2014年7月31日星期四

1z0-061試験過去問、1z0-061復習問題集

あなたの夢は何ですか。あなたのキャリアでいくつかの輝かしい業績を行うことを望まないのですか。きっと望んでいるでしょう。では、常に自分自身をアップグレードする必要があります。IT業種で仕事しているあなたは、夢を達成するためにどんな方法を利用するつもりですか。実際には、IT認定試験を受験して認証資格を取るのは一つの良い方法です。最近、Oracleの1z0-061試験は非常に人気のある認定試験です。あなたもこの試験の認定資格を取得したいのですか。さて、はやく試験を申し込みましょう。JPexamはあなたを助けることができますから、心配する必要がないですよ。

JPexamのOracleの1z0-061トレーニング資料は完璧な資料で、世界的に最高なものです。これは品質の問題だけではなく、もっと大切なのは、JPexamのOracleの1z0-061試験資料は全てのIT認証試験に適用するもので、ITの各領域で使用できます。それはJPexamが受験生の注目を浴びる理由です。受験生の皆様は我々を信じて、依頼しています。これもJPexamが実力を体現する点です。我々の試験トレーニング資料はあなたが買いてから友達に勧めなければならない魅力を持っています。本当に皆様に極大なヘルプを差し上げますから。

JPexamが提供しておりますのは専門家チームの研究した問題と真題で弊社の高い名誉はたぶり信頼をうけられます。安心で弊社の商品を使うために無料なサンブルをダウンロードしてください。

1z0-061試験番号:1z0-061問題集
試験科目:Oracle Database 12c: SQL Fundamentals
最近更新時間:2014-07-31
問題と解答:全75問 1z0-061 資格認定
100%の返金保証。1年間の無料アップデート。

>>詳しい紹介はこちら

 

弊社が提供した問題集がほかのインターネットに比べて問題のカーバ範囲がもっと広くて対応性が強い長所があります。JPexamが持つべきなIT問題集を提供するサイトでございます。

JPexam提供した商品の品質はとても良くて、しかも更新のスピードももっともはやくて、もし君はOracleの1z0-061の認証試験に関する学習資料をしっかり勉強して、成功することも簡単になります。

あなたに最大の利便性を与えるために、JPexamは様々なバージョンの教材を用意しておきます。PDF版の1z0-061問題集は読みやすくて、忠実に試験の問題を再現することができます。テストエンジンとして、ソフトウェア版の1z0-061問題集はあなたの試験の準備についての進捗状況をテストするために利用することができます。もし試験の準備を十分にしたかどうかを確認したいなら、ソフトウェア版の1z0-061問題集を利用して自分のレベルをテストしてください。従って、すぐに自分の弱点や欠点を識別することができ、正しく次の1z0-061学習内容を手配することもできます。

もし弊社のOracleの1z0-061認証試験について問題集に興味があったら、購入するまえにインターネットで弊社が提供した無料な部分問題集をダウンロードして、君の試験に役に立つかどうかのを自分が判断してください。それにJPexamは一年の無料な更新のサービスを提供いたします。

購入前にお試し,私たちの試験の質問と回答のいずれかの無料サンプルをダウンロード:http://www.jpexam.com/1z0-061_exam.html

NO.1 View the Exhibit for the structure of the student and faculty tables.
You need to display the faculty name followed by the number of students handled by the faculty at
the base location.
Examine the following two SQL statements:
Which statement is true regarding the outcome?
A. Only statement 1 executes successfully and gives the required result.
B. Only statement 2 executes successfully and gives the required result.
C. Both statements 1 and 2 execute successfully and give different results.
D. Both statements 1 and 2 execute successfully and give the same required result.
Answer: D

Oracle合格率   1z0-061クラムメディア   1z0-061認証試験   1z0-061通信

NO.2 You need to create a table for a banking application. One of the columns in the table has the
following requirements:
1. You want a column in the table to store the duration of the credit period.
2) The data in the column should be stored in a format such that it can be easily added and
subtracted with date data type without using conversion functions.
3) The maximum period of the credit provision in the application is 30 days.
4) The interest has to be calculated for the number of days an individual has taken a credit for.
Which data type would you use for such a column in the table?
A. DATE
B. NUMBER
C. TIMESTAMP
D. INTERVAL DAY TO SECOND
E. INTERVAL YEAR TO MONTH
Answer: D

