1Z0-051 | Oracle 1Z0-051 Exam Dumps 2021

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Free demo questions for Oracle 1Z0-051 Exam Dumps Below:

NEW QUESTION 1
View the Exhibit and examine the structure of the PRODUCTS, SALES, and SALE_SUMMARY tables.
SALE_VW is a view created using the following command:
SQL>CREATE VIEW sale_vw AS
SELECT prod_id, SUM(quantity_sold) QTY_SOLD
FROM sales GROUP BY prod_id;
You issue the following command to add a row to the SALE_SUMMARY table:
SQL>INSERT INTO sale_summary
SELECT prod_id, prod_name, qty_sold FROM sale_vw JOIN products
USING (prod_id) WHERE prod_id = 16;
What is the outcome?

  • A. It executes successfull
  • B. It gives an error because a complex view cannot be used to add data into the SALE_SUMMARY tabl
  • C. It gives an error because the column names in the subquery and the SALE_SUMMARY table do not matc
  • D. It gives an error because the number of columns to be inserted does not match with the number of columns in the SALE_SUMMARY tabl

Answer: D

NEW QUESTION 2
Which statements are true regarding the FOR UPDATE clause in a SELECT statement? (Choose all that apply.)

  • A. It locks only the columns specified in the SELECT lis
  • B. It locks the rows that satisfy the condition in the SELECT statemen
  • C. It can be used only in SELECT statements that are based on a single tabl
  • D. It can be used in SELECT statements that are based on a single or multiple table
  • E. After it is enforced by a SELECT statement, no other query can access the same rows until a COMMIT or ROLLBACK is issue

Answer: BD

Explanation:
FOR UPDATE Clause in a SELECT Statement Locks the rows in the EMPLOYEES table where job_id is SA_REP. Lock is released only when you issue a ROLLBACK or a COMMIT. If the SELECT statement attempts to lock a row that is locked by another user, the database waits until the row is available, and then returns the results of the SELECT statement. FOR UPDATE Clause in a SELECT Statement When you issue a SELECT statement against the database to query some records, no locks are placed on the selected rows. In general, this is required because the number of records locked at any given time is (by default) kept to the absolute minimum: only those records that have been changed but not yet committed are locked. Even then, others will be able to read those records as they appeared before the change (the “before image” of the data). There are times, however, when you may want to lock a set of records even before you change them in your program. Oracle offers the FOR UPDATE clause of the SELECT statement to perform this locking. When you issue a SELECT...FOR UPDATE statement, the relational database management system (RDBMS) automatically obtains exclusive row-level locks on all the rows identified by the SELECT statement, thereby holding the records “for your changes only.” No one else will be able to change any of these records until you perform a ROLLBACK or a COMMIT. You can append the optional keyword NOWAIT to the FOR UPDATE clause to tell the Oracle server not to wait if the table has been locked by another user. In this case, control will be returned immediately to your program or to your SQL Developer environment so that you can perform other work, or simply wait for a period of time before trying again. Without the NOWAIT clause, your process will block until the table is available, when the locks are released by the other user through the issue of a COMMIT or a ROLLBACK command.

NEW QUESTION 3
Which two statements are true regarding views? (Choose two.)

  • A. A sub query that defines a view cannot include the GROUP BY clause
  • B. A view is created with the sub query having the DISTINCT keyword can be updated
  • C. A Data Manipulation Language (DML) operation can be performed on a view that is created with the sub query having all the NOT NULL columns of a table
  • D. A view that is created with the sub query having the pseudo column ROWNUM keyword cannot be updated

Answer: CD

Explanation:
Rules for Performing DML Operations on a View You cannot add data through a view if the view includes: Group functions A GROUP BY clause The DISTINCT keyword The pseudocolumn ROWNUM keyword Columns defined by expressions NOT NULL columns in the base tables that are not selected by the view

NEW QUESTION 4
Examine the structure of the INVOICE table: Exhibit:
1Z0-051 dumps exhibit
Which two SQL statements would execute successfully? (Choose two.)

