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Title: [Dec-2023 Newly Released] 1z0-071 Dumps for Oracle PL/SQL Developer Certified Associate Certified [Q29-Q53]

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 [Dec-2023 Newly Released] 1z0-071 Dumps for Oracle PL/SQL Developer Certified Associate Certified
Updated Verified 1z0-071 dumps Q&As - 100% Pass


QUESTION 29View the Exhibit and examine the data in the PRODUCT_INFORMATION table.Which two tasks would require subqueries? (Choose two.)
 displaying all the products whose minimum list prices are more than average list price of products having the status orderable
 displaying the total number of products supplied by supplier 102071 and having product status OBSOLETE
 displaying the number of products whose list prices are more than the average list price
 displaying all supplier IDs whose average list price is more than 500
 displaying the minimum list price for each product status
QUESTION 30Examine the description of the CUSTOMERStable:You want to display details of all customers who reside in cities starting with the letter D followed by at least two characters.Which query can be used?
 SELECT * FROM customers WHERE city = ‘D_%’;
 SELECT * FROM customers WHERE city LIKE ‘D_’;
 SELECT * FROM customers WHERE city LIKE ‘D_%’;
 SELECT * FROM customers WHERE city = ‘%D_’;
QUESTION 31Which two statements are true about a full outer join? (Choose two.)
 It includes rows that are returned by an inner join.
 It returns only unmatched rows from both tables being joined.
 It includes rows that are returned by a Cartesian product.
 It returns matched and unmatched rows from both tables being joined.
 The Oracle join operator (+) must be used on both sides of the join condition in the WHEREclause.
Explanation/Reference: https://www.w3resource.com/oracle/joins/oracle-full-outer-join.phpQUESTION 32Which two statements are true about sequences crated in a single instance Oracle database?
 The numbers generated by an explicitly defined sequence can only be used to insert data in one table.
 DELETE <sequencename> would remove a sequence from the database.
 CURRVAL is used to refer to the most recent sequence number that has been generated for a particular sequence.
 When the MAXVALUE limit for a sequence is reached, it can be increased by using the ALTER SEQUENCE statement.
 When the database instance shuts down abnormally, sequence numbers that have been cached but not used are available again when the instance is restarted.
QUESTION 33Examine this partial command:Which two clauses are required for this command to execute successfully?
 the LOCATION clause
 the access driver TYPE clause
 the REJECT LIMIT clause
 the DEFAULT DIRECTORY clause
 the ACCESS PARAMETERS clause
QUESTION 34View the Exhibit and examine the details of the PRODUCT_INFORMATION table. (Choose two.)Evaluate this SQL statement:SELECT TO_CHAR (list_price, ‘$9,999’)From product_information;Which two statements are true regarding the output?
 A row whose LIST_PRICE column contains value 11235.90 would be displayed as #######.
 A row whose LIST_PRICE column contains value 1123.90 would be displayed as $1,123.
 A row whose LIST_PRICE column contains value 1123.90 would be displayed as $1,124.
 A row whose LIST_PRICE column contains value 11235.90 would be displayed as $1,123.
QUESTION 35Examine the commands used to create DEPARTMENT_DETAILSand 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?SELECT course_id, department_id, FROM department_details d RIGHT OUTER JOIN
 course_details c USING (department_id)SELECT c.course_id, d.department_id FROM course_details c RIGHT OUTER JOIN
 .department_details d ON (c.depatrment_id=d.department_id)SELECT c.course_id, d.department_id FROM course_details c FULL OUTER JOIN
 department_details d ON (c.department_id=d. department_id)SELECT c.course_id, d.department_id FROM course_details c FULL OUTER JOIN
 department_details d ON (c.department_id<>d. department_id)
QUESTION 36View the Exhibit and examine the structure of CUSTOMERS table.Using the CUSTOMERS table, you need to generate a report that shows an increase in the credit limit by 15% for all customers. Customers whose credit limit has not been entered should have the message “Not Available” displayed.Which SQL statement would produce the required result?
