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Preparation Guide for Oracle 1Z0-071: Oracle Database SQL Exam
Introduction
Oracle is an American international information technology corporation located in California. Oracle is a multinational organisation that designs enterprise software systems. Oracle offers a range of cloud-based software and systems, as well as hardware and resources that support companies expand their operations. The emphasis of Oracle apps is on data analytics, market research, IT processes management, business resource planning, security and new technology. Oracle is well regarded for its hardware and software databases. The Oracle Database is one of the most common corporate database goods, a relational database management system (DBMS).
Oracle also designs and creates software and technology for middle-level database management systems, human capital management systems (HCMs), client relationship management systems (CRMs), company resource planning systems (ERPs) and supply chain management systems (SCM). Oracle Database, also referred to as Oracle DBMS, is a multi-model database management framework. It is an extensively used database to conduct database analysis and data retrieval functions. Different providers make Oracle Database usable on the server, premises or a hybrid software implementation. This guide gives the complete understanding of 1Z0-071 dumps and 1Z0-071 practice exams.
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The price of the Oracle 1Z0-071 Exam is $245 USD. For more information related to the Oracle 1Z0-071 Exam please visit Oracle website.
NEW QUESTION 69
Which statement is true about an inner join specified in the WHERE clause of a query?
- A. It must have primary-key and foreign key constraints defined on the columns used in the join condition.
- B. It is applicable for only equijoin conditions.
- C. It requires the column names to be the same in all tables used for the join conditions.
- D. It is applicable for equijoin and nonequijoin conditions.
Answer: D
NEW QUESTION 70
Examine the structure of the SALES table. (Choose two.)
Examine this statement:
Which two statements are true about the SALES1 table? (Choose two.)
- A. It has PRIMARY KEY and UNIQUE constraints on the selected columns which had those constraints in the SALES table.
- B. It will not be created because of the invalid WHERE clause.
- C. It will have NOT NULL constraints on the selected columns which had those constraints in the SALES table.
- D. It will not be created because the column-specified names in the SELECT and CREATE TABLE clauses do not match.
- E. It is created with no rows.
Answer: C,E
NEW QUESTION 71
Examine the description of the PROMOTIONStable:
You want to display the unique promotion costs in each promotion category.
Which two queries can be used? (Choose two.)
SELECT DISTINCT promo_category || ' has ' || promo_cost AS COSTS FROM
- A. ORDER BY 1;
SELECT promo_cost, promo_category FROM promotions ORDER BY 1; - B. promotions ORDER BY 1;
SELECT DISTINCT promo_category, promo_cost FROM promotions ORDER BY 1; - C. SELECT promo_category, DISTINCT promo_cost FROM promotions ORDER BY 2;
- D. SELECT DISTINCT promo_cost ||' in '|| DISTINCT promo_category FROM promotions
- E.
Answer: C,D
NEW QUESTION 72
Which three statements are true about sequences in a single instance Oracle database? (Choose three.)
- A. A sequence can issue duplicate values
- B. Sequences can always have gaps
- C. A sequence number that was allocated can be rolled back if a transaction fails
- D. A sequence can only be dropped by a DBA
- E. A sequence's unallocated cached value are lost if the instance shuts down
- F. Two or more tables cannot have keys generated from the same sequence
Answer: B,C,E
Explanation:
Explanation/Reference: https://docs.oracle.com/cd/B19306_01/server.102/b14200/statements_6015.htm
NEW QUESTION 73
View the Exhibit and examine the details of the PRODUCT_INFORMATIONtable.
Exhibit
You must display PRODUCT_NAME from the table where the CATEGORY_IDcolumn has values 12or 13, and the SUPPLIER_IDcolumn has the value 102088.
You executed this SQL statement:
Which statement is true regarding the execution?
- A. It would not execute because the entire WHEREclause is not enclosed within parentheses.
- B. It would execute and return the desired result.
- C. It would execute but would return no rows.
- D. It would not execute because the same column has been used twice with the ANDlogical operator.
Answer: C
NEW QUESTION 74
Which statement is true regarding the default behavior of the ORDER BY clause?
- A. Numeric values are displayed in descending order if they have decimal positions.
- B. NULLs are not included in the sort operation.
- C. Only columns that are specified in the SELECT list can be used in the ORDER BY clause.
- D. In a character sort, the values are case-sensitive.
Answer: D
NEW QUESTION 75
The following are the steps for a correlated subquery, listed in random order:
1. The WHEREclause of the outer query is evaluated.
2. The candidate row is fetched from the table specified in the outer query.
3. The procedure is repeated for the subsequent rows of the table, till all the rows are processed.
4. Rows are returned by the inner query, after being evaluated with the value from the candidate row in the outer query.
Identify the option that contains the steps in the correct sequence in which the Oracle server evaluates a correlated subquery.
