第二，專注，為了做好我們決定完成的事情，必須放棄所有不重要的機會。第三，人們的確會用表面來判斷一個東西的好壞，我們或許擁有最優秀最高品質的產品，但如果以粗製濫造的方式展示出來，自然會被列為粗製濫造的產品，如果以既有創意又很專業的方式呈現，那麼我們將得到最高的效果。Goldmile-Infobiz Oracle的1z0-071考古題考試培訓資料就是這樣成功的培訓資料，舍它其誰? 因為這是一個可以保證一次通過考試的資料。這個考古題的高合格率已經被廣大考生證明了。 Goldmile-Infobiz之所以能幫助每個IT人士，是因為它能證明它的能力。
它就是Goldmile-Infobiz的1z0-071考古題考古題。
經過我們確認之后，就會處理您的請求，這樣客戶擁有足夠的保障放心購買我們的Oracle 1z0-071 - Oracle Database SQL考古題考古題。 这个考古題是IT业界的精英们研究出来的，是一个难得的练习资料。這個考古題的命中率很高，合格率可以達到100%。
通過Oracle 1z0-071考古題的考試是不簡單的，選擇合適的培訓是你成功的第一步，選擇好的資訊來源是你成功的保障，而Goldmile-Infobiz的產品是有很好的資訊來源保障。如果你選擇了Goldmile-Infobiz的產品不僅可以100%保證你通過Oracle 1z0-071考古題認證考試，還可以為你提供長達一年的免費更新。
Oracle 1z0-071考古題 - Goldmile-Infobiz提供的考試練習題的答案是非常準確的。
你在擔心如何通過可怕的Oracle的1z0-071考古題考試嗎？不用擔心，有Goldmile-Infobiz Oracle的1z0-071考古題考試培訓資料在手，任何IT考試認證都變得很輕鬆自如。我們Goldmile-Infobiz Oracle的1z0-071考古題考試培訓資料是Oracle的1z0-071考古題考試認證準備的先鋒。
我相信不論在哪個行業工作的人都希望自己有很好的職業前景。當然在競爭激烈的IT行業裏面也不例外。
1z0-071 PDF DEMO:
QUESTION NO: 1
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: 2
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
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
View the exhibit and examine the structure of the SALES, CUSTOMERS, PRODUCTS and TIMES tables.
The PROD_ID column is the foreign key in the SALES table referencing the PRODUCTS table.
The CUST_ID and TIME_ID columns are also foreign keys in the SALES table referencing the
CUSTOMERS and TIMES tables, respectively.
Examine this command:
CREATE TABLE new_sales (prod_id, cust_id, order_date DEFAULT SYSDATE)
AS
SELECT prod_id, cust_id, time_id
FROM sales;
Which statement is true?
A. The NEW_SALES table would not get created because the DEFAULT value cannot be specified in the column definition.
B. The NEW_SALES table would not get created because the column names in the CREATE TABLE command and the SELECT clause do not match.
C. The NEW_SALES table would get created and all the NOT NULL constraints defined on the selected columns from the SALES table would be created on the corresponding columns in the NEW_SALES table.
D. The NEW_SALES table would get created and all the FOREIGN KEY constraints defined on the selected columns from the SALES table would be created on the corresponding columns in the
NEW_SALES table.
Answer: C
QUESTION NO: 5
The user SCOTT who is the owner of ORDERS and ORDER_ITEMS tables issues this GRANT command:
GRANT ALL
ON orders, order_items
TO PUBLIC;
What must be done to fix the statement?
A. Separate GRANT statements are required for the ORDERS and ORDER_ITEMS tables.
B. PUBLIC should be replaced with specific usernames.
C. ALL should be replaced with a list of specific privileges.
D. WITH GRANT OPTION should be added to the statement.
Answer: A
Explanation:
http://docs.oracle.com/javadb/10.8.3.0/ref/rrefsqljgrant.html
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Updated: May 28, 2022