Oracle関節   1z0-061受験記   1z0-061問題集   1z0-061合格率   1z0-061

NO.3 View the Exhibit and evaluate the structure and data in the CUST_STATUS table.
You issue the following SQL statement:
Which statement is true regarding the execution of the above query?
A. It produces an error because the AMT_SPENT column contains a null value.
B. It displays a bonus of 1000 for all customers whose AMT_SPENT is less than CREDIT_LIMIT.
C. It displays a bonus of 1000 for all customers whose AMT_SPENT equals CREDIT_LIMIT, or
AMT_SPENT is null.
D. It produces an error because the TO_NUMBER function must be used to convert the result of the
NULLIF function before it can be used by the NVL2 function.
Answer: C

Oracle練習   1z0-061   1z0-061認定   1z0-061   1z0-061練習問題
Explanation:
The NULLIF Function The NULLIF function tests two terms for equality. If they are equal the function
returns a null, else it returns the first of the two terms tested. The NULLIF function takes two
mandatory parameters of any data type. The syntax is NULLIF(ifunequal, comparison_term), where
the parameters ifunequal and comparison_term are compared. If they are identical, then NULL is
returned. If they differ, the ifunequal parameter is returned.

NO.4 View the Exhibit and examine the structure of the product, component, and PDT_COMP
tables.
In product table, PDTNO is the primary key.
In component table, COMPNO is the primary key.
In PDT_COMP table, <PDTNO, COMPNO) is the primary key, PDTNO is the foreign key referencing
PDTNO in product table and COMPNO is the foreign key referencing the COMPNO in component
table.
You want to generate a report listing the product names and their corresponding component names,
if the component names and product names exist.
Evaluate the following query:
SQL>SELECT pdtno, pdtname, compno, compname
FROM product _____________ pdt_comp
USING (pdtno) ____________ component USING (compno)
WHERE compname IS NOT NULL;
Which combination of joins used in the blanks in the above query gives the correct output?
A. JOIN; JOIN
B. FULL OUTER JOIN; FULL OUTER JOIN
C. RIGHT OUTER JOIN; LEFT OUTER JOIN
D. LEFT OUTER JOIN; RIGHT OUTER JOIN
Answer: C

Oracle割引   1z0-061 PDF   1z0-061   1z0-061体験

NO.5 Examine the structure proposed for the transactions table:
Which two statements are true regarding the creation and storage of data in the above table
structure?
A. The CUST_STATUS column would give an error.
B. The TRANS_VALIDITY column would give an error.
C. The CUST_STATUS column would store exactly one character.
D. The CUST_CREDIT_LIMIT column would not be able to store decimal values.
E. The TRANS_VALIDITY column would have a maximum size of one character.
F. The TRANS_DATE column would be able to store day, month, century, year, hour, minutes,
seconds, and fractions of seconds
Answer: B,C

Oracle認定試験   1z0-061過去   1z0-061認定証
Explanation:
VARCHAR2(size)Variable-length character data (A maximum size must be specified:
minimum size is 1; maximum size is 4, 000.)
CHAR [(size)] Fixed-length character data of length size bytes (Default and minimum size
is 1; maximum size is 2, 000.)
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.)
DATE Date and time values to the nearest second between January 1, 4712 B.C., and
December 31, 9999 A.D.

NO.6 Which normal form is a table in if it has no multi-valued attributes and no partial
dependencies?
A. First normal form
B. Second normal form
C. Third normal form
D. Fourth normal form
Answer: B

Oracle問題集   1z0-061教育   1z0-061認定試験   1z0-061教本

NO.7 Evaluate the following SQL statement:
Which statement is true regarding the outcome of the above query?
A. It executes successfully and displays rows in the descending order of PROMO_CATEGORY .
B. It produces an error because positional notation cannot be used in the order by clause with set
operators.
C. It executes successfully but ignores the order by clause because it is not located at the end of the
compound statement.
D. It produces an error because the order by clause should appear only at the end of a compound
query-that is, with the last select statement.
Answer: D

Oracleスクール   1z0-061   1z0-061割引   1z0-061   1z0-061問題

NO.8 Which three tasks can be performed using SQL functions built into Oracle Database?
A. Displaying a date in a nondefault format
B. Finding the number of characters in an expression
C. Substituting a character string in a text expression with a specified string
D. Combining more than two columns or expressions into a single column in the output
Answer: A,B,C

Oracle合格率   1z0-061対策   1z0-061   1z0-061

没有评论:

发表评论