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Introduction to Oracle9i: SQL

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1z0-007题库部分演示（DEMO）

1. What does the FORCE option for creating a view do?
A.creates a view with constraints
B.creates a view even if the underlying parent table has constraints
C.creates a view in another schema even if you don't have privileges
D.creates a view regardless of whether or not the base tables exist
Answer: D

2. What are two reasons to create synonyms? (Choose two.)
A.You have too many tables.
B.Your tables are too long.
C.Your tables have difficult names.
D.You want to work on your own tables.
E.You want to use another schema's tables.
F.You have too many columns in your tables.
Answer: CE

3. The STUDENT_GRADES table has these columns:
STUDENT_ID NUMBER(12)
SEMESTER_END DATE
GPA NUMBER(4,3)
The registrar requested a report listing the students' grade point averages (GPA) sorted from highest grade point average to lowest.
Which statement produces a report  that  displays  the student ID and GPA in the   sorted order  requested by the registrar?
A.SELECT student_id, gpa
FROM student_grades
ORDER BY gpa ASC;
B.SELECT student_id, gpa
FROM student_grades
SORT ORDER BY gpa ASC;
C.SELECT student_id, gpa
FROM student_grades
SORT ORDER BY gpa;
D.SELECT student_id, gpa
FROM student_grades
ORDER BY gpa;
E.SELECT student_id, gpa
FROM student_grades
SORT ORDER BY gpa DESC;
F.SELECT student_id, gpa
FROM student_grades
ORDER BY gpa DESC;
Answer: F

4. In which three cases would you use the USING clause? (Choose three.)
A.You want to create a nonequijoin.
B.The tables to be joined have multiple NULL columns.
C.The tables to be joined have columns of the same name and different data types.
D.The tables to be joined have columns with the same name and compatible data types.
E.You want to use a NATURAL join, but you want to restrict the number of columns in the join condition.
Answer: CDE

5. The CUSTOMERS table has these columns:
CUSTOMER_ID NUMBER(4) NOT NULL
CUSTOMER_NAME VARCHAR2(100) NOT NULL
STREET_ADDRESS VARCHAR2(150)
CITY_ADDRESS VARCHAR2(50)
STATE_ADDRESS VARCHAR2(50)
PROVINCE_ADDRESS VARCHAR2(50)
COUNTRY_ADDRESS VARCHAR2(50)
POSTAL_CODE VARCHAR2(12)
CUSTOMER_PHONE VARCHAR2(20)
The CUSTOMER_ID column is the primary key for the table.
You   need   to   determine   how   dispersed   your   customer   base   is.   Which   expression   finds   the   number   of   different countries represented in the CUSTOMERS table?
A.COUNT(UPPER(country_address))
B.COUNT(DIFF(UPPER(country_address)))
C.COUNT(UNIQUE(UPPER(country_address)))
D.COUNT DISTINCT UPPER(country_address)
E.COUNT(DISTINCT (UPPER(country_address)))
Answer: E

6. Click the Exhibit button and examine the data in the EMPLOYEES table.

Which three subqueries work? (Choose three.)
A.SELECT *
FROM employees
where salary > (SELECT MIN(salary)
FROM employees
GROUP BY department_id);
B.SELECT *
FROM employees
WHERE salary = (SELECT AVG(salary)
FROM employees
GROUP BY department_id);
C.SELECT distinct department_id
FROM employees
WHERE salary > ANY (SELECT AVG(salary)
FROM employees
GROUP BY department_id);
D.SELECT department_id
FROM employees
WHERE salary > ALL (SELECT AVG(salary)
FROM employees
GROUP BY department_id);
E.SELECT last_name
FROM employees
WHERE salary > ANY (SELECT MAX(salary)
FROM employees
GROUP BY department_id);
F.SELECT department_id
FROM employees
WHERE salary > ALL (SELECT AVG(salary)
FROM employees
GROUP BY AVG(SALARY));
Answer: CDE
7. A SELECT statement can be used to perform these three functions:
1. Choose rows from a table.
2. Choose columns from a table.
3. Bring together data that is stored in different tables by creating a link between them.
Which set of keywords describes these capabilities?
A.difference, projection, join
B.selection, projection, join
C.selection, intersection, join
D.intersection, projection, join
E.difference, projection, product
Answer: B

8. Evaluate this SQL statement:
SELECTe.EMPLOYEE_ID,e.LAST_NAME,e.DEPARTMENT_ID,
d.DEPARTMENT_NAME FROM EMPLOYEES e, DEPARTMENTS d
WHERE e.DEPARTMENT_ID = d.DEPARTMENT_ID;
In the statement, which capabilities of a SELECT statement are performed?
A.selection, projection, join
B.difference, projection, join
C.selection, intersection, join
D.intersection, projection, join
E.difference, projection, product
Answer: A

