あなたは自分の職場の生涯にユニークな挑戦に直面していると思いましたら、Oracleの1Z1-051の認定試験に合格することが必要になります。IT-Passports.comはOracleの1Z1-051の認定試験を真実に、全面的に研究したサイトです。IT-Passports.com のユニークなOracleの1Z1-051の認定試験の問題と解答を利用したら、試験に合格することがたやすくなります。IT-Passports.comは認証試験の専門的なリーダーで、最全面的な認証基準のトレーニング方法を追求して、１００パーセントの成功率を保証します。IT-Passports.comのOracleの1Z1-051の試験問題と解答は当面の市場で最も徹底的かつ正確かつ最新な模擬テストです。それを利用したら、初めに試験を受けても、合格する自信を持つようになります。
この人材が多い社会で、人々はずっと自分の能力を高めていますが、世界で最先端のIT専門家に対する需要が継続的に拡大しています。ですから、Oracleの1Z1-051認定試験に受かりたい人が多くなります。しかし、試験に受かるのは容易なことではないです。実は良いトレーニング資料を選んだら試験に合格することは不可能ではないです。IT-Passports.comが提供したOracleの1Z1-051試験トレーニング資料はあなたが試験に合格することを助けられます。IT-Passports.comのトレーニング資料は大勢な受験生に証明されたもので、国際的に他のサイトをずっと先んじています。Oracleの1Z1-051認定試験に合格したいのなら、IT-Passports.comが提供したOracleの1Z1-051トレーニング資料をショッピングカートに入れましょう。
1Z1-051認定試験に合格することは難しいようですね。試験を申し込みたいあなたは、いまどうやって試験に準備すべきなのかで悩んでいますか。そうだったら、下記のものを読んでください。いま1Z1-051試験に合格するショートカットを教えてあげますから。あなたを試験に一発合格させる素晴らしい1Z1-051試験に関連する参考書が登場しますよ。それはIT-Passports.comの1Z1-051問題集です。気楽に試験に合格したければ、はやく試しに来てください。
試験番号：1Z1-051問題集
試験科目：Oracle 「Oracle Database: SQL Fundamentals I」
問題と解答：全254問
このインターネットが普及された時代に、どのような情報を得るのが非常に簡単なことだということを我々はよく知っていますが、品質と適用性の欠如が問題です。インターネットでOracleの1Z1-051試験トレーニング資料を探す人がたくさんいますが、どれが信頼できるか良く分からないです。ここで私はIT-Passports.comのOracleの1Z1-051試験トレーニング資料を勧めたいです。この資料はインターネットでのクリック率と好評率が一番高いです。IT-Passports.comはOracleの1Z1-051試験トレーニング資料の一部の問題と解答を無料に提供しますから、あなたは試用してから買うかどうかを決めることができます。
現在でOracleの1Z1-051試験を受かることができます。IT-Passports.comにOracleの1Z1-051試験のフルバージョンがありますから、最新のOracleの1Z1-051のトレーニング資料をあちこち探す必要がないです。IT-Passports.comを利用したら、あなたはもう最も良いOracleの1Z1-051のトレーニング資料を見つけたのです。弊社の質問と解答を安心にご利用ください。あなたはきっとOracleの1Z1-051試験に合格できますから。
Oracleの1Z1-051認定試験は実は技術専門家を認証する試験です。 Oracleの1Z1-051認定試験はIT人員が優れたキャリアを持つことを助けられます。優れたキャリアを持ったら、社会と国のために色々な利益を作ることができて、国の経済が継続的に発展していることを進められるようになります。全てのIT人員がそんなにられるとしたら、国はぜひ強くなります。Pass4のOracleの1Z1-051試験トレーニング資料はIT人員の皆さんがそんな目標を達成できるようにヘルプを提供して差し上げます。Pass4のOracleの1Z1-051試験トレーニング資料は１００パーセントの合格率を保証しますから、ためらわずに決断してPass4を選びましょう。
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.it-passports.com/1Z1-051.html
NO.1 Examine the structure of the PROMOTIONS table:
name Null Type
PROMO_ID NOT NULL NUMBER(6)
PROMO_NAME NOT NULL VARCHAR2(30)
PROMO_CATEGORY NOT NULL VARCHAR2(30)
PROMO_COST NOT NULL NUMBER(10,2)
The management wants to see a report of unique promotion costs in each promotion category.
Which query would achieve the required result?
