Updated May 23, 2022 1z0-071 Exam Dumps – PDF Questions and Testing Engine [Q84-Q99]

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Updated May 23, 2022 1z0-071 Exam Dumps – PDF Questions and Testing Engine

New (2022) Oracle 1z0-071 Exam Dumps

NEW QUESTION 84
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?

The EXECUTION_DATETIME must be of INTERVAL DAY TO SECOND data type.

The EXECUTION_DATETIME must be of TIMESTAMP data type.

The EXECUTION_DATETIME must be of DATE data type.

The COMPONENT_IDmust be of ROWID data type.

The COMPONENT_IDmust be of VARCHAR2 data type.

The COMPONENT_IDcolumn must be of CHAR data type.

NEW QUESTION 85
Examine the structure of the EMPLOYEES table.

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 LEFT 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;

NEW QUESTION 86
You issue the following command to drop the PRODUCTStable:
SQL > DROP TABLE products;
Which three statements are true about the implication of this command? (Choose three.)

All data along with the table structure is deleted.

A pending transaction in the session is committed.

All indexes on the table remain but they are invalidated.

All views and synonyms on the table remain but they are invalidated.

All data in the table is deleted but the table structure remains.

NEW QUESTION 87
Examine this SELECT statement and view the Exhibit to see its output:

SELECT constraints_name, constraints_type, search_condition, r_constraint_name, delete_rule, status,
FROM user_constraints
WHERE table_name = ‘ORDERS’;
Which two statements are true about the output? (Choose two.)

The DELETE_RULE column indicates the desired state of related rows in the child table when the corresponding row is deleted from the parent table.

The R_CONSTRAINT_NAME column contains an alternative name for the constraint.

In the second column, ‘c’ indicates a check constraint.

The STATUS column indicates whether the table is currently in use.

NEW QUESTION 88
The first DROP operation is performed on PRODUCTS table using the following command:
DROP TABLE products PURGE;
Then you performed the FLASHBACK operation by using the following command:
FLASHBACK TABLE products TO BEFORE DROP;
Which statement describes the outcome of the FLASHBACK command?

It recovers only the table structure.

It recovers the table structure, data, and the indexes.

It recovers the table structure and data but not the related indexes.

It is not possible to recover the table structure, data, or the related indexes.

NEW QUESTION 89
Which two statements are true regarding the DELETEand TRUNCATEcommands? (Choose two.)

DELETEcan be used to remove rows from only one table in one statement.

DELETEcan be used to remove rows from multiple tables in one statement.

DELETEcan be used to remove rows only for tables that are parents for a child table that has a referential integrity constraint referring to the parent.

DELETEcan be used to remove data from specific columns as well as complete rows.

DELETEand TRUNCATEcan be used for tables that are parents for a child table that has a referential integrity constraint having an ON DELETErule.

NEW QUESTION 90
Examine this SELECT statement and view the Exhibit to see its output:

SELECT constraints_name, constraints_type, search_condition, r_constraints_name, delete_rule, status, FROM user_constraints WHERE table_name = ‘ORDERS’; Which two statements are true about the output? (Choose two.)

The DELETE_RULE column indicates the desired state of related rows in the child table when the corresponding row is deleted from the parent table.

The R_CONSTRAINT_NAME column contains an alternative name for the constraint.

In the second column, ‘c’ indicates a check constraint.

The STATUS column indicates whether the table is currently in use.

NEW QUESTION 91
Examine the description of the ORDER_ITEMS table:

Examine this incomplete query:
SELECT DISTINCT quantity * unit_price total_paid FROM order_items ORDER BY <clause>;
Which two can replace <clause> so the query completes successfully?

quantity

quantity, unit_price

total_paid

product_id

quantity * unit_price

NEW QUESTION 92
Evaluate the following two queries:

Which statement is true regarding the above two queries?

Performance would improve in query 2 only if there are null values in the CUST_CREDIT_LIMIT column.

There would be no change in performance.

Performance would degrade in query 2.

Performance would improve in query 2.

NEW QUESTION 93
The PRODUCTStable has the following structure.

