Display elements in list with for-each - scheme

Question

I have 1 big list of smaller 3-element-lists that look like:
( ("001" "Bob" 80) ("002" "Sam" 85) ("003" "Aaron" 94) etc . . .)
I'm trying to create something like:
No.1: ID=001, Name=’’Bob’’, Grade=80
No.2: ID=002, Name=’’Sam’’, Grade=85
No.3: ID=003, Name=’’Aaron’’, Grade=94
I only have access to display and for-each (no "for" or "printf" functions)
I've been trying to create a for-each function that takes the list and:
pseudo-code:
for-each list in list
display "ID=(car list)"
display "Name ="(cadr list)" "
etc
Any help would be greatly appreciated!

Answer 1

So, your interpreter doesn't have printf after all? that's a shame. We can get the desired output by hand, it's a bit cumbersome but this should work on most Scheme interpreters, notice that an extra procedure is required for keeping track of the index:
(define lst
'(("001" "Bob" 80) ("002" "Sam" 85) ("003" "Aaron" 94)))
(define (add-index lst)
(let loop ((lst lst) (idx 1))
(if (null? lst)
'()
(cons (cons idx (car lst))
(loop (cdr lst) (+ idx 1))))))
(for-each (lambda (e)
(display "No.")
(display (car e))
(display ": ID=")
(display (cadr e))
(display ", Name=’’")
(display (caddr e))
(display "’’, Grade=")
(display (cadddr e))
(newline))
(add-index lst))
It prints the desired result:
No.1: ID=001, Name=’’Bob’’, Grade=80
No.2: ID=002, Name=’’Sam’’, Grade=85
No.3: ID=003, Name=’’Aaron’’, Grade=94

Answer 2

Here's another version. It avoids construction of a temporary list.
(define lst
'(("001" "Bob" 80) ("002" "Sam" 85) ("003" "Aaron" 94)))
(define (print-list lst)
(define (display-one-item item index)
(display "No.")
(display index)
(display ": ID=")
(display (car item))
(display ", Name=’’")
(display (cadr item))
(display "’’, Grade=")
(display (caddr item))
(newline))
(define (helper in index)
(if (not (null? in))
(begin
(display-one-item (car in) index)
(helper (cdr in) (+ index 1))
)))
(helper lst 0))
(print-list lst)