A. SELECT DISTINCT promo_cost, promo_category FROM promotions;
B. SELECT promo_category, DISTINCT promo_cost FROM promotions;
C. SELECT DISTINCT promo_cost, DISTINCT promo_category FROM promotions;
D. SELECT DISTINCT promo_category, promo_cost FROM promotions ORDER BY 1;
Answer: D
Oracle 1Z1-051問題集 1Z1-051 1Z1-051
NO.2 Which statement is true regarding the INTERSECT operator?
A. It ignores NULL values.
B. Reversing the order of the intersected tables alters the result.
C. The names of columns in all SELECT statements must be identical.
D. The number of columns and data types must be identical for all SELECT statements in the query.
Answer: D
Oracle認定資格 1Z1-051 1Z1-051
NO.3 Which two statements are true regarding the USING and ON clauses in table joins? (Choose two.)
A. Both USING and ON clauses can be used for equijoins and nonequijoins.
B. A maximum of one pair of columns can be joined between two tables using the ON clause.
C. The ON clause can be used to join tables on columns that have different names but compatible data
types.
D. The WHERE clause can be used to apply additional conditions in SELECT statements containing the
ON or the USING clause.
Answer: CD
Oracle 1Z1-051参考書 1Z1-051
NO.4 You need to produce a report where each customer's credit limit has been incremented by $1000. In
the output, t he customer's last name should have the heading Name and the incremented credit limit
should be labeled New Credit Limit. The column headings should have only the first letter of each word in
uppercase .
Which statement would accomplish this requirement?
A. SELECT cust_last_name Name, cust_credit_limit + 1000
"New Credit Limit"
FROM customers;
B. SELECT cust_last_name AS Name, cust_credit_limit + 1000
AS New Credit Limit
FROM customers;
C. SELECT cust_last_name AS "Name", cust_credit_limit + 1000
AS "New Credit Limit"
FROM customers;
D. SELECT INITCAP(cust_last_name) "Name", cust_credit_limit + 1000
INITCAP("NEW CREDIT LIMIT")
FROM customers;
Answer: C
Oracle過去問 1Z1-051認定試験 1Z1-051
NO.5 Examine the structure of the SHIPMENTS table:
name Null Type
PO_ID NOT NULL NUMBER(3)
PO_DATE NOT NULL DATE
SHIPMENT_DATE NOT NULL DATE
SHIPMENT_MODE VARCHAR2(30)
SHIPMENT_COST NUMBER(8,2)
You want to generate a report that displays the PO_ID and the penalty amount to be paid if the
SHIPMENT_DATE is later than one month from the PO_DATE. The penalty is $20 per day.
Evaluate the following two queries:
SQL> SELECT po_id, CASE
WHEN MONTHS_BETWEEN (shipment_date,po_date)>1 THEN
TO_CHAR((shipment_date - po_date) * 20) ELSE 'No Penalty' END PENALTY
FROM shipments;
SQL>SELECT po_id, DECODE
(MONTHS_BETWEEN (po_date,shipment_date)>1,
TO_CHAR((shipment_date - po_date) * 20), 'No Penalty') PENALTY
FROM shipments;
Which statement is true regarding the above commands?
A. Both execute successfully and give correct results.
B. Only the first query executes successfully but gives a wrong result.
C. Only the first query executes successfully and gives the correct result.
D. Only the second query executes successfully but gives a wrong result.
E. Only the second query executes successfully and gives the correct result.
Answer: C
Oracle 1Z1-051参考書 1Z1-051参考書 1Z1-051 1Z1-051
NO.6 Evaluate the following query:
SELECT INTERVAL '300' MONTH,
INTERVAL '54-2' YEAR TO MONTH,
INTERVAL '11:12:10.1234567' HOUR TO SECOND
FROM dual;
What is the correct output of the above query?
A. +25-00 , +54-02, +00 11:12:10.123457
B. +00-300, +54-02, +00 11:12:10.123457
C. +25-00 , +00-650, +00 11:12:10.123457
D. +00-300 , +00-650, +00 11:12:10.123457
Answer: A
Oracle過去問 1Z1-051 1Z1-051 1Z1-051参考書
NO.7 You need to extract details of those products in the SALES table where the PROD_ID column
contains the string '_D123'.
Which WHERE clause could be used in the SELECT statement to get the required output?
A. WHERE prod_id LIKE '%_D123%' ESCAPE '_'
B. WHERE prod_id LIKE '%\_D123%' ESCAPE '\'
C. WHERE prod_id LIKE '%_D123%' ESCAPE '%_'
D. WHERE prod_id LIKE '%\_D123%' ESCAPE '\_'
Answer: B
Oracle 1Z1-051 1Z1-051認定試験 1Z1-051 1Z1-051
NO.8 Which two statements are true regarding single row functions? (Choose two.)