- A. 2, 1, 4, 3
- B. 4, 1, 2, 3
- C. 2, 4, 1, 3
- D. 4, 2, 1, 3
Answer: C
Explanation:
Explanation/Reference:
References:
http://rajanimohanty.blogspot.co.uk/2014/01/correlated-subquery.html
NEW QUESTION 76
Which two are true about the WITH GRANT OPTION clause?
- A. The grantee can grant the object privilege to any user in the database, with of without including this option.
- B. The grantee must have the GRANT ANY OBJECT PRIVILEGE system prvilege to use this option.
- C. It can be used when granting privileges to roles.
- D. It can be used to pass on privileges to other users by the grantee.
- E. It can be used for system and object privileges.
- F. It cannot be used to pass on privileges to PUBLIC by the grantee.
Answer: A,D
NEW QUESTION 77
Examine this statement:
SELECT cust_id, cust^last_ndma "Last Name"
FROM customers
WHERE country_id - 10 UNION
SELECT cust_ID_CUST_NO, cust_last_name
PROM customers
WHERE country__id = 30
Identify three order by clauses, any one of which will complete the query successfully.
- A. ORDER BY "CUST^NO
- B. ORDER BY CUST_NO
- C. ORDER BY 2,
- D. ORDER BY 2. cust _id
- E. ORDER BY "Last Name
Answer: C,D,E
NEW QUESTION 78
Which three actions can you perform only with system privileges?
- A. Access flat files via a database, which are stored in an operating system directory.
- B. Use the WITH GRANT OPTION clause.
- C. Execute a procedure in another schema.
- D. Truncate a table in another schema.
- E. Log in to a database.
- F. Query any table in a database.
Answer: B,E,F
NEW QUESTION 79
Examine the data in the ENPLOYEES table:
Which statement will compute the total annual compensation tor each employee?
- A. SECECT last_namo, (menthy_salary + monthly_commission_pct) * 12 AS annual_comp
FROM employees; - B. SELCECT last_namo, (monthly_salary * 12) + (monthly_commission_pct * 12) AS annual_comp
FROM employees - C. SELCECT last_namo, (monthly_salary * 12) + (menthy_salary * 12 * NVL
(monthly_commission_pct, 0)) AS annual_comp FROM employees - D. SELCECT last_namo, (monthly_salary * 12) + (menthy_salary * 12 * monthly_commission_pct)
AS annual_comp FROM employees
Answer: C
NEW QUESTION 80
View the Exhibit and examine the structure of the CUSTOMERS and CUST_HISTORY tables.
The CUSTOMERS table contains the current location of all currently active customers.
The CUST_HISTORY table stores historical details relating to any changes in the location of all current as well as previous customers who are no longer active with the company.
You need to find those customers who have never changed their address.
Which SET operator would you use to get the required output?
- A. INTERSECT
- B. MINUS
- C. UNION ALL
- D. UNION
Answer: B
NEW QUESTION 81
Which two are true about a SQL statement using SEToperators such as UNION? (Choose two.)
- A. The data type of each column returned by the second query must be implicitly convertible to the data type of the corresponding column returned by the first query.
- B. The names and number of columns must be identical for all SELECT statements in the query.
- C. The number, but not names, of columns must be identical for all SELECT statements in the query.
- D. The data type group of each column returned by the second query must match the data type group of the corresponding column returned by the first query.
- E. The data type of each column returned by the second query must exactly match the data type of the corresponding column returned by the first query.
Answer: D,E
NEW QUESTION 82
Examine the structure of the PROGRAMS table:
Which two SQL statements would execute successfully?
- A. SELECT NVL (ADD_MONTHS (END_DATE,1) SYSDATE) FROM programs;
- B. SELECT NVL (MONTHS_BETWEEN (start_date, end_date), 'Ongoing') FROM
programs; - C. SELECT NVL (TO_CHAR (MONTHS_BETWEEN (start-date, end_date)), 'Ongoing') FROM programs
- D. SELECT TO_DATE (NVL (SYSDATE-END_DATE, SYSDATE)) FROM programs;
Answer: A,C
NEW QUESTION 83
View the exhibit and examine the structure of ORDERS and CUSTOMERS tables.
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? Assume there exists only one row with CUST_LAST_NAME as Roberts and CREDIT_LIMIT as 600.