 SELECT NVL (TO CHAR(cust_credit_limit * .15), ‘Not Available’) “NEW CREDIT”FROM customers;
 SELECT TO_CHAR (NVL(cust_credit_limit * .15), ‘Not Available’) “NEW CREDIT”FROM customers;
 SELECT NVL(cust_credit_limit * .15), ‘Not Available’) “NEW CREDIT”FROM customers;
 SELECT NVL(cust_credit_limit), ‘Not Available’) “NEW CREDIT”FROM customers;
QUESTION 37View the Exhibit and examine the structure of the PROMOTIONS table.Evaluate the following SQL statement:Which statement is true regarding the outcome of the above query?
 It produces an error because subqueries cannot be used with the CASE expression.
 It shows COST_REMARK for all the promos in the promo category ‘TV’.
 It shows COST_REMARK for all the promos in the table.
 It produces an error because the subquery gives an error.
QUESTION 38You 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
 (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
 (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
 (employee_id NUMBER CONSTRAINT emp_id_pk PRIMARY KEY,login_id NUMBER UNIQUE,employee_name VARCHAR2(25),hire_date DATE);CREATE TABLE employees
 (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
 (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));
QUESTION 39View the exhibit and examine the description of the EMPLOYEES table. (Choose two.)You executed this SQL statement:SELECT first_name, department_id, salaryFROM employeesORDER BY department_id, first_name, salary desc;Which two statements are true regarding the result? (Choose two.)
 The values in the SALARY column would be returned in descending order for all employees having the same value in the DEPARTMENT_ID and FIRST_NAME column.
 The values in the FIRST_NAME column would be returned in ascending order for all employees having the same value in the DEPARTMENT_ID column.
 The values in the SALARY column would be returned in descending order for all employees having the same value in the DEPARTMENT_ID column.
 The values in all columns would be returned in descending order.
 The values in the FIRST_NAME column would be returned in descending order for all employees having the same value in the DEPARTMENT_ID column.
QUESTION 40Which two tasks can be performed by using Oracle SQL statements?
 changing the password for an existing database user
 connecting to a database instance
 querying data from tables across databases
 starting up a database instance
 executing operating system (OS) commands in a session
References:http://www.techonthenet.com/oracle/password.phphttps://docs.oracle.com/cd/B28359_01/server.111/b28324/tdpii_distdbs.htmQUESTION 41Evaluate the following SQL statement:SELECT product_name || ‘it’s not available for order’FROM product_informationWHERE product_status = ‘obsolete’;You received the following error while executing the above query:ERRORORA-01756: quoted string not properly terminatedWhat would you do to execute the query successfully?
 Remove the single quotation marks enclosing the character literal string in the SELECTclause
 Use the escape character to negate the single quotation mark within the literal character string in the SELECTclause
 Enclose the character literal string in the SELECT clause within double quotation marks
 Use the Oracle (q) operator and delimiter to allow the use of a single quotation mark within the literal character string in the SELECTclause
Explanation/Reference:References:http://docs.oracle.com/cd/B19306_01/server.102/b14200/sql_elements003.htmQUESTION 42Examine the following query:SQL> SELECT prod_id, amount_soldFROM salesORDER BY amount_soldFETCH FIRST 5 PERCENT ROWS ONLY;What is the output of this query?
 It displays 5 percent of the products with the highest amount sold.
 It displays the first 5 percent of the rows from the SALEStable.
 It displays 5 percent of the products with the lowest amount sold.
 It results in an error because the ORDER BYclause should be the last clause.
ExplanationExplanation/Reference:References:https://oracle-base.com/articles/12c/row-limiting-clause-for-top-n-queries-12cr1QUESTION 43You must create a SALES table with these column specifications and data types: (Choose the best answer.) SALESID: Number STOREID: Number ITEMID: Number QTY: Number, should be set to 1 when no value is specified SLSDATE: Date, should be set to current date when no value is specified PAYMENT: Characters up to 30 characters, should be set to CASH when no value is specified Which statement would create the table?
