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1Z0-071考題 & 1Z0-071學習筆記 - 1Z0-071題庫分享 - Royalholidayclubbed
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1z0-071 PDF DEMO:
QUESTION NO: 1
Which two statements are true regarding multiple-row subqueries? (Choose two.)
A. They can contain group functions.
B. They use the < ALL operator to imply less than the maximum.
C. They should not be used with the NOT IN operator in the main query if NULL is likely to be a part of the result of the subquery.
D. They can be used to retrieve multiple rows from a single table only.
E. They always contain a subquery within a subquery.
Answer: A,C
QUESTION NO: 2
Which two statements are true about INTERVAL data types?
A. INTERVAL YEAR TO MONTH columns only support monthly intervals within a single year.
B. The YEAR field in an INTERVAL YEAR TO MONTH column must be a positive value.
C. INTERVAL DAY TO SECOND columns support fractions of seconds.
D. The value in an INTERVAL DAY TO SECOND column can be copied into an INTERVAL YEAR TO
MONTH column.
E. INTERVAL YEAR TO MONTH columns only support monthly intervals within a range of years.
F. INTERVAL YEAR TO MONTH columns support yearly intervals.
Answer: C,F
QUESTION NO: 3
A non-correlated subquery can be defined as __________. (Choose the best answer.)
A. A set of sequential queries, all of which must always return a single value.
B. A set of sequential queries, all of which must return values from the same table.
C. A set of one or more sequential queries in which generally the result of the inner query is used as the search value in the outer query.
D. A SELECT statement that can be embedded in a clause of another SELECT statement only.
Answer: C
QUESTION NO: 4
Examine the structure of the EMPLOYEES table:
There is a parent/child relationship between EMPLOYEE_ID and MANAGER_ID.
You want to display the name, joining date, and manager for all employees. Newly hired employees are yet to be assigned a department or a manager. For them, 'No Manager' should be displayed in the MANAGER column.
Which SQL query gets the required output?
A. SELECT e.last_name, e.hire_date, NVL(m.last_name, 'No Manager') ManagerFROM employees e
RIGHT OUTER JOIN employees mON (e.manager_id = m.employee_id);
B. SELECT e.last_name, e.hire_date, NVL(m.last_name, 'No Manager') ManagerFROM employees e
JOIN employees mON (e.manager_id = m.employee_id);
C. SELECT e.last_name, e.hire_date, NVL(m.last_name, 'No Manager') ManagerFROM employees e
NATURAL JOIN employees mON (e.manager_id = m.employee_id).
D. SELECT e.last_name, e.hire_date, NVL(m.last_name, 'No Manager') ManagerFROM employees e
LEFT OUTER JOIN employees mON (e.manager_id = m.employee_id);
Answer: D
QUESTION NO: 5
View the Exhibit and examine the structure of the CUSTOMERS table.
Evaluate the following SQL statement:
Which statement is true regarding the outcome of the above query?
A. It returns an error because WHERE and HAVING clauses cannot be used in the same SELECT statement.
B. It returns an error because the BETWEEN operator cannot be used in the HAVING clause.
C. It returns an error because WHERE and HAVING clauses cannot be used to apply conditions on the same column.
D. It executes successfully.
Answer: D
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Updated: May 28, 2022
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