9. Evaluate this SQL statement:
SELECT e.employee_id, (.15* e.salary) + (.5 * e.commission_pct)
+ (s.sales_amount * (.35 * e.bonus)) AS CALC_VALUE
FROM employees e, sales s
WHERE e.employee_id = s.emp_id;
What will happen if you remove all the parentheses from the calculation?
A.The value displayed in the CALC_VALUE column will be lower.
B.The value displayed in the CALC_VALUE column will be higher.
C.There will be no difference in the value displayed in the CALC_VALUE column.
D.An error will be reported.
Answer: C

10. Which SQL statement generates the alias Annual Salary for the calculated column SALARY*12?
A.SELECT ename, salary*12 'Annual Salary'
FROM employees;
B.SELECT ename, salary*12 "Annual Salary"
FROM employees;
C.SELECT ename, salary*12 AS Annual Salary
FROM employees;
D.SELECT ename, salary*12 AS INITCAP("ANNUAL SALARY")
FROM employees
Answer: B

11. Evaluate this SQL statement:
SELECT ename, sal, 12*sal+100
FROM emp;
The SAL column stores the monthly salary of the employee. Which change must be made to the above syntax to calculate the annual compensation as "monthly salary plus a monthly bonus of $100, multiplied by 12"?
A.No change is required to achieve the desired results.
B.SELECT ename, sal, 12*(sal+100)
FROM emp;
C.SELECT ename, sal, (12*sal)+100
FROM emp;
D.SELECT ename, sal+100,*12
FROM emp;
Answer: B

12. The CUSTOMERS table has these columns:
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,
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;
Answer: D

13. Which two are attributes of iSQL*Plus? (Choose two.)
A.iSQL*Plus commands cannot be abbreviated.
B.iSQL*Plus commands are accessed from a browser.
C.iSQL*Plus commands are used to manipulate data in tables.
D.iSQL*Plus commands manipulate table definitions in the database.
E.iSQL*Plus is the Oracle proprietary interface for executing SQL statements.
Answer: BE

14. Which is an iSQL*Plus command?
A.INSERT
B.UPDATE
C.SELECT
D.DESCRIBE
E.DELETE
F.RENAME
Answer: D

15. Which are iSQL*Plus commands? (Choose all that apply.)
A.INSERT
B.UPDATE
C.SELECT
D.DESCRIBE
E.DELETE
F.RENAME
Answer: D

16. Which two statements are true about constraints? (Choose two.)
A.The UNIQUE constraint does not permit a null value for the column.
B.A UNIQUE index gets created for columns with PRIMARY KEY and UNIQUE constraints.
C.The PRIMARY KEY and FOREIGN KEY constraints create a UNIQUE index.
D.The NOT NULL constraint ensures that null values are not permitted for the column.
Answer: BD

17. Which three statements correctly describe the functions and use of constraints? (Choose three.)
A.Constraints provide data independence.
B.Constraints make complex queries easy.
C.Constraints enforce rules at the view level.
D.Constraints enforce rules at the table level.
E.Constraints prevent the deletion of a table if there are dependencies.
F.Constraints prevent the deletion of an index if there are dependencies.
Answer: CDE

18. Which SQL statement defines a FOREIGN KEY constraint on the DEPTNO column of the EMP table?
A.CREATE TABLE EMP
(empno NUMBER(4),
ename VARCHAR2(35),
deptno NUMBER(7,2) NOT NULL,
CONSTRAINT emp_deptno_fk FOREIGN KEY deptno
REFERENCES dept deptno);
B.CREATE TABLE EMP
(empno NUMBER(4),
ename VARCHAR2(35),
deptno NUMBER(7,2)
CONSTRAINT emp_deptno_fk REFERENCES dept (deptno));
C.CREATE TABLE EMP
(empno NUMBER(4),
ename VARCHAR2(35),
deptno NUMBER(7,2) NOT NULL,
CONSTRAINT emp_deptno_fk REFERENCES dept (deptno)
FOREIGN KEY (deptno));
D.CREATE TABLE EMP
(empno NUMBER(4),
ename VARCHAR2(35),
deptno NUMBER(7,2) FOREIGN KEY
CONSTRAINT emp_deptno_fk REFERENCES dept (deptno));
Answer: B

