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Preparing for the Oracle 1z1-071 certification exam requires a strong understanding of SQL concepts and principles. Candidates should have experience working with Oracle databases and be familiar with SQL syntax and commands. There are many resources available to help candidates prepare for the exam, including study guides, practice exams, and training courses.
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- Systems Administrator
- Database Architect
- Senior Database Administrator
- Systems Engineer
- Database Administrator (DBA)
- Database Developer
- Data Scientist
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NEW QUESTION # 144
Examine the structure of the BOOKS_TRANSACTIONS table:
Examine the SQL statement:
Which statement is true about the outcome?
- A. It displays details for members who have borrowed before today's date with either RM as TRANSACTION_TYPE or MEMBER_ID as A101 and A102.
- B. It displays details only for members who have borrowed before today with RM as TRANSACTION_TYPE.
- C. It displays details for only members A101 and A102 who have borrowed before today with RM TRANSACTION_TYPE.
- D. It displays details for members who have borrowed before today with RM as TRANSACTION_TYPE and the details for members A101 or A102.
Answer: D
NEW QUESTION # 145
View the Exhibit and examine the structure in the EMPLOYEES tables.
Evaluate the following SQL statement:
SELECT employee_id, department_id
FROM employees
WHERE department_id= 50 ORDER BY department_id
UNION
SELECT employee_id, department_id
FROM employees
WHERE department_id= 90
UNION
SELECT employee_id, department_id
FROM employees
WHERE department_id= 10;
What would be the outcome of the above SQL statement?
- A. The statement would execute successfully and display all the rows in the ascending order of DEPARTMENT_ID.
- B. The statement would not execute because the ORDER BY clause should appear only at the end of the SQL statement, that is, in the last SELECT statement.
- C. The statement would not execute because the positional notation instead of the column name should be used with the ORDER BY clause.
- D. The statement would execute successfully but it will ignore the ORDER BY clause and display the rows in random order.
Answer: B
NEW QUESTION # 146
In the PROMOTIONS table, the PROMO_ BEGIN_DATE column is of data type and the default date format is DD-MON-RR Which two statements are true about expressions using PROMO_ BEGIN_DATE in a query?
- A. TODATE(PROMO BEGIN_DATE *5) will return a date
- B. PROMO_ BEGIN_DATE - 5 will return a date
- C. PROMO_ BEGIN_DATE - SYSDATE will return a number
- D. TONUMBER (PROMO BEGIN_DATE) - 5 will return a number
- E. PROMO_ BEGIN_DATE - SYSDATE will return an error
Answer: B,C
NEW QUESTION # 147
View the Exhibit and examine the structure of ORDERS and CUSTOMERS tables.
There is only one customer with the cust_last_name column having value Roberts. Which INSERT statement should be used to add a row into the ORDERS table for the customer whose CUST_LAST_NAME is Roberts and CREDIT_LIMIT is 600?
- A. INSERT INTO ordersVALUES(1, '10-mar-2007', 'direct',(SELECT customer_idFROM customersWHERE cust_last_name='Roberts' ANDcredit_limit=600), 1000);
- B. INSERT INTO orders (order_id, order_data, order_mode,(SELECT customer_idFROM customersWHERE cust_last_name='Roberts' ANDcredit_limit=600), order_total)VALUES(1,
'10-mar-2007', 'direct', &&customer_id, 1000); - C. INSERT INTO orders (order_id, order_data, order_mode,(SELECT customer_idFROM customersWHERE cust_last_name='Roberts' ANDcredit_limit=600), order_total)VALUES (1,
'10-mar-2007', 'direct', &customer_id, 1000). - D. INSERT INTO(SELECT o.order_id, o.order_date, o.order_mode, c.customer_id, o.order_totalFROM orders o, customers cWHERE o.customer_id = c.customer_idAND c.cust_last_name='Roberts' AND c.credit_limit=600 )VALUES (1, '10-mar-2007', 'direct', (SELECT customer_idFROM customersWHERE cust_last_name='Roberts' ANDcredit_limit=600), 1000);
Answer: A
NEW QUESTION # 148
View the Exhibit and examine the structure of the CUSTOMERS table.