Evaluate the following two SQL statements:
SQL>SELECT prod_id, NVL2 (prod_expiry_date, prod_expiry_date + 15, ‘ ‘) FROM products; SQL>SELECT prod_id, NVL (prod_expiry_date, prod_expiry_date + 15) FROM products; Which statement is true regarding the outcome?

Both the statements execute and give different results

Only the second SQL statement executes successfully

Both the statements execute and give the same result

Only the first SQL statement executes successfully

NEW QUESTION 94
Which two statements are true about views?

Views can be updated without the need to re-grant privileges on the view.

Views can be indexed.

The with check clause prevents certain rows from being displayed when querying the view.

The with check clause prevents certain rows from being updated or inserted in the underlying table through the view.

Tables in the defining query of a view must always exist in order to create the view.

NEW QUESTION 95
Evaluate the following SQL statement:

Which statement is true regarding the outcome of the above query?

It executes successfully and displays rows in the descending order of PROMO_CATEGORY.

It produces an error because positional notation cannot be used in the ORDER BY clause with SET operators.

It executes successfully but ignores the ORDER BY clause because it is not located at the end of the compound statement.

It produces an error because the ORDER BY clause should appear only at the end of a compound query-that is, with the last SELECT statement.

NEW QUESTION 96
You must create a SALES table with these column specifications and data types:
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? (Choose the best answer.)

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’);

NEW QUESTION 97
Evaluate the following SQL statements that are issued in the given order: CREATE TABLE emp
(emp_no NUMBER(2) CONSTRAINT emp_emp_no_pk PRIMARY KEY,
enameVARCHAR2(15),
salary NUMBER(8,2),
mgr_no NUMBER(2) CONSTRAINT emp_mgr_fk REFERENCES emp);
ALTER TABLE emp
DISABLE CONSTRAINT emp_emp_no_pk CASCADE;
ALTER TABLE emp
ENABLE CONSTRAINT emp_emp_no_pk;
What would be the status of the foreign key EMP_MGR_FK?

It would be automatically enabled and deferred.

It would be automatically enabled and immediate.

It would remain disabled and has to be enabled manually using the ALTER TABLE command.

It would remain disabled and can be enabled only by dropping the foreign key constraint and re-creating it.

NEW QUESTION 98
Which two statements are true regarding the COUNT function? (Choose two.)

A SELECT statement using the COUNT function with a DISTINCT keyword cannot have a WHERE clause.

COUNT(DISTINCT inv_amt) returns the number of rows excluding rows containing duplicates and NULLs in the INV_AMT column.

COUNT(cust_id) returns the number of rows including rows with duplicate customer IDs and NULLs in the CUST_ID column.

COUNT(*) returns the number of rows including duplicate rows and rows containing NULL value in any column.

The COUNT function can be used only for CHAR, VARCHAR2, and NUMBER data types.

NEW QUESTION 99
Examine the data in the CUST_NAME column of the CUSTOMERS table.
CUST_NAME
——————-
Renske Ladwig
Jason Mallin
Samuel McCain
Allan MCEwen
Irene Mikkilineni
Julia Nayer
You need to display customers’ second names where the second name starts with “Mc” or “MC”.
Which query gives the required output?

SELECT SUBSTR(cust_name, INSTR(cust_name,’ ‘)+1)FROM customersWHERE SUBSTR(cust_name, INSTR(cust_name,’ ‘)+1) LIKE INITCAP(‘MC%’);

SELECT SUBSTR(cust_name, INSTR(cust_name,’ ‘)+1)FROM customersWHERE INITCAP(SUBSTR(cust_name, INSTR(cust_name,’ ‘)+1))=’Mc’;

SELECT SUBSTR(cust_name, INSTR(cust_name,’ ‘)+1)FROM customersWHERE INITCAP(SUBSTR(cust_name, INSTR(cust_name,’ ‘)+1)) LIKE ‘Mc%’;

SELECT SUBSTR(cust_name, INSTR(cust_name,’ ‘)+1)FROM customersWHERE INITCAP(SUBSTR(cust_name, INSTR(cust_name,’ ‘)+1)) = INITCAP(‘MC%’);

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