  • A. SELECT inv_no,NVL2(inv_date,'Pending','Incomplete') FROM invoice;
  • B. SELECT inv_no,NVL2(inv_amt,inv_date,'Not Available') FROM invoice;
  • C. SELECT inv_no,NVL2(inv_date,sysdate-inv_date,sysdate) FROM invoice;
  • D. SELECT inv_no,NVL2(inv_amt,inv_amt*.25,'Not Available') FROM invoice;

Answer: AC

Explanation:
The NVL2 Function
The NVL2 function provides an enhancement to NVL but serves a very similar purpose. It evaluates whether a column or expression of any data type is null or not. 5-6 The NVL function If the first term is not null, the second parameter is returned, else the third parameter is returned. Recall that the NVL function is different since it returns the original term if it is not null. The NVL2 function takes three mandatory parameters. Its syntax is NVL2(original, ifnotnull, ifnull), where original represents the term being tested. Ifnotnull is returned if original is not null, and ifnull is returned if original is null. The data types of the ifnotnull and ifnull parameters must be compatible, and they cannot be of type LONG. They must either be of the same type, or it must be possible to convert ifnull to the type of the ifnotnull parameter. The data type returned by the NVL2 function is the same as that of the ifnotnull parameter.

NEW QUESTION 5
View the Exhibit and examine the structure of the PRODUCTS table.
You want to display only those product names with their list prices where the list price is at least double the minimum price. The report should start with the product name having the maximum list price satisfying this
condition.
Evaluate the following SQL statement:
SQL>SELECT prod_name,prod_list_price FROM products WHERE prod_list_price >= 2 * prod_min_price
Which ORDER BY clauses can be added to the above SQL statement to get the correct output?
(Choose all that apply.)
1Z0-051 dumps exhibit

  • A. ORDER BY prod_list_price DESC, prod_name;
  • B. ORDER BY (2*prod_min_price)DESC, prod_name;
  • C. ORDER BY prod_name, (2*prod_min_price)DESC;
  • D. ORDER BY prod_name DESC, prod_list_price DESC;
  • E. ORDER BY prod_list_price DESC, prod_name DESC;

Answer: AE

Explanation:
Using the ORDER BY Clause The order of rows that are returned in a query result is undefined. The ORDER BY clause can be used to sort the rows. However, if you use the ORDER BY clause, it must be the last clause of the SQL statement. Further, you can specify an expression, an alias, or a column position as the sort condition. Syntax SELECT expr FROM table [WHERE condition(s)] [ORDER BY {column, expr, numeric_position} [ASC|DESC]]; In the syntax: ORDER BY specifies the order in which the retrieved rows are displayed ASC orders the rows in ascending order (This is the default order.)
DESC orders the rows in descending order If the ORDER BY clause is not used, the sort order is undefined, and the Oracle server may not fetch rows in the same order for the same query twice. Use the ORDER BY clause to display the rows in a specific order. Note: Use the keywords NULLS FIRST or NULLS LAST to specify whether returned rows containing null values should appear first or last in the ordering sequence.

NEW QUESTION 6
View the Exhibit and examine the structure of the PROMOTIONS table. Evaluate the following SQL statement:
1Z0-051 dumps exhibit
The above query generates an error on execution.
Which clause in the above SQL statement causes the error?
1Z0-051 dumps exhibit

  • A. WHERE
  • B. SELECT
  • C. GROUP BY
  • D. ORDER BY

Answer: C

NEW QUESTION 7
The EMPLOYEES table contains these columns:
EMPLOYEE_ID NUMBER(4)
ENAME VARCHAR2 (25)
JOB_ID VARCHAR2(10)
Which SQL statement will return the ENAME, length of the ENAME, and the numeric position of the letter "a" in the ENAME column, for those employees whose ENAME ends with a the letter "n"?

  • A. SELECT ENAME, LENGTH(ENAME), INSTR(ENAME, 'a') FROM EMPLOYEES WHERE SUBSTR(ENAME, -1, 1) = 'n';
  • B. SELECT ENAME, LENGTH(ENAME), INSTR(ENAME, ,-1,1) FROM EMPLOYEES WHERE SUBSTR(ENAME, -1, 1) = 'n';
  • C. SELECT ENAME, LENGTH(ENAME), SUBSTR(ENAME, -1,1) FROM EMPLOYEES WHERE INSTR(ENAME, 1, 1) = 'n';
  • D. SELECT ENAME, LENGTH(ENAME), SUBSTR(ENAME, -1,1) FROM EMPLOYEES WHERE INSTR(ENAME, -1, 1) = 'n';

