee_id = m.employee_id)WHERE e.employee_id = 123;
D. SELECT m.last_name, e.manager_idFROM employees e LEFT OUTER JOIN employees mon (e.manager_id = m.manager_id)WHERE e.employee_id = 123;
Correct Answer: B
Question 9:
Examine the commands used to create DEPARTMENT_DETAILS and COURSE_DETAILS:
You want to generate a report that shows all course IDs irrespective of whether they have corresponding department IDs or not but no department IDs if they do not have any courses. Which SQL statement must you use?
A. SELECT course_id, department_id, FROM department_details d RIGHT OUTER JOIN course_details c USING (department_id)
B. SELECT c.course_id, d.department_id FROM course_details c RIGHT OUTER JOIN .department_details d ON (c.depatrment_id=d.department_id)
C. SELECT c.course_id, d.department_id FROM course_details c FULL OUTER JOIN department_details d ON (c.department_id=d. department_id)
D. SELECT c.course_id, d.department_id FROM course_details c FULL OUTER JOIN department_details d ON (c.department_idd. department_id)
Correct Answer: C
Question 10:
Examine the commands used to create the DEPARTMENT_DETAILS and the COURSE-DETAILS tables:
SQL> CREATE TABLE DEPARTMfiNT_D AILS
DEPARTMENT_ID NUMBER PRIMARY KEY ,
DEPARTMEHT_NAME VARCHAR2(50) ,
HOD VARCHAP2(50));
SQL> CREATE TABLE COURSE-DETAILS
(COURSE ID NUMBER PRIMARY KEY ,
COURS_NAME VARCHAR2 (50) ,
DEPARTMEHT_ID NUMBER REFERENCES DEPARTMENT_DETAIL
You want to generate a list of all department IDs along with any course IDs that may have been assigned to them.
Which SQL statement must you use?
A. SELECT d.departranc_id, c.cours玙id FROM cource_deatils c LEFT OUTER JOIN departmnt_details d ON (c.dapartmsnt_id=d.departtnent_id);
B. SELECT d.department_id, c. course_id FROM dapartment_details d RIGHT OUTER JOIN course_dotails c ON (c.depattnient_id=d.department_id) ;
C. SELECT d.department id. ccours_id FROM department_details d RIGHT OUTER JOIN course_details c ON (d.department_id);
D. SELECT d.department_id, c.course_id FROM department_details d LEFT OUTER JOIN course_details c ON (d.department___id).- (DEPARTMENT_ID) ;
Correct Answer: D
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