Oracle Database: SQL Fundamentals Iはベンダー公式の認定資格として、スキルの証明に高い信頼性があります。CertJukenの1Z0-051練習問題は公式の出題範囲に沿って構成されており、資格取得への近道となります。

Oracle 1Z0-051 試験概要:

認定ベンダー:	Oracle
試験名:	Oracle Database: SQL Fundamentals I
試験番号:	1Z0-051
合格点:	60%
関連資格:	Oracle Certified Associate (OCA) Oracle PL/SQL Developer Oracle Certified Associate (OCA) Oracle Database 11g Administrator
対応言語:	英語
試験時間:	120 分
試験形式:	複数選択（正しい選択肢を複数選ぶ）, 単一選択
出題数:	64-66
受験料:	US$95
認定の有効期間:	N/A（試験は廃止・置き換え済み）
サンプル問題:
受験方法:	Pearson VUE の試験会場での受験、またはオンライン監視受験（過去の実績に基づく）
前提条件:	正式な前提条件はなし。基礎的な SQL の知識を推奨
公式シラバスのURL:	https://education.oracle.com/

Oracle 1Z0-051 試験シラバストピック:

1Z0-051試験で受験者がよく抱く疑問

Oracle Database: SQL Fundamentals I 認定 1Z0-051 試験問題：

問題 #1

Which two statements are true regarding constraints? (Choose two.)

A. A foreign key cannot contain NULL values.
B. A constraint can be disabled even if the constraint column contains data.
C. A column with the UNIQUE constraint can contain NULL values.
D. All constraints can be defined at the column level as well as the table level.
E. A constraint is enforced only for the INSERT operation on a table.

問題 #2

You work as a database administrator at ABC.com. You study the exhibit carefully. Exhibit:

Which two SQL statements would execute successfully? (Choose two.)

A. UPDATE promotions SET promo_cost = promo_cost+ 100 WHERE TO_CHAR(promo_end_date, 'yyyy') > '2000';
B. SELECT promo_begin_date FROM promotions WHERE TO_CHAR(promo_begin_date,'mon dd yy')='jul 01 98';
C. SELECT TO_CHAR(promo_begin_date,'dd/month')
FROM promotions
WHERE promo_begin_date IN (TO_DATE('JUN 01 98'), TO_DATE('JUL 01 98'));
D. UPDATE promotions SET promo_cost = promo_cost+ 100 WHERE promo_end_date > TO_DATE(SUBSTR('01-JAN-2000',8));

問題 #3

View the Exhibits and examine PRODUCTS and SALES tables.

You issue the following query to display product name and the number of times the product has been sold:
SQL>SELECT p.prod_name, i.item_cnt
FROM (SELECT prod_id, COUNT(*) item_cnt
FROM sales
GROUP BY prod_id) i RIGHT OUTER JOIN products p
ON i.prod_id = p.prod_id;
What happens when the above statement is executed?

A. The statement produces an error because ITEM_CNT cannot be displayed in the outer query.
B. The statement executes successfully and produces the required output.
C. The statement produces an error because the GROUP BY clause cannot be used in a subquery in the FROM clause.
D. The statement produces an error because a subquery in the FROM clause and outer-joins cannot be used together.

問題 #4

CUSTOMER_ID NUMBER(4) NOT NULL
CUSTOMER_NAME VARCHAR2(100) NOT NULL
CUSTOMER_ADDRESS VARCHAR2(150)
CUSTOMER_PHONE VARCHAR2(20)
You need to produce output that states "Dear Customer customer_name, ".
The customer_name data values come from the CUSTOMER_NAME column in the CUSTOMERS table.
Which statement produces this output?

A. SELECT "Dear Customer", customer_name || ',' FROM customers;
B. SELECT 'Dear Customer ' || customer_name || ',' FROM customers;
C. SELECT "Dear Customer " || customer_name || "," FROM customers;
D. SELECT dear customer, customer_name, FROM customers;
E. SELECT 'Dear Customer ' || customer_name ',' FROM customers;
F. SELECT 'Dear Customer ' || customer_name || ',' || FROM customers;

問題 #5

View the Exhibit and examine the description for the CUSTOMERS table.

You want to update the CUST_INCOME_LEVEL and CUST_CREDIT_LIMIT columns for the customer with the CUST_ID 2360. You want the value for the CUST_INCOME_LEVEL to have the same value as that of the customer with the CUST_ID 2560 and the CUST_CREDIT_LIMIT to have the same value as that of the customer with CUST_ID 2566.
Which UPDATE statement will accomplish the task?

A. UPDATE customers SET (cust_income_level,cust_credit_limit) = (SELECT cust_income_level, cust_credit_limit FROM customers WHERE cust_id=2560 AND cust_id=2566) WHERE cust_id=2360;
B. UPDATE customers SET (cust_income_level,cust_credit_limit) = (SELECT cust_income_level, cust_credit_limit
FROM customers
WHERE cust_id IN(2560, 2566)
WHERE cust_id=2360;
C. UPDATE customers SET cust_income_level = (SELECT cust_income_level FROM customers WHERE cust_id = 2560), cust_credit_limit = (SELECT cust_credit_limit FROM customers WHERE cust_id = 2566) WHERE cust_id=2360;
D. UPDATE customers SET (cust_income_level,cust_credit_limit) = (SELECT cust_income_level, cust_credit_limit FROM customers WHERE cust_id=2560 OR cust_id=2566) WHERE cust_id=2360;

解説：