How to handle both integer and string from raw_input?

Question

Still trying to understand python. It is so different than php.

I set the choice to integer, the problem is on my menu I need to use letters as well.

How c

Answer 1

I believe your code can be simplified a bit:

def main():
    while 1:
        print
        print "  Draw a Shape"
        print "  ============"
        print
        print "  1 - Draw a triangle"
        print "  2 - Draw a square"
        print "  3 - Draw a rectangle"
        print "  4 - Draw a pentagon"
        print "  5 - Draw a hexagon"
        print "  6 - Draw an octagon"
        print "  7 - Draw a circle"
        print
        print "  D - Display what was drawn"
        print "  X - Exit"
        print
        choice = raw_input('  Enter your choice: ').strip().upper()
        if choice == 'X':
            break
        elif choice == 'D':
            log.show_log()
        elif choice in ('1', '2', '3', '4', '5', '6', '7'):
            my_shape_num = h_m.how_many()
            if my_shape_num is None:
                continue
            elif my_shape_num == 1:
                d_s.start_point(0, 0)
            else:
                d_s.start_point()
            if choice == '1':
                d_s.draw_triangle(my_shape_num)
            elif choice == '2':
                d_s.draw_square(my_shape_num)
            elif choice == '3':
                d_s.draw_rectangle(my_shape_num)
            elif choice == '4':
                d_s.draw_pentagon(my_shape_num)
            elif choice == '5':
                d_s.draw_hexagon(my_shape_num)
            elif choice == '6':
                d_s.draw_octagon(my_shape_num)
            elif choice == '7':
                d_s.draw_circle(my_shape_num)
            d_s.t.end_fill()
        else:
            print
            print '  Invalid choice: ' + choice
            print