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Searching in nested list
Posted: Fri Oct 02, 2009 10:47 am
by Fritz
I have converted xml-document to a big list. It looks like:
Code: Select all
<group>
<group>
<element>
</group>
<element>
</group>
List:
Code: Select all
(("group" ("group" ("element" (("code" "1")))) ("element" (("code" "2")))))
How can I find an element inside the list?
returns nil.
Posted: Fri Oct 02, 2009 12:05 pm
by Jeff
Here is a tutorial on drilling down into xml content in newlisp:
http://www.artfulcode.net/articles/working-xml-newlisp/
Here is a tutorial on using find-all with xml (among others) in newlisp:
http://www.artfulcode.net/articles/usin ... -find-all/
Posted: Fri Oct 02, 2009 12:29 pm
by Lutz
'assoc' and 'find-all' only search on the top level of the list. But the '("code" "2") part is deeply nested into the list.
Use 'ref', 'ref-all' in this case and to change imformation use 'set-ref' and 'set-ref-all'
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> (ref '("code" "2") xml-list)
(0 2 1 0)
> (set-ref '("code" "2") xml-list '("code" "99"))
(("group" ("group" ("element" (("code" "1")))) ("element" (("code" "99")))))
> xml-list
(("group" ("group" ("element" (("code" "1")))) ("element" (("code" "99")))))
>
(0 2 1 0) is called and
index vector. If you cut off indices from the right side you can retrieve the higher level elements of 'xml-list'
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> (nth '(0 2 1 0) xml-list)
("code" "99")
> (nth '(0 2 1) xml-list)
(("code" "99"))
> (nth '(0 2) xml-list)
("element" (("code" "99")))
you can also use 'setf' on the returned reference of 'nth' to change 'xml-list':
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> (setf (nth '(0 2 1 0) xml-list) '("code" "111"))
("code" "111")
> xml-list
(("group" ("group" ("element" (("code" "1")))) ("element" (("code" "111")))))
>
PS: here is a whole chapter about "Modifying and searching lists":
http://www.newlisp.org/downloads/CodePa ... html#toc-6
Posted: Fri Oct 02, 2009 12:36 pm
by Fritz
Thanx! "ref" was what is need!