Chapter 4 SQL in PLSQL

1)	Create a table called CHAP4 with two columns ; one is ID (a number) and the second is NAME , which is a varchar2(20).
A1:	Answer: Your answer should look similar to the following: PROMPT Creating Table 'CHAP4' CREATE TABLE chap4 (id NUMBER, name VARCHAR2(20));
2)	Create a sequence called CHAP4_SEQ that increments by units of 5.
A2:	Answer: Your answer should look similar to the following: PROMPT Creating Sequence 'CHAP4_SEQ' CREATE SEQUENCE chap4_seq NOMAXVALUE NOMINVALUE NOCYCLE NOCACHE;
3)	Write a PL/SQL block that performs the following in this order: Declares 2 variables , one for the v_name and one for v_id. The v_name variable can be used throughout the block for holding the name that will be inserted; realize that the value will change in the course of the block. The block then inserts into the table the name of the student that is enrolled in the most classes and uses a sequence for the ID; afterward there is SAVEPOINT A. Then the student with the least enrollments is inserted; afterward there is SAVEPOINT B. Then the instructor who is teaching the maximum number of courses is inserted in the same way. Afterward there is SAVEPOINT C. Using a SELECT INTO statement, hold the value of the instructor in the variable v_id. Undo the instructor insert by use of rollback. Insert the instructor teaching the least amount of courses but do not use the sequence to generate the ID; instead use the value from the first instructor whom you have since undone. Now insert the instructor teaching the most number of courses and use the sequence to populate his ID. Add DBMS_OUTPUT throughout the block to display the values of the variables as they change. (This is good practice for debugging.)
A3:	Answer: Your answer should look similar to the following: DECLARE v_name student.last_name%TYPE; v_id student.student_id%TYPE; BEGIN BEGIN -- A second block is used to capture the possibility of -- multiple students meeting this requirement. -- The exception section will handles this situation SELECT s.last_name INTO v_name FROM student s, enrollment e WHERE s.student_id = e.student_id HAVING COUNT(*) = (SELECT MAX(COUNT(*)) FROM student s, enrollment e WHERE s.student_id = e.student_id GROUP BY s.student_id) GROUP BY s.last_name; EXCEPTION WHEN TOO_MANY_ROWS THEN v_name := 'Multiple Names'; END; INSERT INTO CHAP4 VALUES (CHAP4_SEQ.NEXTVAL, v_name); SAVEPOINT A; BEGIN SELECT s.last_name INTO v_name FROM student s, enrollment e WHERE s.student_id = e.student_id HAVING COUNT(*) = (SELECT MIN(COUNT(*)) FROM student s, enrollment e WHERE s.student_id = e.student_id GROUP BY s.student_id) GROUP BY s.last_name; EXCEPTION WHEN TOO_MANY_ROWS THEN v_name := 'Multiple Names'; END; INSERT INTO CHAP4 VALUES (CHAP4_SEQ.NEXTVAL, v_name); SAVEPOINT B; BEGIN SELECT i.last_name INTO v_name FROM instructor i, section s WHERE s.instructor_id = i.instructor_id HAVING COUNT(*) = (SELECT MAX(COUNT(*)) FROM instructor i, section s WHERE s.instructor_id = i.instructor_id GROUP BY i.instructor_id) GROUP BY i.last_name; EXCEPTION WHEN TOO_MANY_ROWS THEN v_name := 'Multiple Names'; END; SAVEPOINT C; BEGIN SELECT instructor_id INTO v_id FROM instructor WHERE last_name = v_name; EXCEPTION WHEN NO_DATA_FOUND THEN v_id := 999; END; INSERT INTO CHAP4 VALUES (v_id, v_name); ROLLBACK TO SAVEPOINT B; BEGIN SELECT i.last_name INTO v_name FROM instructor i, section s WHERE s.instructor_id = i.instructor_id HAVING COUNT(*) = (SELECT MIN(COUNT(*)) FROM instructor i, section s WHERE s.instructor_id = i.instructor_id GROUP BY i.instructor_id) GROUP BY i.last_name; EXCEPTION WHEN TOO_MANY_ROWS THEN v_name := 'Multiple Names'; END; INSERT INTO CHAP4 VALUES (v_id, v_name); BEGIN SELECT i.last_name INTO v_name FROM instructor i, section s WHERE s.instructor_id = i.instructor_id HAVING COUNT(*) = (SELECT MAX(COUNT(*)) FROM instructor i, section s WHERE s.instructor_id = i.instructor_id GROUP BY i.instructor_id) GROUP BY i.last_name; EXCEPTION WHEN TOO_MANY_ROWS THEN v_name := 'Multiple Names'; END; INSERT INTO CHAP4 VALUES (CHAP4_SEQ.NEXTVAL, v_name); END;