Create the necessary table , add data , calculate the median by hand, run
the function. Use both an odd and an even number of grades. Also do it
with no grades.
Jim
"KDankwah" <kdankwah_at_aol.comnojunk> wrote in message
news:20011214224550.10834.00000295_at_mb-fa.aol.com...
> Can someone plese tell me how to test the function below I want to know if
it
> works.
>
> CREATE OR REPLACE FUNCTION Median_Grade
> (p_course_no SECTION.course_no%TYPE,
> p_section_no SECTION.section_no%TYPE,
> p_grade_type_cd GRADE_TYPE.grade_type_code%TYPE)
> RETURN GRADE.numeric_grade%TYPE
> AS
> CURSOR c_grade IS
> SELECT g.numeric_grade, ROWNUM
> FROM GRADE g, SECTION s
> WHERE g.section_id = s.section_id
> AND s.section_no = p_section_no
> AND s.course_no = p_course_no
> AND g.grade_type_code = p_grade_type_cd
> ORDER BY numeric_grade;
> ---Table to hold out grades------------------------------------
> TYPE t_grade_list_type IS TABLE OF
> c_grade%ROWTYPE
> INDEX BY BINARY_INTEGER;
> t_grade_list t_grade_list_type;
> BEGIN
> -- Loop PL/SQL table with list of grades
> FOR r_grade IN c_grade
> LOOP
> t_grade_list(NVL(t_grade_list.COUNT, 0) +
> 1).numeric_grade := r_grade.numeric_grade;
> END LOOP;
> ----Median calculation starts here-------------------
> IF MOD(t_grade_list.COUNT,2) = 0
> THEN
> --If grades in table is even
> -- take two middle and average the grade with this return-----------:
> RETURN
> (t_grade_list(t_grade_list.COUNT / 2).numeric_grade
> + t_grade_list((t_grade_list.COUNT / 2) +
> 1).numeric_grade) / 2;
> ELSE
> RETURN t_grade_list(TRUNC(t_grade_list.COUNT / 2, 0)
> + 1).numeric_grade;
> END IF;
> END Median_Grade;
> /
>
>
Received on Fri Dec 14 2001 - 22:49:25 CST
