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Code d'Examen: 1Z1-051
Nom d'Examen: Oracle (Oracle Database: SQL Fundamentals I)
Questions et réponses: 254 Q&As
Code d'Examen: 1Z0-805
Nom d'Examen: Oracle (Upgrade to Java SE 7 Programmer)
Questions et réponses: 90 Q&As
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NO.1 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
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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
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NO.3 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
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NO.4 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
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NO.5 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
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NO.6 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
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NO.7 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
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NO.8 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
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