A. They a ccept only a single argument.
B. They c an be nested only to two levels.
C. Arguments can only be column values or constants.
D. They a lways return a single result row for every row of a queried table.
E. They c an return a data type value different from the one that is referenced.
Answer: DE
Oracle 1Z1-051 1Z1-051参考書 1Z1-051 1Z1-051
NO.9 View the Exhibit to examine the description for the SALES table.
Which views can have all DML operations performed on it? (Choose all that apply.)
A. CREATE VIEW v3
AS SELECT * FROM SALES
WHERE cust_id = 2034
WITH CHECK OPTION;
B. CREATE VIEW v1
AS SELECT * FROM SALES
WHERE time_id <= SYSDATE - 2*365
WITH CHECK OPTION;
C. CREATE VIEW v2
AS SELECT prod_id, cust_id, time_id FROM SALES
WHERE time_id <= SYSDATE - 2*365
WITH CHECK OPTION;
D. CREATE VIEW v4
AS SELECT prod_id, cust_id, SUM(quantity_sold) FROM SALES
WHERE time_id <= SYSDATE - 2*365
GROUP BY prod_id, cust_id
WITH CHECK OPTION;
Answer: AB
Oracle過去問 1Z1-051 1Z1-051認定資格 1Z1-051認定証
NO.10 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, which references the PRODUCTS table.
Similarly, the CUST_ID and TIME_ID columns are also foreign keys in the SALES table referencing the
CUSTOMERS and TIMES tables, respectively.
Evaluate the following CREATE TABLE 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 regarding the above command?
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 get created and all the NOT NULL constraints defined on the specified
columns would be passed to the new table.
C. The NEW_SALES table would not get created because the column names in the CREATE TABLE
command and the SELECT clause do not match.
D. The NEW_SALES table would get created and all the FOREIGN KEY constraints defined on the
specified columns would be passed to the new table.
Answer: B
Oracle過去問 1Z1-051 1Z1-051
NO.11 View the Exhibit and examine the structure of the CUSTOMERS table.
Which two tasks would require subqueries or joins to be executed in a single statement? (Choose two.)
A. listing of customers who do not have a credit limit and were born before 1980
B. finding the number of customers, in each city, whose marital status is 'married'
C. finding the average credit limit of male customers residing in 'Tokyo' or 'Sydney'
D. listing of those customers whose credit limit is the same as the credit limit of customers residing in the
city 'Tokyo'
E. finding the number of customers, in each city, whose credit limit is more than the average credit limit of
all the customers
Answer: DE
Oracle認定資格 1Z1-051過去問 1Z1-051認定証
NO.12 Which three statements are true regarding the data types in Oracle Database 10g/11g? (Choose
three.)
A. Only one LONG column can be used per table.
B. A TIMESTAMP data type column stores only time values with fractional seconds.
C. The BLOB data type column is used to store binary data in an operating system file.
D. The minimum column width that can be specified for a VARCHAR2 data type column is one.
E. The value for a CHAR data type column is blank-padded to the maximum defined column width.
Answer: ADE
Oracle過去問 1Z1-051参考書 1Z1-051過去問 1Z1-051
NO.13 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: ADE
Oracle認定証 1Z1-051参考書 1Z1-051 1Z1-051 1Z1-051
NO.14 View the Exhibit and examine the structure of the PRODUCTS table.
All products have a list price.
You issue the following command to display the total price of each product after a discount of 25% and a
tax of 15% are applied on it. Freight charges of $100 have to be applied to all the products.
SQL>SELECT prod_name, prod_list_price -(prod_list_price*(25/100))
+(prod_list_price -(prod_list_price*(25/100))*(15/100))+100
AS "TOTAL PRICE"
FROM products;
What would be the outcome if all the parenthese s are removed from the above statement?
A. It produces a syntax error.
B. The result remains unchanged.
C. The total price value would be lower than the correct value.
D. The total price value would be higher than the correct value.
Answer: B
Oracle認証試験 1Z1-051認定試験 1Z1-051 1Z1-051参考書 1Z1-051認証試験 1Z1-051練習問題
NO.15 View the Exhibit and examine the structure of the PRODUCTS table.
You need to generate a report in the following format:
CATEGORIES
5MP Digital Photo Camera's category is Photo
Y Box's category is Electronics
Envoy Ambassador's category is Hardware
Which two queries would give the required output? (Choose two.)