 CREATE TABLE Sales(SALESID NUMBER (4),STOREID NUMBER (4),ITEMID NUMBER (4),QTY NUMBER DEFAULT = 1,SLSDATE DATE DEFAULT SYSDATE,PAYMENT VARCHAR2(30) DEFAULT = “CASH”);
 CREATE TABLE Sales(SALESID NUMBER (4),STOREID NUMBER (4),ITEMID NUMBER (4),QTY NUMBER DEFAULT = 1,SLSDATE DATE DEFAULT ‘SYSDATE’,PAYMENT VARCHAR2(30) DEFAULT CASH);
 CREATE TABLE Sales(SALESID NUMBER (4),STOREID NUMBER (4),ITEMID NUMBER (4),qty NUMBER DEFAULT = 1,SLSDATE DATE DEFAULT SYSDATE,PAYMENT VARCHAR2(30) DEFAULT = “CASH”);
 Create Table sales(salesid NUMBER (4),Storeid NUMBER (4),Itemid NUMBER (4),QTY NUMBER DEFAULT 1,Slsdate DATE DEFAULT SYSDATE,payment VARCHAR2(30) DEFAULT ‘CASH’);
QUESTION 44Which two statements execute successfully?
 SELECT TO_ DATE(‘2019-DEC-25 15:30’, ‘YYYY-MON-DD HH24:MI’, ‘NLS_ DATE_ LANGUAGE=AMERICAN’ ) FROM DUAL;
 SELECT TO_CHAR(‘2019-DEC-25 15:30″, YY-MON-D HH24:M2’, ‘NLS_DATE LANGUAGE =AMERICAN’)FROM DUAL;
 SELECT TO _DATE (TO_ CHAR (‘2019-DEC-25 03:30’, ‘YYYY-MON-DD HH12:MI’))FROM DUAL;
 SELECT TO _ CHAR (TO_ DATE (‘2019-DEC-25 03:30′,’YYYY-MON-DD HH12:MI’))FROM DUAL
 SELECT TO _ CHAR (‘2019-DEC-25 15:30′.’YYYY-MON-DD HH24:MI’)FROM DUAL
QUESTION 45Which three statements are true regarding the data types? (Choose three.)
 The minimum column width that can be specified for a VARCHAR2data type column is one.
 Only one LONGcolumn can be used per table.
 A TIMESTAMPdata type column stores only time values with fractional seconds.
 The BLOBdata type column is used to store binary data in an operating system file.
 The value for a CHARdata type column is blank-padded to the maximum defined column width.
QUESTION 46Examine the description of the EMPLOYEES table:Which query requires explicit data type conversion?
 SELECT join_date FROM employees WHERE join_date > ’10-02-2018′;
 SELECT salary + ‘120.50’ FROM employees;
 SELECT SUBSTR(join_date, 1, 2) – 10 FROM employees;
 SELECT join_date + ’20’ FROM employees;
 SELECT join_date || ‘ ‘ || salary FROM employees;
QUESTION 47Which three are key components of an Entity Relationship Model?
 a table
 an attribute
 a unique identifier
 an activity
 a relationship
 an entity
QUESTION 48Examine 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 NULLDATEJOB_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?
 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;
 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;
 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;
 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;
QUESTION 49These are the steps for a correlated subquery, listed in random order:1. The WHEREclause of the outer query is evaluated.2. A candidate row is fetched from the table specified in the outer query.3. This is repeated for the subsequent rows of the table, until 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.Which is the correct sequence in which the Oracle server evaluates a correlated subquery?