19. Which view should a user query to display the columns associated with the constraints on a table owned by the user?
A.USER_CONSTRAINTS
B.USER_OBJECTS
C.ALL_CONSTRAINTS
D.USER_CONS_COLUMNS
E.USER_COLUMNS
Answer: D

20. You need to design a student registration database that contains several tables storing academic information. The STUDENTS  table stores information about a   student. The STUDENT_GRADES table stores information about the student's grades. Both of the tables have a column named STUDENT_ID. The STUDENT_ID column in
the STUDENTS table is a primary key. You need to create a foreign key on the STUDENT_ID column of the STUDENT_GRADES table that points to the STUDENT_ID column of the STUDENTS table. Which statement creates the foreign key?
A.CREATE TABLE student_grades
(student_id NUMBER(12),
semester_end DATE,
gpa NUMBER(4,3),
CONSTRAINT student_id_fk REFERENCES (student_id)
FOREIGN KEY students(student_id));
B.CREATE TABLE student_grades
(student_id NUMBER(12),
semester_end DATE,
gpa NUMBER(4,3),
student_id_fk FOREIGN KEY (student_id)
REFERENCES students(student_id));
C.CREATE TABLE student_grades
(student_id NUMBER(12),
semester_end DATE,
gpa NUMBER(4,3),
CONSTRAINT FOREIGN KEY (student_id)
REFERENCES students(student_id));
D.CREATE TABLE student_grades
(student_id NUMBER(12),
semester_end DATE,
gpa NUMBER(4,3),
CONSTRAINT student_id_fk FOREIGN KEY (student_id)
REFERENCES students(student_id));
Answer: D

21. You need to modify the STUDENTS table to add a primary key on the STUDENT_ID column. The table is currently empty. Which statement accomplishes this task?
A.ALTER TABLE students
ADD PRIMARY KEY student_id;
B.ALTER TABLE students
ADD CONSTRAINT PRIMARY KEY (student_id);
C.ALTER TABLE students
ADD CONSTRAINT stud_id_pk PRIMARY KEY student_id;
D.ALTER TABLE students
ADD CONSTRAINT stud_id_pk PRIMARY KEY (student_id);
E.ALTER TABLE students
MODIFY CONSTRAINT stud_id_pk PRIMARY KEY (student_id);
Answer: D

22. Which statement adds a constraint that ensures the CUSTOMER_NAME column of the CUSTOMERS table holds a value?
A.ALTER TABLE customers
ADD CONSTRAINT cust_name_nn CHECK customer_name IS NOT NULL;
B.ALTER TABLE customers
MODIFY CONSTRAINT cust_name_nn CHECK customer_name IS NOT NULL;
C.ALTER TABLE customers
MODIFY customer_name CONSTRAINT cust_name_nn NOT NULL;
D.ALTER TABLE customers
MODIFY customer_name CONSTRAINT cust_name_nn IS NOT NULL;
E.ALTER TABLE customers
MODIFY name CONSTRAINT cust_name_nn NOT NULL;
F.ALTER TABLE customers
ADD CONSTRAINT cust_name_nn CHECK customer_name NOT NULL;
Answer: C

23. Which two statements about views are true? (Choose two.)
A.A view can be created as read only.
B.A view can be created as a join on two or more tables.
C.A view cannot have an ORDER BY clause in the SELECT statement.
D.A view cannot be created with a GROUP BY clause in the SELECT statement.
E.A view must have aliases defined for the column names in the SELECT statement.
Answer: AB

24. You need to perform certain data manipulation operations through a view called EMP_DEPT_VU, which you previously created. You want to look at the definition of the view (the SELECT statement on which the view was created.)
How do you obtain the definition of the view?
A.Use the DESCRIBE command on the EMP_DEPT_VU view.
B.Use the DEFINE VIEW command on the EMP_DEPT_VU view.
C.Use the DESCRIBE VIEW command on the EMP_DEPT_VU view.
D.Query the USER_VIEWS data dictionary view to search for the EMP_DEPT_VU view.
E.Query the USER_SOURCE data dictionary view to search for the EMP_DEPT_VU view.
F.Query the USER_OBJECTS data dictionary view to search for the EMP_DEPT_VU view.
Answer: D