A. SELECT prod_name q'''s category is ' prod_category CATEGORIES
FROM products;
B. SELECT prod_name q'['s ]'category is ' prod_category CATEGORIES
FROM products;
C. SELECT prod_name q'\'s\' ' category is ' prod_category CATEGORIES
FROM products;
D. SELECT prod_name q'<'s >' 'category is ' prod_category CATEGORIES
FROM products;
Answer: CD
Oracle認証試験 1Z1-051練習問題 1Z1-051認証試験
NO.16 View the Exhibit; e xamine the structure of the PROMOTIONS table.
Each promotion has a duration of at least seven days .
Your manager has asked you to generate a report, which provides the weekly cost for each promotion
done to l date.
Which query would achieve the required result?
A. SELECT promo_name, promo_cost/promo_end_date-promo_begin_date/7
FROM promotions;
B. SELECT promo_name,(promo_cost/promo_end_date-promo_begin_date)/7
FROM promotions;
C. SELECT promo_name, promo_cost/(promo_end_date-promo_begin_date/7)
FROM promotions;
D. SELECT promo_name, promo_cost/((promo_end_date-promo_begin_date)/7)
FROM promotions;
Answer: D
Oracle 1Z1-051 1Z1-051 1Z1-051 1Z1-051認定試験 1Z1-051
NO.17 Evaluate the following query:
SQL> SELECT promo_name q'{'s start date was }' promo_begin_date
AS "Promotion Launches"
FROM promotions;
What would be the outcome of the above query?
A. It produces an error because flower braces have been used.
B. It produces an error because the data types are not matching.
C. It executes successfully and introduces an 's at the end of each promo_name in the output.
D. It executes successfully and displays the literal " {'s start date was } " for each row in the output.
Answer: C
Oracle練習問題 1Z1-051 1Z1-051 1Z1-051過去問 1Z1-051
NO.18 Using the CUSTOMERS table, you need to generate a report that shows 50% of each credit
amount in each income level. The report should NOT show any repeated credit amounts in each income
level.
Which query would give the required result?
A. SELECT cust_income_level, DISTINCT cust_credit_limit * 0.50
AS "50% Credit Limit"
FROM customers;
B. SELECT DISTINCT cust_income_level, DISTINCT cust_credit_limit * 0.50
AS "50% Credit Limit"
FROM customers;
C. SELECT DISTINCT cust_income_level ' ' cust_credit_limit * 0.50
AS "50% Credit Limit"
FROM customers;
D. SELECT cust_income_level ' ' cust_credit_limit * 0.50 AS "50% Credit Limit"
FROM customers;
Answer: C
Oracle認証試験 1Z1-051 1Z1-051
NO.19 View the Exhibit and examine the data in the CUSTOMERS table.
Evaluate the following query:
SQL> SELECT cust_name AS "NAME", cust_credit_limit/2 AS MIDPOINT,MIDPOINT+100 AS "MAX
LOWER LIMIT"
FROM customers;
The above query produces an error on execution.
What is the reason for the error?
A. An alias cannot be used in an expression.
B. The a lias NAME should not be enclosed with in double quotation marks .
C. The MIDPOINT+100 expression gives an error because CUST_CREDIT_LIMIT contains NULL
values.
D. The a lias MIDPOINT should be enclosed with in double quotation marks for the
CUST_CREDIT_LIMIT/2 expression .
Answer: A
Oracle 1Z1-051認定試験 1Z1-051 1Z1-051認定証
NO.20 View the E xhibit and examine the data in the EMPLOYEES table.
You want to generate a report showing the total compensation paid to each employee to date.
You issue the following query:
SQL>SELECT ename ' joined on ' hiredate
', the total compensation paid is '
TO_CHAR(ROUND(ROUND(SYSDATE-hiredate)/365) * sal + comm)
"COMPENSATION UNTIL DATE"
FROM employees;
What is the outcome?
A. It generates an error because the alias is not valid.
B. It executes successfully and gives the correct output.
C. It executes successfully but does not give the correct output.
D. It generates an error because the usage of the ROUND function in the expression is not valid.
E. It generates an error because the concatenation operator can be used to combine only two items.
Answer: C
Oracle 1Z1-051 1Z1-051 1Z1-051 1Z1-051練習問題
Oracleの1Z1-051認定試験と言ったら、人々は迷っています。異なる考えがありますが、要約は試験が大変難しいことです。Oracleの1Z1-051認定試験は確かに難しい試験ですが、IT-Passports.com を選んだら、これは大丈夫です。IT-Passports.comのOracleの1Z1-051試験トレーニング資料は受験生としてのあなたが欠くことができない資料です。それは受験生のために特別に作成したものですから、１００パーセントの合格率を保証します。信じないになら、IT-Passports.comのサイトをクリックしてください。購入する人々が大変多いですから、あなたもミスしないで速くショッピングカートに入れましょう。
没有评论:
发表评论