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1z0-061 PDF DEMO:

QUESTION NO: 1
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

QUESTION NO: 2
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: 3
View the Exhibit and examine the structure and data in the INVOICE table.
Which two statements are true regarding data type conversion in expressions used in queries?
A. inv_amt ='0255982': requires explicit conversion
B. inv_date > '01-02-2008': uses implicit conversion
C. CONCAT (inv_amt, inv_date): requires explicit conversion
D. inv_date = '15-february-2008': uses implicit conversion
E. inv_no BETWEEN '101' AND '110': uses implicit conversion
Answer: D,E
Explanation:
In some cases, the Oracle server receives data of one data type where it expects data of a different data type.
When this happens, the Oracle server can automatically convert the data to the expected data type.
This data type conversion can be done implicitly by the Oracle server or explicitly by the user.
Explicit data type conversions are performed by using the conversion functions. Conversion functions convert a value from one data type to another. Generally, the form of the function names follows the convention data type TO data type. The first data type is the input data type and the second data type is the output.
Note: Although implicit data type conversion is available, it is recommended that you do the explicit data type conversion to ensure the reliability of your SQL statements.

QUESTION NO: 4
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: 5
You need to create a table with the following column specifications:
1 . Employee ID (numeric data type) for each employee
2 . Employee Name (character data type) that stores the employee name
3 . Hire date, which stores the date of joining the organization for each employee
4 . Status (character data type), that contains the value 'active1 if no data is entered
5 . Resume (character large object [CLOB] data type), which contains the resume submitted by the employee Which is the correct syntax to create this table?
A. Option A
B. Option B
C. Option C
D. Option D
Answer: D
Explanation:
CLOB Character data (up to 4 GB)
NUMBER [(p, s)] Number having precision p and scale s (Precision is the total number of decimal digits and scale is the number of digits to the right of the decimal point; precision can range from 1 to
38, and scale can range from -84 to 127.)

そして、Fortinet NSE6_FSW-7.2試験参考書の問題は本当の試験問題とだいたい同じことであるとわかります。 EC-COUNCIL 312-39 - では、この試験に合格するためにどのように試験の準備をしているのですか。 Fortinet NSE7_SDW-7.2 - この試験に合格すれば君の専門知識がとても強いを証明し得ます。 そして、NewValidDumpsのOracleのCompTIA N10-008問題集を購入したら、私たちは一年間で無料更新サービスを提供することができます。 きみはOracleのJuniper JN0-683認定テストに合格するためにたくさんのルートを選択肢があります。

Updated: May 28, 2022