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QUESTION NO: 1
The CUSTOMERS table has these columns:
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
QUESTION NO: 2
Which SQL statement accepts user input for the columns to be displayed, the table name, and WHERE condition?
A. SELECT &1, "&2"FROM &3WHERE last_name = '&4';
B. SELECT &1, '&2'FROM &3WHERE '&last_name = '&4' ';
C. SELECT &1, &2FROM &3WHERE last_name = '&4';
D. SELECT &1, '&2'FROM EMPWHERE last_name = '&4';
Answer: C
Explanation:
In a WHERE clause, date and characters values must be enclosed within single quotation marks.
Sample of the correct syntax
SELECT EMPLOYEE_ID, &COLUMN_NAME
FROM EMPLOYEES
Incorrect Answers :
A: Incorrect use of " symbol
B: Incorrect use of ' symbol
D: No input for table name as EMP has been use in the statement.
Refer: Introduction to Oracle9i: SQL, Oracle University Student Guide, Producing Readable Output with iSQL*PLUS, p. 7-8
QUESTION NO: 3
Which SQL statements would display the value 1890.55 as $1, 890.55? (Choose three.)
A. SELECT TO_CHAR(1890.55, '$0G000D00')FROM DUAL;
B. SELECT TO_CHAR(1890.55, '$9, 999V99')FROM DUAL;
C. SELECT TO_CHAR(1890.55, '$99, 999D99')FROM DUAL;
D. SELECT TO_CHAR(1890.55, '$99G999D00')FROM DUAL;
E. SELECT TO_CHAR(1890.55, '$99G999D99')FROM DUAL;
Answer: A,D,E
QUESTION NO: 4
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 list.
B. It locks the rows that satisfy the condition in the SELECT statement.
C. It can be used only in SELECT statements that are based on a single table.
D. It can be used in SELECT statements that are based on a single or multiple tables.
E. After it is enforced by a SELECT statement, no other query can access the same rows until a
COMMIT or ROLLBACK is issued.
Answer: B,D
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.
QUESTION NO: 5
EMPLOYEES and DEPARTMENTS data:
EMPLOYEES
DEPARTMENTS
On the EMPLOYEES table, EMPLOYEE_ID is the primary key. MGR_ID is the ID managers and refers to the EMPLOYEE_ID.
On the DEPARTMENTS table DEPARTMENT_ID is the primary key.
Evaluate this UPDATE statement.
UPDATE employees
SET mgr_id
. (SELECT mgr_id
. FROM. employees
. WHERE dept_id
. (SELECT department_id
. FROM departments
. WHERE department_name = 'Administration')),
. Salary = (SELECT salary
. . FROM employees
. . WHERE emp_name = 'Smith')
WHERE job_id = 'IT_ADMIN';
What happens when the statement is executed?
A. The statement executes successfully, leaves the manager ID as the existing value, and changes the salary to 4000 for the employees with ID 103 and 105.
B. The statement executes successfully, changes the manager ID to NULL, and changes the salary to
4000 for the employees with ID 103 and 105.
C. The statement executes successfully, changes the manager ID to NULL, and changes the salary to
3000 for the employees with ID 103 and 105.
D. The statement fails because there is more than one row matching the employee name Smith.
E. The statement fails because there is more than one row matching the IT_ADMIN job ID in the
EMPLOYEES table.
F. The statement fails because there is no 'Administration' department in the DEPARTMENTS table.
Answer: D
Explanation:
'=' is use in the statement and sub query will return more than one row.
Employees table has 2 row matching the employee name Smith.
The update statement will fail.
Incorrect Answers :
A: The Update statement will fail no update was done.
B: The update statement will fail no update was done.
C: The update statement will fail no update was done.
E: The update statement will fail but not due to job_it='IT_ADMIN'
F: The update statement will fail but not due to department_id='Administration' Refer: Introduction to Oracle9i: SQL, Oracle University Student Guide, Sub queries, p. 6-12
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Updated: May 28, 2022
試験コード:1z1-061
試験名称:Oracle Database 12c: SQL Fundamentals
最近更新時間:2024-10-31
問題と解答:全 340 問
Oracle 1z1-061 復習範囲
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試験コード:1z1-061
試験名称:Oracle Database 12c: SQL Fundamentals
最近更新時間:2024-10-31
問題と解答:全 340 問
Oracle 1z1-061 試験内容
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試験コード:1z1-061
試験名称:Oracle Database 12c: SQL Fundamentals
最近更新時間:2024-10-31
問題と解答:全 340 問
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