Answer: A

Explanation:
INSTR is a character function return the numeric position of a named string.
INSTR(NAMED,’a’)
Incorrect Answer:
BDid not return a numeric position for ‘a’.
CDid not return a numeric position for ‘a’.
DDid not return a numeric position for ‘a’.
Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 3-8

NEW QUESTION 8
View the Exhibit and examine the description for the CUSTOMERS table.
1Z0-051 dumps exhibit
You want to update the CUST_INCOME_LEVEL and CUST_CREDIT_LIMIT columns for the customer with the CUST_ID 2360. You want the value for the CUST_INCOME_LEVEL to have the same value as that of the customer with the CUST_ID 2560 and the CUST_CREDIT_LIMIT to have the same value as that of the customer with CUST_ID 2566.
Which UPDATE statement will accomplish the task?

  • A. UPDATE customers SET cust_income_level = (SELECT cust_income_level FROM customers WHERE cust_id = 2560), cust_credit_limit = (SELECT cust_credit_limit FROM customers WHERE cust_id = 2566) WHERE cust_id=2360;
  • B. UPDATE customers SET (cust_income_level,cust_credit_limit) = (SELECT cust_income_level, cust_credit_limit FROM customers WHERE cust_id=2560 OR cust_id=2566) WHERE cust_id=2360;
  • C. UPDATE customers SET (cust_income_level,cust_credit_limit) = (SELECT cust_income_level, cust_credit_limit FROM customers WHERE cust_id IN(2560, 2566) WHERE cust_id=2360;
  • D. UPDATE customers SET (cust_income_level,cust_credit_limit) = (SELECT cust_income_level, cust_credit_limit FROM customers WHERE cust_id=2560 AND cust_id=2566) WHERE cust_id=2360;

Answer: A

Explanation:
Updating Two Columns with a Subquery
You can update multiple columns in the SET clause of an UPDATE statement by writing
multiple subqueries. The syntax is as follows:
UPDATE table
SET column =
(SELECT column
FROM table
WHERE condition)
[ ,
column =
(SELECT column
FROM table
WHERE condition)]
[WHERE condition ] ;

NEW QUESTION 9
The CUSTOMERS table has these columns:
1Z0-051 dumps exhibit
The CUSTOMER_ID column is the primary key for the table.
You need to determine how dispersed your customer base is.
Which expression finds the number of different countries represented in the CUSTOMERS table?

  • A. COUNT(UPPER(country_address))
  • B. COUNT(DIFF(UPPER(country_address)))
  • C. COUNT(UNIQUE(UPPER(country_address)))
  • D. COUNT DISTINCT UPPER(country_address)
  • E. COUNT(DISTINCT (UPPER(country_address)))

Answer: E

NEW QUESTION 10
In which four clauses can a sub query be used? (Choose four.)

  • A. in the INTO clause of an INSERT statement
  • B. in the FROM clause of a SELECT statement
  • C. in the GROUP BY clause of a SELECT statement
  • D. in the WHERE clause of a SELECT statement
  • E. in the SET clause of an UPDATE statement
  • F. in the VALUES clause of an INSERT statement

Answer: ABDE

Explanation:
A: a sub query is valid on the INTO clause of an ISERT Statement
B: a sub query can be used in the FROM clause of a SELECT statement
D: a sub query can be used in the WHERE clause of a SELECT statement,
E: a sub query can be used in the SET clauses of an UPDATE statement,
Incorrect Answer:
Csub query cannot be used
F: is incorrect.
Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 6-5

NEW QUESTION 11
Which two statements are true regarding sub queries? (Choose two.)

  • A. A sub query can retrieve zero or more row
  • B. Only two sub queries can be placed at one leve
  • C. A sub query can be used only in SQL query statement
  • D. A sub query can appeal* on either side of a comparison operato
  • E. There is no limit on the number of sub query levels in the WHERE clause of a SELECT statemen

Answer: AD

NEW QUESTION 12
Which statement is true regarding synonyms?