- A. 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_id AND c.cust_last_name='Roberts' AND c.credit_limit=600)VALUES (1,'10-mar-2007', 'direct', (SELECT customer_idFROM customersWHERE cust_last_name='Roberts' AND credit_limit=600), 1000);
- B. INSERT INTO orders (order_id, order_date, order_mode,(SELECT customer_idFROM customersWHERE cust_last_name='Roberts' AND credit_limit=600), order_total)VALUES (1,'10-mar-2007', 'direct', &&customer_id, 1000);
- C. INSERT INTO ordersVALUES (1,'10-mar-2007', 'direct',(SELECT customer_idFROM customersWHERE cust_last_name='Roberts' AND credit_limit=600), 1000);
- D. INSERT INTO orders (order_id, order_date, order_mode,(SELECT customer_idFROM customersWHERE cust_last_name='Roberts' AND credit_limit=600), order_total)VALUES (1,'10-mar-2007', 'direct', &customer_id, 1000);
Answer: C
NEW QUESTION 84
Which three actions can you perform only with system privileges?
- A. Use the WITH GRANT OPTION clause.
- B. Access flat files via a database, which are stored in an operating system directory.
- C. Execute a procedure in another schema.
- D. Truncate a table in another schema.
- E. Log in to a database.
- F. Query any table in a database.
Answer: C,E,F
NEW QUESTION 85
You are designing the structure of a table in which two columns have the specifications:
COMPONENT_ID- must be able to contain a maximum of 12 alphanumeric characters and uniquely identify the row EXECUTION_DATETIME- contains Century, Year, Month, Day, Hour, Minute, Second to the maximum precision and is used for calculations and comparisons between components.
Which two options define the data types that satisfy these requirements most efficiently?
- A. The COMPONENT_IDmust be of VARCHAR2 data type.
- B. The COMPONENT_IDcolumn must be of CHAR data type.
- C. The EXECUTION_DATETIME must be of DATE data type.
- D. The EXECUTION_DATETIME must be of TIMESTAMP data type.
- E. The EXECUTION_DATETIME must be of INTERVAL DAY TO SECOND data type.
- F. The COMPONENT_IDmust be of ROWID data type.
Answer: B,C
NEW QUESTION 86
Examine the data in the CUSTOMERStable:
You want to list all cities that have more than one customer along with the customer details.
Evaluate the following query:
Which two JOIN options can be used in the blank in the above query to give the correct output? (Choose two.) LEFT OUTER JOIN
- A. JOIN
- B. NATURAL JOIN
- C. FULL OUTER JOIN
- D. RIGHT OUTER JOIN
Answer: B,C
NEW QUESTION 87
Which two are true about the WITH GRANT OPTION clause?
- A. It can be used when granting privileges to roles.
- B. The grantee can grant the object privilege to any user in the database, with or without including this option.
- C. It can be used to pass on privileges to other users by the grantee.
- D. The grantee must have the GRANT ANY OBJECT PRIVILEGE system privilege to use this option.
- E. It can be used for system and object privileges.
- F. It cannot be used to pass on privileges to PUBLIC by the grantee.
Answer: C,E
NEW QUESTION 88
SCOTT is a user in the database.
Evaluate the commands issued by the DBA:
Which statement is true regarding the execution of the above commands?
- A. Statement 3 would not execute because role and system privileges cannot be granted together in a single GRANT statement.
- B. Statement 2 would not execute because system privileges and object privileges cannot be granted together in a single GRANT command.
- C. Statement 1 would not execute because the IDENTIFIED BY <password> clause is missing.
- D. Statement 1 would not execute because the WITH GRANT option is missing.
Answer: B
NEW QUESTION 89
View the Exhibit and examine the structure of the ORDERS table.
Which UPDATE statement is valid?
- A. UPDATE ordersSET order_date = '12-mar-2007',AND order_total
TO_NUMBER(NULL)WHERE order_id = 2455; - B. UPDATE ordersSET order_date = '12-mar-2007',order_total = NULLWHERE order_id
2455; - C. UPDATE ordersSET order_date = TO_DATE('12-mar-2007','dd-mon-yyyy'),SET order_total = TO_NUMBER (NULL)WHERE order_id = 2455;
- D. UPDATE ordersSET order_date = '12-mar-2007',order_total IS NULLWHERE order_id
= 2455;
Answer: B
NEW QUESTION 90
You create a table named 123.
Which statement runs successfully?
- A. SELECT * FROM '123';
- B. SELECT * FROM TABLE (123) ;
- C. SELECT * FROM V'123V';
- D. SELECT * FROM "123";
Answer: D
NEW QUESTION 91
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 TO_CHAR(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 NVL(cust_credit_limit*.15,'Not Available') "NEW CREDIT"FROM customers;
Answer: C
NEW QUESTION 92
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