Using the CUSTOMERS table, you must generate a report that displays a credit limit increase of 15% for all customers.
Customers with no credit limit should have "Not Available" displayed.
Which SQL statement would produce the required result?
- A. SELECT NVL (cust_credit_limit*.15, 'Not Available') "NEW CREDIT" FROM customers
- B. SELECT NVL (cust_credit_limit, 'Not Available')*.15 "NEW CREDIT" FROM customers
- C. SELECT NVL (TO_CHAR(cust_credit_limit*.15), 'Not Available') "NEW CREDIT" FROM customers
- D. SELECT TO_CHAR(NVL(cust_credit_limit*.15), 'Not Available')) "NEW CREDIT" FROM customers
Answer: A
NEW QUESTION # 149
View the exhibit and examine the structure in ORDERSand ORDER_ITEMStables.
You need to create a view that displays the ORDER_ID, ORDER_DATE, and the total number of items in each order.
Which CREATEVIEWstatement would create the views successfully?
- A. 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; - 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 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)
" 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; - 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
WHITH CHECK OPTION;
Answer: C
NEW QUESTION # 150
Examine the data in the COLORS table:
Examine the data in the BRICKS table:
Which two queries return all the rows from COLORS?
- A. CASE is a function and DECODE is not.
- B. AIl conditions evaluated using DECODE can also be evaluated using CASE.
- C. Both CASE and DECODE are functions.
- D. Neither CASE nor DECODE is a function.
- E. All conditions evaluated using CASE can also be evaluated using DECODE.
- F. DECODE is a function and CASE is not.
Answer: E,F
NEW QUESTION # 151
View the Exhibit and examine the structure of the ORDERS table.
The columns ORDER_MODE and ORDER TOTAL have the default values'direct "and respectively.
Which two INSERT statements are valid? (Choose two.)
- A. INSERT INTO orders (order_id, order_date, order mode,customer_id, order_total) VALUES (1, TO_DATE (NULL),'online',101, NULL) ;
- B. INSERT INTO orders VALUES (1, '09-mar-2007', 'online',' ',1000);
- C. INSERT INTO orders (order id, order_date, order mode, order_total)VALUES (1,'10-mar-2007','online', 1000)
- D. INSERT INTO (SELECT order_id, order date, customer_id FROM orders) VALUES (1, '09-mar-2007",101);
- E. INSERT INTO orders VALUES('09-mar-2007',DEFAULT,101, DEFALLT);
Answer: D,E
NEW QUESTION # 152
Examine the create table statements for the stores and sales tables.
SQL> CREATE TABLE stores(store_id NUMBER(4) CONSTRAINT store_id_pk PRIMARY KEY, store_name VARCHAR2(12), store_address VARCHAR2(20), start_date DATE); SQL> CREATE TABLE sales(sales_id NUMBER(4) CONSTRAINT sales_id_pk PRIMARY KEY, item_id NUMBER(4), quantity NUMBER(10), sales_date DATE, store_id NUMBER(4), CONSTRAINT store_id_fk FOREIGN KEY(store_id) REFERENCES stores(store_id)); You executed the following statement:
SQL> DELETE from stores
WHERE store_id=900;
The statement fails due to the integrity constraint error:
ORA-02292: integrity constraint (HR.STORE_ID_FK) violated
Which three options ensure that the statement will execute successfully?
- A. Use CASCADE keyword with DELETE statement.
- B. Create the foreign key in the SALES table on SALES_ID column with on DELETE CASCADE option.
- C. DELETE the rows with STORE_ID = 900 from the SALES table and then delete rows from STORES table.
- D. Disable the primary key in the STORES table.
- E. Disable the FOREIGN KEY in SALES table and then delete the rows.
Answer: B,C,E
Explanation:
Explanation/Reference:
NEW QUESTION # 153
View the exhibit and examine the structure of the PROMOTIONS table.
You have to generate a report that displays the promo name and start date for all promos that started after the last promo in the 'INTERNET' category.
Which query would give you the required output?