25. Examine the structure of the EMP_DEPT_VU view:
Column Name Type Remarks
EMPLOYEE_ID NUMBER From the EMPLOYEES table
EMP_NAME VARCHAR2(30) From the EMPLOYEES table
JOB_ID VARCHAR2(20) From the EMPLOYEES table
SALARY NUMBER From the EMPLOYEES table
DEPARTMENT_ID NUMBER From the DEPARTMENTS table
DEPT_NAME VARCHAR2(30) From the DEPARTMENTS table
Which SQL statement produces an error?
A.SELECT *
 FROM emp_dept_vu;
B.SELECT department_id, SUM(salary)
  FROM emp_dept_vu
 GROUP BY department_id;
C. SELECT department_id, job_id, AVG(salary)
  FROM emp_dept_vu
   GROUP BY department_id, job_id;
D.SELECT job_id, SUM(salary)
 FROM emp_dept_vu
WHERE department_id IN (10,20)
 GROUP BY job_id
  HAVING SUM(salary) > 20000;
E.None of the statements produce an error; all are valid.
Answer: E

26. What is necessary for your query on an existing view to execute successfully?
A.The underlying tables must have data.
B.You need SELECT privileges on the view.
C.The underlying tables must be in the same schema.
D.You need SELECT privileges only on the underlying tables.
Answer: B

27. Examine the structure of the EMPLOYEES table:
Column name Data type Remarks
EMPLOYEE_ID NUMBER NOT NULL, Primary Key
EMP_NAME VARCHAR2(30)
JOB_ID VARCHAR2(20) NOT NULL
SAL NUMBER
MGR_ID NUMBER References EMPLOYEE_ID column
DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID column
of the DEPARTMENTS table
You need to create a view called EMP_VU that allows the users to insert rows through the view. Which SQL statement, when used to create the EMP_VU view, allows the users to insert rows?
A.CREATE VIEW emp_vu AS
 SELECT employee_id, emp_name,
  department_id
 FROM employees
  WHERE mgr_id IN (102, 120);
B.CREATE VIEW emp_vu AS
 SELECT employee_id, emp_name, job_id,
department_id
 FROM employees
  WHERE mgr_id IN (102, 120);
C.CREATE VIEW emp_vu AS
 SELECT department_id, SUM(sal) TOTALSAL
  FROM employees
 WHERE mgr_id IN (102, 120)
 GROUP BY department_id;
D.CREATE VIEW emp_vu AS
 SELECT employee_id, emp_name, job_id,
  DISTINCT department_id
  FROM employees;
Answer: B

28. You need to create a view EMP_VU. The view should allow the users to manipulate the records of only the employees that are working for departments 10 or 20.
Which SQL statement would you use to create the view EMP_VU?
A.CREATE VIEW emp_vu AS
 SELECT *
 FROM employees
  WHERE department_id IN (10,20);
B.CREATE VIEW emp_vu AS
SELECT *
  FROM employees
 WHERE department_id IN (10,20)
  WITH READ ONLY;
C.CREATE VIEW emp_vu AS
 SELECT *
 FROM employees
 WHERE department_id IN (10,20)
 WITH CHECK OPTION;
D.CREATE FORCE VIEW emp_vu AS
  SELECT *
FROM employees
 WHERE department_id IN (10,20);
E.CREATE FORCE VIEW emp_vu AS
  SELECT *
  FROM employees
  WHERE department_id IN (10,20)
  NO UPDATE;
Answer: C

29. What is true about updates through a view?
A.You cannot update a view with group functions.
B.When you update a view group functions are automatically computed.
C.When you update a view only the constraints on the underlying table will be in effect.
D.When you update a view the constraints on the views always override the constraints on the underlying tables.
Answer: A

30. Examine the structure of the EMPLOYEES table:
EMPLOYEE_ID NUMBER NOT NULL, Primary Key
EMP_NAME VARCHAR2(30)
JOB_ID NUMBER
SAL NUMBER
MGR_ID NUMBER References EMPLOYEE_ID column
DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID column
of the DEPARTMENTS table
You   created   a   sequence   called   EMP_ID_SEQ   in   order   to   populate   sequential   values   for   the   EMPLOYEE_ID
column of the EMPLOYEES table.
Which two statements regarding the EMP_ID_SEQ sequence are true? (Choose two.)
A.You cannot use the EMP_ID_SEQ sequence to populate the JOB_ID column.
B.The EMP_ID_SEQ sequence is invalidated when you modify the EMPLOYEE_ID column.
C.The EMP_ID_SEQ sequence is not affected by modifications to the
EMPLOYEES table.
D.Any other column of NUMBER data type in your schema can use the EMP_ID_SEQ sequence.
E.The EMP_ID_SEQ sequence is dropped automatically when you drop the EMPLOYEES table.
F.The EMP_ID_SEQ sequence is dropped automatically when you drop the EMPLOYEE_ID column.
Answer: CD

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