Online Access Free Oracle.1z0-071.v2025-09-16.q232 Practice Test (Page 26)

1z0-071 Exam Question 121

Examine the data in the CUST_NAME column of the CUSTOMERS table.

You want to extract only those customer names that have three names and display the * symbol in place of the first name as follows:

Which two queries give the required output?

A.SELECT LPAD(SUBSTR(cust_name, INSTR(cust_name, ' ')),LENGTH(cust_name),'*') "CUST NAME"FROM customersWHERE INSTR(cust_name, ' ',1,2)<>0;

B.SELECT LPAD(SUBSTR(cust_name, INSTR(cust_name, ' ')),LENGTH(cust_name),'*') "CUST NAME"FROM customersWHERE INSTR(cust_name, ' ',-1,2)<>0;

C.SELECT LPAD(SUBSTR(cust_name, INSTR (cust_name ' ')),LENGTH(cust_name) - INSTR(cust_name, ' '), '*') "CUST NAME"FROM customersWHERE INSTR(cust_name, ' ',1,2)<>0;

D.SELECT LPAD(SUBSTR(cust_name, INSTR (cust_name ' ')),LENGTH(cust_name) - INSTR(cust_name, ' '), '*') "CUST NAME"FROM customersWHERE INSTR(cust_name, ' ',1,-2)<>0;

1z0-071 Exam Question 122

View the exhibit and examine the description of the PRODUCT_INFORMATION table.
Which SQL statement would retrieve from the table the number of products having LIST_PRICE as NULL?

A.SELECT COUNT (list_price)FROM product_informationWHERE list_price i= NULL

B.SELECT COUNT (DISTINCT list_price)FROM product_informationWHERE list_price is NULL

C.SELECT COUNT (NVL(list_price, 0))FROM product_informationWHERE list_price is NULL

D.SELECT COUNT (list_price)FROM product_informationWHERE list_price is NULL

1z0-071 Exam Question 123

Which two statement are true regarding table joins available in the Oracle Database server? (Choose two.)

A.You can use the USING clause to join tables on more than one column.

B.You can explicitly provide the join condition with a NATURAL JOIN.

C.You can use the ON clause to specify multiple conditions while joining tables.

D.You can use the JOIN clause to join only two tables.

1z0-071 Exam Question 124

View the Exhibit and examine the structure of the EMPLOYEES table.
You want to display all employees and their managers having 100 as the MANAGER_ID.
You want the output in two columns: the first column would have the LAST_NAME of the managers and the second column would have LAST_NAME of the employees.
Which SQL statement would you execute?

A.SELECT m.last_name "Manager", e.last_name "Employee"
FROM employees m JOIN employees e
ON m.employee_id = e.manager_id
WHERE m.manager_id=100;

B.SELECT m.last_name "Manager", e.last_name "Employee"
FROM employees m JOIN employees e
WHERE m.employee_id = e.manager_id AND e.managerjd=100;

C.SELECT m.last_name "Manager", e.last_name "Employee"
FROM employees m JOIN employees e
ON m.employee_id = e.manager_id
WHERE e.managerjd=100;

D.SELECT m.last_name "Manager", e.last_name "Employee"
FROM employees m JOIN employees e
ON e.employee_id = m.manager_id
WHERE m.manager_id=100;

1z0-071 Exam Question 125

View the exhibit and examine the structure 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_vuAS SELECT o.order_id, o.order_date,
COUNT(i.line_item_id)FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id,o.order_date;

B.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 iON (o.order_id = i.order_id)GROUP BY o.order_id, o.order_date;

C.CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date,
COUNT(i.line_item_id)||'NO OF ITEMS'FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id,o.order_dateWHITH CHECK OPTION;

D.CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date,
COUNT(i.line_item_id)"NO OF ITEMS"FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id,o.order_date;

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1z0-071 Exam Question 121

1z0-071 Exam Question 122

1z0-071 Exam Question 123

1z0-071 Exam Question 124

1z0-071 Exam Question 125

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