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Oracle Database 1z1-061 あなた準備しましたか。

真剣にNewValidDumpsのOracle 1z1-061 - Oracle Database 12c: SQL Fundamentals絶対合格問題集を勉強する限り、受験したい試験に楽に合格することができるということです。 あなたがする必要があるのは、問題集に出るすべての問題を真剣に勉強することです。この方法だけで、試験を受けるときに簡単に扱うことができます。

がむしゃらに試験に関連する知識を勉強しているのですか。それとも、効率が良い試験1z1-061絶対合格参考書を使っているのですか。Oracleの認証資格は最近ますます人気になっていますね。

Oracle 1z1-061絶対合格 - まだ何を待っていますか。

1z1-061絶対合格認定試験は現在で本当に人気がある試験ですね。まだこの試験の認定資格を取っていないあなたも試験を受ける予定があるのでしょうか。確かに、これは困難な試験です。しかし、難しいといっても、高い点数を取って楽に試験に合格できないというわけではないです。では、まだ試験に合格するショートカットがわからないあなたは、受験のテクニックを知りたいですか。今教えてあげますよ。それはNewValidDumpsの1z1-061絶対合格問題集を利用することです。

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1z1-061 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
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
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
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: 5
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

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Updated: May 28, 2022