- A. SELECT promo_name, promo_begin_date FROM promotionsWHERE promo_begin_date > ALL (SELECT promo_begin_dateFROM promotionsWHERE promo_category = 'INTERNET');
- B. SELECT promo_name, promo_begin_date FROM promotionsWHERE promo_begin_date> ALL (SELECT MAX (promo_begin_date)FROM promotions) ANDpromo_category= 'INTERNET';
- C. SELECT promo_name, promo_begin_date FROM promotionsWHERE promo_begin_date IN (SELECT promo_begin_dateFROM promotionsWHERE promo_category= 'INTERNET');
- D. SELECT promo_name, promo_begin_date FROM promotionsWHERE promo_begin_date> ANY (SELECT promo_begin_dateFROM promotionsWHERE promo_category= 'INTERNET');
Answer: A
NEW QUESTION # 154
Examine the command:
What does ON DELETE CASCADE imply?
- A. When the BOOKStable is dropped, all the rows in the BOOK_TRANSACTIONStable are deleted but the table structure is retained.
- B. When a value in the BOOKS.BOOK_IDcolumn is deleted, the corresponding value is updated in the BOOKS_TRANSACTIONS.BOOK_IDcolumn.
- C. When the BOOKStable is dropped, the BOOK_TRANSACTIONStable is dropped.
- D. When a row in the BOOKStable is deleted, the rows in the BOOK_TRANSACTIONStable whose BOOK_ID matches that of the deleted row in the BOOKStable are also deleted.
Answer: D
Explanation:
Explanation/Reference:
NEW QUESTION # 155
Examine this statement:
SELECT 1 AS id, ' John' AS first name
FROM DUAL
UNION
SELECT 1 , ' John' AS name
FROM DUAL
ORDER BY 1;
What is returned upon execution?
- A. 0 rows
- B. an error
- C. 2 rows
- D. 1 row
Answer: D
NEW QUESTION # 156
Examine the commands used to create DEPARTMENT_DETAILSand COURSE_DETAILS:
SQL>CREATE TABLE DEPARTMENT_DETAILS
( DEPARTMENT_ID NUMBER PRIMARY KEY,
DEPARTMENT_NAME VARCHAR2(50),
HOD VARCHAR2(50));
SQL>CREATE TABLE COURSE_DETAILS
(COURSE_ID NUMBER PRIMARY KEY,
COURSE_NAME VARCHAR2(50),
DEPARTMENT_ID VARCHAR2(50));
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.department_id, c.course_id FROM course_details c LEFT OUTER JOIN department_details d ON (c.department_id=d. department_id);
- B. SELECT d.department_id, c.course_id FROM department_details d RIGHT OUTER JOIN course_details c ON (d.department_id=c. department_id);
- C. SELECT d.department_id, c.course_id FROM department_details d LEFT OUTER JOIN course_details c ON (d.department_id=c. department_id);
- D. SELECT d.department_id, c.course_id FROM department_details d RIGHT OUTER JOIN course_details c ON (c.department_id=d. department_id);
Answer: C
NEW QUESTION # 157
Which three statements are true about single-row functions? (Choose three.)
- A. They can be used only in the WHERE clause of a SELECT statement
- B. They can be nested to any level
- C. The argument can be a column name, variable, literal or an expression
- D. They return a single result row per table
- E. They can accept only one argument
- F. The data type returned can be different from the data type of the argument
Answer: C,F
NEW QUESTION # 158
Which two statements are true about Entity Relationships?
- A. A one-to-one relationship is always a self-referencing relationship
- B. A many-to-many relationship can be implemented only by using foreign keys
- C. A one-to-many relationship in one direction is a one-to-one relationship in the other direction
- D. A table name can be specified just once when selecting data from a table having a selfreferencing relationship
- E. A Relationship can be mandatory for both entities
Answer: B,E
NEW QUESTION # 159
You must create a table EMPLOYEES in which the values in the columns EMPLOYEES_ID and LOGIN_ID must be unique and not null.
