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Code d'Examen: 1Z0-051
Nom d'Examen: Oracle (Oracle Database: SQL Fundamentals I)
Questions et réponses: 292 Q&As
Code d'Examen: 1Z0-451
Nom d'Examen: Oracle (Oracle SOA Foundation Practitioner)
Questions et réponses: 121 Q&As
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NO.1 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
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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 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.4 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
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NO.5 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
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NO.6 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.7 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.8 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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