Answer: Your answer should look similar to the following: SET SERVEROUTPUT ON DECLARE v_student_id NUMBER := &sv_student_id; v_first_name VARCHAR2(30) := '&sv_first_name'; v_last_name VARCHAR2(30) := '&sv_last_name'; v_zip CHAR(5) := '&sv_zip'; v_name VARCHAR2(50); BEGIN SELECT first_name||' '||last_name INTO v_name FROM student WHERE student_id = v_student_id; DBMS_OUTPUT.PUT_LINE ('Student '||v_name|| ' is a valid student'); EXCEPTION WHEN NO_DATA_FOUND THEN DBMS_OUTPUT.PUT_LINE ('This student does not exist, and will be '|| 'added to the STUDENT table'); INSERT INTO student (student_id, first_name, last_name, zip, registration_date, created_by, created_date, modified_by, modified_date) VALUES (v_student_id, v_first_name, v_last_name, v_zip, SYSDATE, USER, SYSDATE, USER, SYSDATE); COMMIT; END; This script accepts a value for student's ID from a user. For a given student ID, it determines the student's name via the SELECT INTO statement and displays it on the screen. If the value provided by the user is not a valid student ID, the control of execution is passed to the exception-handling section of the block, where the NO_DATA_FOUND exception is raised. As a result, the message 'This student does not exist…' is displayed on the screen, and a new record is inserted in the STUDENT table. To test this script fully, consider running it for two values of student ID. Only one value should correspond to an existing student ID. It is important to note that a valid zipcode should be provided for both runs. Why do you think this is necessary? When 319 is provided for the student ID (it is a valid student ID), this exercise produces the following output: Enter value for sv_student_id: 319 old 2: v_student_id NUMBER := &sv_student_id; new 2: v_student_id NUMBER := 319; Enter value for sv_first_name: John old 3: v_first_name VARCHAR2(30) := '&sv_first_name'; new 3: v_first_name VARCHAR2(30) := 'John'; Enter value for sv_last_name: Smith old 4: v_last_name VARCHAR2(30) := '&sv_last_name'; new 4: v_last_name VARCHAR2(30) := 'Smith'; Enter value for sv_zip: 07421 old 5: v_zip CHAR(5) := '&sv_zip'; new 5: v_zip CHAR(5) := '07421'; Student George Eakheit is a valid student PLSQL procedure successfully completed. Notice that the name displayed by the script does not correspond to the name entered at runtime. Why do you think this occurs? When 555 is provided for the student ID (it is not a valid student ID), this exercise produces the following output: Enter value for sv_student_id: 555 old 2: v_student_id NUMBER := &sv_student_id; new 2: v_student_id NUMBER := 555; Enter value for sv_first_name: John old 3: v_first_name VARCHAR2(30) := '&sv_first_name'; new 3: v_first_name VARCHAR2(30) := 'John'; Enter value for sv_last_name: Smith old 4: v_last_name VARCHAR2(30) := '&sv_last_name'; new 4: v_last_name VARCHAR2(30) := 'Smith'; Enter value for sv_zip: 07421 old 5: v_zip CHAR(5) := '&sv_zip'; new 5: v_zip CHAR(5) := '07421'; This student does not exist, and will be added to the STUDENT table PLSQL procedure successfully completed. Next, you can select this new record from the STUDENT table as follows: SELECT student_id, first_name, last_name FROM student WHERE student_id = 555; STUDENT_ID FIRST_NAME LAST_NAME ---------- ------------------------- ---------------- 555 John Smith | 
