ShikenPASSの問題集でOracle 1Z0-051 最新知識試験の認定資格を取ろう

11g SQL基礎I [11g SQLI]
試験番号:1Z0-051
受験のための準備
Oracle University おすすめ研修サービス
研修コース：Oracle Database 11g: 入門 SQL基礎 I
テスト内容チェックリスト
単一行関数
変換関数および条件式の使用
グループ関数を使用したデータの集計
複数の表のデータの表示
副問合せによる問合せの解決方法
集合演算子の使用方法
データ操作
DDL文を使用した表の作成および管理
その他のスキーマ・オブジェクトの作成

NO.1 Here is the structure and data of the CUST_TRANS table: Exhibit:
Dates are stored in the default date format dd-mm-rr in the CUST_TRANS table.
Which three SQL statements would execute successfully? (Choose three.)
A. SELECT transamt FROM cust_trans WHERE custno > '11';
B. SELECT * FROM cust_trans WHERE transdate='01-JANUARY-07';
C. SELECT custno + 'A' FROM cust_trans WHERE transamt > 2000;
D. SELECT * FROM cust_trans WHERE transdate = '01-01-07';
E. SELECT transdate + '10' FROM cust_trans;
Answer: A,B,E

1Z0-051 技術   

NO.2 Examine the data in the CUST_NAME column of the CUSTOMERS table.
CUST_NAME
Lex De Haan
Renske Ladwig
Jose Manuel Urman
Jason Mallin
You want to extract only those customer names that have three names and display the * symbol in
place of the
first name as follows:
CUST NAME
*** De Haan
**** Manuel Urman
Which two queries give the required output? (Choose two.)
A. SELECT LPAD(SUBSTR(cust_name,INSTR(cust_name,' ')),LENGTH(cust_name),'*')
"CUST NAME" FROM customers
WHERE INSTR(cust_name, ' ',1,2)<>0;
B. SELECT LPAD(SUBSTR(cust_name,INSTR(cust_name,' ')),LENGTH(cust_name)-
INSTR(cust_name,''),'*') "CUST NAME"
FROM customers
WHERE INSTR(cust_name, ' ',-1,-2)<>0;
C. SELECT LPAD(SUBSTR(cust_name,INSTR(cust_name,' ')),LENGTH(cust_name),'*')
"CUST NAME" FROM customers
WHERE INSTR(cust_name, ' ',-1,2)<>0;
D. SELECT LPAD(SUBSTR(cust_name,INSTR(cust_name,' ')),LENGTH(cust_name)-
INSTR(cust_name,' '),'*') "CUST NAME"
FROM customers
WHERE INSTR(cust_name, ' ',1,2)<>0 ;
Answer: A,C

1Z0-051 指導   

NO.3 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 DISTINCT cust_income_level ' ' cust_credit_limit * 0.50 AS "50% Credit Limit"
FROM customers;
B. SELECT cust_income_level, DISTINCT cust_credit_limit * 0.50 AS "50% Credit Limit"
FROM customers;
C. SELECT cust_income_level ' ' cust_credit_limit * 0.50 AS "50% Credit Limit" FROM
customers;
D. SELECT DISTINCT cust_income_level, DISTINCT cust_credit_limit * 0.50 AS "50%
Credit Limit"
FROM customers;
Answer: A

1Z0-051 対象者   
Explanation:
Duplicate Rows Unless you indicate otherwise, SQL displays the results of a query without eliminating
the duplicate rows. To eliminate duplicate rows in the result, include the DISTINCT keyword in the
SELECT clause immediately after the SELECT keyword. You can specify multiple columns after the
DISTINCT qualifier. The DISTINCT qualifier affects all the selected columns, and the result is every
distinct combination of the columns.

NO.4 Examine the structure of the TRANSACTIONS table:
Name Null Type
TRANS_ID NOT NULL NUMBER(3)
CUST_NAME VARCHAR2(30)
TRANS_DATE TIMESTAMP
TRANS_AMT NUMBER(10,2)
You want to display the date, time, and transaction amount of transactions that where done before
12 noon.
The value zero should be displayed for transactions where the transaction amount has not been
entered.
Which query gives the required result?
A. SELECT TO_CHAR(trans_date,'dd-mon-yyyy hh24:mi:ss'),
NVL(TO_CHAR(trans_amt,' $99999999D99'),0)
FROM transactions
WHERE TO_CHAR(trans_date,'hh24') < 12;
B. SELECT TO_CHAR(trans_date,'dd-mon-yyyy hh24:mi:ss'),
COALESCE(TO_NUMBER(trans_amt,' $99999999.99'),0)
FROM transactions
WHERE TO_DATE(trans_date,'hh24') < 12;
C. SELECT TO_CHAR(trans_date,'dd-mon-yyyy hh24:mi:ss'),
TO_CHAR(trans_amt,'$99999999D99')
FROM transactions
WHERE TO_NUMBER(TO_DATE(trans_date,'hh24')) < 12 AND
COALESCE(trans_amt,NULL)<>NULL;
D. SELECT TO_DATE (trans_date,'dd-mon-yyyy hh24:mi:ss'),
NVL2(trans_amt,TO_NUMBER(trans_amt,'$99999999.99'), 0)
FROM transactions
WHERE TO_DATE(trans_date,'hh24') < 12;
Answer: A

1Z0-051 問題   

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