 2, 1, 4, 3
 4, 1, 2, 3
 4, 2, 1, 3
 2, 4, 1, 3
Explanation/Reference:References:http://rajanimohanty.blogspot.co.uk/2014/01/correlated-subquery.htmlQUESTION 50Examine the structure of the EMPLOYEEStable.You must display the details of employees who have manager with MANAGER_ID 100, who were hired in the past 6 months and who have salaries greater than 10000.Which query would retrieve the required result?SELECT last_name, hire_date, salary
 FROM employeesWHERE salary > 10000UNION ALLSELECT last_name, hire_date, salaryFROM employeesWHERE manager_ID = (SELECT employee_id FROM employees WHERE employee_id = 100) INTERSECT SELECT last_name, hire_date, salary FROM employees WHERE hire_date > SYSDATE-180; SELECT last_name, hire_date, salary
 FROM employeesWHERE manager_id = (SELECT employee_id FROM employees WHERE employee_id = 100) UNION ALL (SELECT last_name, hire_date, salary FROM employees WHERE hire_date > SYSDATE-180 INTERSECT SELECT last_name, hire_date, salary FROM employees WHERE salary > 10000); SELECT last_name, hire_date, salary
 FROM employeesWHERE manager_id = (SELECT employee_id FROM employees WHERE employee_id‘100’)UNIONSELECT last_name, hire_date, salaryFROM employeesWHERE hire_date > SYSDATE-180INTERSECTSELECT last_name, hire_date, salaryFROM employeesWHERE salary > 10000;(SELECT last_name, hire_date, salary
 FROM employeesWHERE salary > 10000UNION ALLSELECT last_name, hire_date, salaryFROM employeesWHERE manager_ID = (SELECT employee_id FROM employees WHERE employee_id100))UNIONSELECT last_name, hire_date, salaryFROM employeesWHERE hire_date > SYSDATE-180;
QUESTION 51Which three statements are true about sequences in a single instance Oracle database? (Choose three.)
 A sequence can issue duplicate values
 A sequence’s unallocated cached value are lost if the instance shuts down
 Sequences can always have gaps
 Two or more tables cannot have keys generated from the same sequence
 A sequence can only be dropped by a DBA
 A sequence number that was allocated can be rolled back if a transaction fails
QUESTION 52Which statement is true about transactions?
 A set of Data Manipulation Language (DML) statements executed in a sequence ending with a SAVEPOINT forms a single transaction.
 Each Data Definition Language (DDL) statement executed forms a single transaction.
 A set of DDL statements executed in a sequence ending with a COMMIT forms a single transaction.
 A combination of DDL and DML statements executed in a sequence ending with a COMMIT forms a single transaction.
QUESTION 53View the exhibit and examine the ORDERStable.The ORDERStable contains data and all orders have been assigned a customer ID. Which statement would add a NOTNULLconstraint to the CUSTOMER_IDcolumn?
 ALTER TABLE ordersMODIFY CONSTRAINT orders_cust_id_nn NOT NULL (customer_id);
 ALTER TABLE ordersADD CONSTRAINT orders_cust_id_nn NOT NULL (customer_id);
 ALTER TABLE ordersMODIFY customer_id CONSTRAINT orders_cust_nn NOT NULL (customer_id);
 ALTER TABLE ordersADD customer_id NUMBER(6)CONSTRAINT orders_cust_id_nn NOT NULL;
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Oracle 1z1-071 certification exam is an excellent opportunity for professionals who want to validate their knowledge and skills in SQL programming language. Oracle Database SQL certification exam is globally recognized and highly valued by employers in the IT industry. By passing 1z0-071 exam, candidates can enhance their career prospects and can increase their chances of getting hired for high-paying jobs in the database administration and development field.

Oracle 1z0-071 certification exam is designed to test the skills and knowledge of individuals who want to become Oracle Database SQL Certified Professionals. 1z0-071 exam is intended for individuals who have a basic understanding of SQL, including data manipulation, schema creation, and database administration. 1z0-071 exam is a prerequisite for other Oracle certification exams, including the Oracle Database 12c Administrator Certified Associate and the Oracle Database 12c Administrator Certified Professional.
 

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