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1z0-061 PDF DEMO:

QUESTION NO: 1
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.)

QUESTION NO: 2
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: 3
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: 4
The STUDENT_GRADES table has these columns:
STUDENT_ID NUMBER(12)
SEMESTER_END DATE
GPA NUMBER(4, 3)
Which statement finds the highest grade point average (GPA) per semester?
A. SELECT MAX(gpa) FROM student_grades WHERE gpa IS NOT NULL;
B. SELECT (gpa) FROM student_grades GROUP BY semester_end WHERE gpa IS NOT NULL;
C. SELECT MAX(gpa) FROM student_grades WHERE gpa IS NOT NULL GROUP BY
semester_end;
D. SELECT MAX(gpa) GROUP BY semester_end WHERE gpa IS NOT NULL FROM
student_grades;
E. SELECT MAX(gpa) FROM student_grades GROUP BY semester_end WHERE gpa IS NOT NULL;
Answer: C
Explanation:
For highest gpa value MAX function is needed, for result with per semester GROUP BY clause is needed Incorrect answer:
A: per semester condition is not included
B: result would not display the highest gpa value
D: invalid syntax error
E: invalid syntax error
Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 5-7

QUESTION NO: 5
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

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Updated: May 28, 2022

1Z0-061コンポーネント - 1Z0-061模擬対策問題、Oracle Database 12C: SQL Fundamentals

PDF問題と解答

試験コード:1z0-061
試験名称:Oracle Database 12c: SQL Fundamentals
最近更新時間:2024-05-17
問題と解答:全 340
Oracle 1z0-061 模擬対策

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模擬試験

試験コード:1z0-061
試験名称:Oracle Database 12c: SQL Fundamentals
最近更新時間:2024-05-17
問題と解答:全 340
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オンライン版

試験コード:1z0-061
試験名称:Oracle Database 12c: SQL Fundamentals
最近更新時間:2024-05-17
問題と解答:全 340
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1z0-061 参考書勉強