Lab 15.1 Exercise Answers

15.1.1 Answers

a)	Complete the code for the parameter cursor that was begun in the preceding example. Include a DBMS_OUTPUT line that displays the zipcode , city, and state. This is identical to the process you have already used in a FOR CURSOR loop, only now, when you open the cursor, you pass a parameter.
A1:	Answer: Your block should look like this: -- ch15_17a.sql DECLARE CURSOR c_zip (p_state IN zipcode.state%TYPE) IS SELECT zip, city, state FROM zipcode WHERE state = p_state BEGIN FOR r_zip IN c_zip('NJ') LOOP ... DBMS_OUTPUT.PUT_LINE(r_zip.city ' 'r_zip.zip'); END LOOP; END; To complete the block, the cursor declaration must be surrounded by DECLARE and BEGIN. The cursor is opened by passing the parameter "NJ," and then, for each iteration of the cursor loop, the zipcode and the city are displayed by using the built-in package DBMS_OUTPUT.
b)	The following PL/SQL code is complex. It involves all of the topics covered so far in this chapter. There is a nested cursor with three levels, meaning a grandparent cursor, a parent cursor, and a child cursor. Before running this script, review the code and identify the levels of nesting in the code. When you describe each level of the code, explain what parameters are being passed into the cursor and why. What do you think the result will be from running this statement? -- ch15_1a.sql SET SERVEROUTPUT ON 1 DECLARE 2 CURSOR c_student IS 3 SELECT first_name, last_name, student_id 4 FROM student 5 WHERE last_name LIKE 'J%'; 6 CURSOR c_course 7 (i_student_id IN student.student_id%TYPE) 8 IS 9 SELECT c.description, s.section_id sec_id 10 FROM course c, section s, enrollment e 11 WHERE e.student_id = i_student_id 12 AND c.course_no = s.course_no 13 AND s.section_id = e.section_id; 14 CURSOR c_grade(i_section_id IN section.section_id%TYPE, 15 i_student_id IN student.student_id%TYPE) 16 IS 17 SELECT gt.description grd_desc, 18 TO_CHAR 19 (AVG(g.numeric_grade), '999.99') num_grd 20 FROM enrollment e, 21 grade g, grade_type gt 22 WHERE e.section_id = i_section_id 23 AND e.student_id = g.student_id 24 AND e.student_id = i_student_id 25 AND e.section_id = g.section_id 26 AND g.grade_type_code = gt.grade_type_code 27 GROUP BY gt.description ; 28 BEGIN 29 FOR r_student IN c_student 30 LOOP 31 DBMS_OUTPUT.PUT_LINE(CHR(10)); 32 DBMS_OUTPUT.PUT_LINE(r_student.first_name 33 ' 'r_student.last_name); 34 FOR r_course IN c_course(r_student.student_id) 35 LOOP 36 DBMS_OUTPUT.PUT_LINE ('Grades for course :' 37 r_course.description); 38 FOR r_grade IN c_grade(r_course.sec_id, 39 r_student.student_id) 40 LOOP 41 DBMS_OUTPUT.PUT_LINE(r_grade.num_grd 42 ' 'r_grade.grd_desc); 43 END LOOP; 44 END LOOP; 45 END LOOP; 46 END;
A2:	Answer: The grandparent cursor, c_student , is declared in lines 2 “5. It takes no parameters and is a collection of students with a last name beginning with J. The parent cursor is declared in lines 6 “13. The parent cursor, c_course , takes in the parameter of the student_ID to generate a list of courses taken by that student. The child cursor, c_grade , is declared in lines 14 “27. It takes in two parameters, both the section_id and the student_id . In this way it can generate an average of the different grade types for that student for that course. The grandparent cursor loop begins on line 29, and only the student name is displayed with DBMS_OUTPUT. The parent cursor loop begins on line 35. It takes the parameter of the student_id from the grandparent cursor. Only the description of the course is displayed. The child cursor loop begins on line 40. It takes in the parameter of the section_id from the parent cursor and the student_id from the grandparent cursor. The grades are then displayed. The grandparent cursor loop ends on line 45, the parent cursor on line 44, and, finally, the child on line 43.
c)	Now run the code and see if you were correct. Analyze the code line by line and explain what is being processed and then displayed for each line.
A3:	Answer: The output will be a student name, followed by the courses he or she is taking and the average grade he or she has earned for each grade type. If you did not get the correct answer, try commenting out different sections of the block and see what happens. This will help you to understand what is happening in each step.

15.1.2 Answers

a)

In the example just given, where should the COMMIT be placed? What are the issues involved in deciding where to place a COMMIT in this example?

A1:

Answer: Placing a COMMIT after each update can be costly. But if there are a lot of updates and the COMMIT comes after the block loop, then there is a risk of a rollback segment not being large enough. Normally, the COMMIT would go after the loop, except when the transaction count is high, and then you might want to code something that does a COMMIT for each 10,000 records. If this were part of a large procedure, you may want to put a SAVEPOINT after the loop. Then, if you need to rollback this update at a later point, it would be an easy task.

b)	What do you think will happen if you run the code in this example? After making your analysis, run the code, and then perform a SELECT statement to determine if your guess is correct.
A1:	Answer: The final_grade for all students enrolled in course 135 will be updated to 90. There are two cursors here. One cursor captures the students who are enrolled in course 135 into the active set. The other cursor takes the student_id and the section_id from this active set and selects the corresponding final_grade from the enrollment table and locks the entire enrollment table. The enrollment cursor loop is begun first, and then it passes the student_id and the section_id as an IN parameters for the second cursor loop of the c_grade cursor, which performs the update.
c)	Where should the COMMIT go in the preceding example? Explain the considerations.
A2:	Answer: The COMMIT should go immediately after the update to ensure that each update is committed into the database.

d)	What changes to the database will take place if the preceding example is run? Explain specifically what is being locked as well as when it is locked and when it is released.
A1:	Answer: The phone numbers of students living in Brooklyn are being updated to change the area code to 718. The cursor declaration is only locking the phone column of the student table. The lock is never released because there is no COMMIT or ROLLBACK statement.

15.1.3 Answers

a)	Compare the last two examples. Explain their similarities and differences. What has been altered by using the WHERE CURRENT OF clause? What is the advantage of doing this?
A1:	Answer: These two statements perform the same update. The WHERE CURRENT OF clause allows you to eliminate a match in the UPDATE statement, because the update is being performed for the current record of the cursor only.