  • A. Synonyms can be created only for a table
  • B. Synonyms are used to reference only those tables that are owned by another user
  • C. The DROP SYNONYM statement removes the synonym and the table on which the synonym has been created becomes invalid
  • D. A public synonym and a private synonym can exist with the same name for the same table

Answer: D

NEW QUESTION 13
Which statement describes the ROWID data type?

  • A. Binary data up to 4 gigabyte
  • B. Character data up to 4 gigabyte
  • C. Raw binary data of variable length up to 2 gigabyte
  • D. Binary data stored in an external file, up to 4 gigabyte
  • E. A hexadecimal string representing the unique address of a row in its tabl

Answer: E

Explanation:
The ROWID datatype stores information related to the disk location of table rows. They
also uniquely identify the rows in your table. The ROWID datatype is stored as a
hexadecimal string.
Incorrect Answers
A:It is not a binary data. The ROWID datatype is a hexadecimal string.
B:It is not a character data. The ROWID datatype is a hexadecimal string.
C:It is not a raw binary data. The ROWID datatype is a hexadecimal string.
D:It is not binary data stored in an external file. The ROWID datatype is a hexadecimal
string.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 216
Chapter 5: Creating Oracle Database Objects

NEW QUESTION 14
Which statements are true regarding the WHERE and HAVING clauses in a SELECT statement?
(Choose all that apply.)

  • A. The HAVING clause can be used with aggregate functions in subquerie
  • B. The WHERE clause can be used to exclude rows after dividing them into group
  • C. The WHERE clause can be used to exclude rows before dividing them into group
  • D. The aggregate functions and columns used in the HAVING clause must be specified in the SELECT list of the quer
  • E. The WHERE and HAVING clauses can be used in the same statement only if they are applied to different columns in the tabl

Answer: AC

NEW QUESTION 15
You work as a database administrator at ABC.com. You study the exhibit carefully. Exhibit:
1Z0-051 dumps exhibit
Examine the structure of PRODUCTS table.
Using the PRODUCTS table, you issue the following query to generate the names, current list price and discounted list price for all those products whose list price fails below $10 after a discount of 25% is applied on it.
Exhibit:
1Z0-051 dumps exhibit
The query generates an error.
What is the reason of generating error?

  • A. The column alias should be put in uppercase and enclosed within double quotation marks in the WHERE clause
  • B. The parenthesis should be added to enclose the entire expression
  • C. The column alias should be replaced with the expression in the WHERE clause
  • D. The double quotation marks should be removed from the column alias

Answer: C

Explanation: Note: You cannot use column alias in the WHERE clause.

NEW QUESTION 16
View the Exhibit and examine the data in the PROJ_TASK_DETAILS table.
1Z0-051 dumps exhibit
The PROJ_TASK_DETAILS table stores information about tasks involved in a project and the relation between them.
The BASED_ON column indicates dependencies between tasks. Some tasks do not depend on the completion of any other tasks.
You need to generate a report showing all task IDs, the corresponding task ID they are dependent on, and the name of the employee in charge of the task it depends on.
Which query would give the required result?

  • A. SELECT p.task_id, p.based_on, d.task_in_charge FROM proj_task_details p JOIN proj_task_details d ON (p.based_on = d.task_id);
  • B. SELECT p.task_id, p.based_on, d.task_in_charge FROM proj_task_details p LEFT OUTER JOIN proj_task_details d ON (p.based_on = d.task_id);
  • C. SELECT p.task_id, p.based_on, d.task_in_charge FROM proj_task_details p FULL OUTER JOIN proj_task_details d ON (p.based_on = d.task_id);
  • D. SELECT p.task_id, p.based_on, d.task_in_charge FROM proj_task_details p JOIN proj_task_details d ON (p.task_id = d.task_id);

Answer: B

NEW QUESTION 17
View the Exhibit and examine the structure of the PRODUCTS table.
1Z0-051 dumps exhibit
You want to display the category with the maximum number of items. You issue the following query:
SQL>SELECT COUNT(*),prod_category_id FROM products GROUP BY prod_category_id HAVING COUNT(*) = (SELECT MAX(COUNT(*)) FROM products);
What is the outcome?

  • A. It executes successfully and gives the correct outpu
  • B. It executes successfully but does not give the correct outpu
  • C. It generates an error because the subquery does not have a GROUP BY claus
  • D. It generates an error because = is not valid and should be replaced by the IN operato

Answer: C

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