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Oracle Database: SQL Fundamentals Iの試験情報

Oracle 1Z0-051 試験概要:

1Z0-051試験の出題範囲

Oracle 1Z0-051 試験シラバストピック:

Oracle Database: SQL Fundamentals Iに関する受験前Q&A

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

問題 #1

View the Exhibit and examine the structure of ORDERS and CUSTOMERS tables. There is only one customer with the cus_last_name column having value Roberts. Which INSERT statement should be used to add a row into the ORDERS table for the customer whose CUST_LAST_NAME is Roberts and CREDIT_LIMIT is 600?

A. INSERT INTO(SELECT o.order_id. o.order_date.o.order_modex.customer_id.
o.ordertotal
FROM orders o. customers c
WHERE o.customer_id = c.customerid
AND c.cust_la$t_name-RoberTs' ANDc.credit_liinit=600)
VALUES (L'10-mar-2007\ 'direct'.( SELECT customer_id
FROM customers
WHERE cust_last_iiame='Roberts' AND
credit_limit=600). 1000);
B. INSERT INTO orders VALUES (l.'10-mar-2007\ 'direct'. (SELECT customerid FROM customers WHERE cust_last_iiame='Roberts' AND credit_limit=600). 1000);
C. INSERT INTO orders (order_id.order_date.order_mode.
(SELECT customer_id
FROM customers
WHERE cust_last_iiame='Roberts' AND
credit_limit=600).order_total)
VALUES(l.'10-mar-2007\ 'direct'. &customer_id. 1000):
D. INSERT INTO orders (order_id.order_date.order_mode. (SELECT customer id FROM customers WHERE cust_last_iiame='Roberts' AND redit_limit=600).order_total)
VALUES(L'10-mar-2007'. 'direct', &&customer_id, 1000):

問題 #2

Evaluate the following SQL statement:

Which statement is true regarding the outcome of the above query?

A. It executes successfully but ignores the ORDER BY clause because it is not located at the end of the compound statement
B. It executes successfully and displays rows in the descending order of PROMO_CATEGORY
C. It produces an error because positional notation cannot be used in the ORDER BY clause with SET operators
D. It produces an error because the ORDER BY clause should appear only at the end of a compound query-that is, with the last SELECT statement

問題 #3

Examine the description of the EMPLOYEES table:
EMP_ID NUMBER(4) NOT NULL
LAST_NAME VARCHAR2(30) NOT NULL
FIRST_NAME VARCHAR2(30)
DEPT_ID NUMBER(2)
Which statement produces the number of different departments that have employees with last name Smith?

A. SELECT COUNT (dept_id) FROM employees WHERE last_name='Smith';
B. SELECT COUNT(DISTINCT dept_id) FROM employees WHERE last_name='Smith';
C. SELECT UNIQUE(dept_id) FROM employees WHERE last_name='Smith';
D. SELECT COUNT(*) FROM employees WHERE last_name='Smith';
E. SELECT DISTINCT(COUNT(dept_id)) FROM employees WHERE last_name='Smith';

問題 #4

You need to write a SQL statement that returns employee name, salary, department ID, and maximum salary earned in the department of the employee for all employees who earn less than the maximum salary in their department.
Which statement accomplishes this task?

A. SELECT a.emp_name, a.sal, b.dept_id, MAX(sal) FROM employees a, departments b WHERE a.dept_id = b.dept_id AND a.sal < MAX(sal) GROUP BY b.dept_id;
B. SELECT a.emp_name, a.sal, a.dept_id, b.maxsal FROM employees a, (SELECT dept_id, MAX(sal) maxsal FROM employees GROUP BY dept_id) b WHERE a.dept_id = b.dept_id AND a.sal < b.maxsal;
C. SELECT emp_name, sal, dept_id, maxsal FROM employees, (SELECT dept_id, MAX(sal) maxsal FROM employees GROUP BY dept_id) WHERE a.sal < maxsal;
D. SELECT a.emp_name, a.sal, a.dept_id, b.maxsal FROM employees a WHERE a.sal < (SELECT MAX(sal) maxsal FROM employees b GROUP BY dept_id);

問題 #5

You need to create a table with the following column specifications:
1.Employee ID (numeric data type) for each employee
2.Employee Name (character data type) that stores the employee name
3.Hire date, which stores the date of joining the organization for each employee
4.Status (character data type), that contains the value 'ACTIVE' if no data is entered
5.Resume (character large object [CLOB] data type), which contains the resume submitted by the employee
Which is the correct syntax to create this table?

A. CREATE TABLE 1_EMP
(emp_id NUMBER(4),
emp_name VARCHAR2(25),
start_date DATE,
emp_status VARCHAR2(10) DEFAULT 'ACTIVE',
resume CLOB);
B. CREATE TABLE EMP_1
(emp_id NUMBER(4),
emp_name VARCHAR2(25),
start_date DATE,
emp_status VARCHAR2(10) DEFAULT "ACTIVE",
resume CLOB);
C. CREATE TABLE EMP_1
(emp_id NUMBER(4),
emp_name VARCHAR2(25),
start_date DATE,
e_status VARCHAR2(10) DEFAULT 'ACTIVE',
resume CLOB(200));
D. CREATE TABLE EMP_1
(emp_id NUMBER,
emp_name VARCHAR2(25),
start_date DATE,
emp_status VARCHAR2(10) DEFAULT 'ACTIVE',
resume CLOB);

解説：