Which two SQL statements would create the required table? (Choose two.) CREATE TABLE employees
- A. (employee_id NUMBER,
login_id NUMBER,
employee_name VARCHAR2(25),
hire_date DATE,
CONSTRAINT emp_id_pk PRIMARY KEY (employee_id, login_id));
CREATE TABLE employees - B. (employee_id NUMBER CONSTRAINT emp_id_pk PRIMARY KEY,
login_id NUMBER UNIQUE,
employee_name VARCHAR2(25),
hire_date DATE);
CREATE TABLE employees - C. (employee_id NUMBER,
login_id NUMBER,
employee_name VARCHAR2(100),
hire_date DATE,
CONSTRAINT emp_id_uk UNIQUE (employee_id, login_id);
CONSTRAINT emp_id_nn NOT NULL (employee_id, login_id));
CREATE TABLE employees - D. (employee_id NUMBER CONSTRAINT emp_id_nn NOT NULL,
login_id NUMBER CONSTRAINT login_id_nn NOT NULL,
employee_name VARCHAR2(100),
hire_date DATE,
CONSTRAINT emp_num_id_uk UNIQUE (employee_id, login_id)); - E. (employee_id NUMBER,
login_id NUMBER,
employee_name VARCHAR2(100),
hire_date DATE,
CONSTRAINT emp_id_uk UNIQUE (employee_id, login_id));
CREATE TABLE employees
Answer: A,D
NEW QUESTION # 160
Viev the Exhibit and examine the structure of the PRODUCT INFORMATION and INVENTORIEStables.
You have a requirement from the supplies department to give a list containing PRODUCT _ID,SUPPLIER ID, and QUANTITY_ON HAND for all the products where in QUANTITY ON HAND is lessthan five.
Which two SQL statements can accomplish the task? (Choose two)
- A. SELECT i.product id, i. quantity on hand, pi. supplier id
FROM product information pi JOIN inventories i
ON (pi.product id=i. product id)WHERE quantity on hand < 5; - B. SELECT i. product id, i. quantity .on hand, pi. supplier_id
FROM product_information pi JOIN inventories i
ON (pi. product. id=i. product id) AND quantity on hand < 5; - C. SELECT i. product_id, i. quantity_on hand, pi. supplier id
FROM product information pi JOIN inventories i USING (product id) AND quantity .on hand < 5; - D. SELECT product id, quantity on hand, supplier id
FROM product information
NATURAL JOIN inventories AND quantity .on hand < 5;
Answer: A,B
Explanation:
B: This query correctly joins the two tables on PRODUCT_ID and filters rows where QUANTITY_ON_HAND is less than five. However, the syntax presented here might be incorrect due to improper placement of the condition within the ON clause; it should be in a WHERE clause.
D: This query correctly joins the two tables on PRODUCT_ID and applies the condition in a WHERE clause, which is the proper way to filter rows after performing the join.
Incorrect options:
A: The syntax uses NATURAL JOIN and AND incorrectly; also, conditions should be in a WHERE clause, not combined with the JOIN clause.
C: Similar to A, this option incorrectly places a condition directly in the JOIN clause without using a WHERE clause.
Explanation:
Given the requirement to list PRODUCT_ID, SUPPLIER_ID, and QUANTITY_ON_HAND for products with QUANTITY_ON_HAND less than five:
NEW QUESTION # 161
See the Exhibit and examine the structure of the PROMOTIONS table:
Using the PROMOTIONS table,
you need to find out the average cost for all promos in the range $0-2000 and $2000-5000 in category A.
You issue the following SQL statements:
What would be the outcome?
- A. It generates an error because CASE cannot be used with group functions.
- B. It generates an error because multiple conditions cannot be specified for the WHEN clause.
- C. It generates an error because NULL cannot be specified as a return value.
- D. It executes successfully and gives the required result.
Answer: D
Explanation:
CASE Expression
Facilitates conditional inquiries by doing the work of an IF-THEN-ELSE statement:
CASE expr WHEN comparison_expr1 THEN return_expr1
[WHEN comparison_expr2 THEN return_expr2
WHEN comparison_exprn THEN return_exprn
ELSE else_expr]
END
NEW QUESTION # 162
......
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