list handling

Marcel,
Its impossible to tell what your lists (sgl, bln, el1) originally contained
but the following function will flatten any level of complexity to a single
level list (a flat list)

(defun flatlist (lst) ;;D. C. Broad, Jr.
(cond
((null lst) nil)
((listp (car lst))
(append (flatlist (car lst))
(flatlist (cdr lst))))
(t (cons (car lst) (flatlist (cdr lst))))))

;|Example execution:
$ (flatlist '(((1 2 (3))(4 5))(((6))7)8))
(1 2 3 4 5 6 7 8)
|;

_$

 

Question,

I have;
(append sgl (list (list bln (cdr (assoc 1 el1)) (cddr (assoc (substr bln 1
5) sgl)))))
wich results in;
(("B_PRS" "xxx" "1" "2" "3" "4") ("E_SKL" "xxx" "1" "2" "3" "4") ("E_WCD"
"xxx" "1" "2" "3" "4") ("E_WCD0101" "blabla" ("0.00" "0.0" "00" "032")))

But I want it to be; (without the parentheses)
(("B_PRS" "xxx" "1" "2" "3" "4") ("E_SKL" "xxx" "1" "2" "3" "4") ("E_WCD"
"xxx" "1" "2" "3" "4") ("E_WCD0101" "blabla" "0.00" "0.0" "00" "032"))

Must be simple but I don't know how (yet)

M

 

Try removing one "(List " and the very last ")"

Bob

 

(apply 'append (mapcar 'append '(("B_PRS" "xxx" "1" "2" "3" "4") ("E_SKL" "xxx" "1" "2" "3" "4") ("E_WCD"
"xxx" "1" "2" "3" "4") ("E_WCD0101" "blabla" ("0.00" "0.0" "00" "032")))))

give

("B_PRS" "xxx" "1" "2" "3" "4" "E_SKL" "xxx" "1" "2" "3" "4" "E_WCD" "xxx" "1" "2" "3" "4" "E_WCD0101" "blabla" ("0.00" "0.0" "00" "032"))

 

Perfect Doug!
Just what I needed.

Thanx!

 

Marcel,
Great. You're welcom.
Regards,
Doug