[xslt] (no subject)

[Denis Davidov <den dgtex gmail com>]

Hello. First of all, thanks for libxml and libxslt! I’m working on
software that uses both of the libraries and I found GNOME XML/XSLT
libraries very easy to use. Thanks, guys!

I have several questions and I hope I’ll find some help here. Here they are:

1. I’m trying to pass parameters into XSL, but something goes wrong
and it doesn’t work. Here I’ve published very short code example that
illustrates simple XSL transformation and parameter passing:
http://www.everfall.com/paste/id.php?76x88mgneaad

Unfortunately it doesn’t work. “value-of” element just returns an
empty string. I have this output:

<?xml version="1.0" encoding="utf-8"?>
Param =

What do I do wrong? You can edit code example right on the website
where its published.


2. When I create xsltTransformContext (using xsltNewTransformContext),
I have to specify both XML and XSL documents. When I call
xsltApplyStylesheetUser, I also need to pass both XML and XSL document
pointers. Why do I need to do this two times? Here is an explanation
why do I ask: I register a lot of extension elements and functions,
and perform a lot of transformations on the different documents and
style sheets. I don’t want to register all extensions each time when I
need to transform new XML-XSL pair, I want to reuse XSLT context.
Currently I do it in this way:
http://www.everfall.com/paste/id.php?y22esn1kbwic

Is there any better way to solve this problem?


3. When I perform transformation, I have a lot of such errors in stdout:

element div: error : Name is not from the document dictionnary 'div'

(btw, there is a misspell in “dictionary”). If I’ll specify XHTML
namespace in the XSL as a default namespace, then there are no such
errors. But I don’t want to specify it in XSL. I want to know:
- How I can specify my own dictionary using libxslt API?
- How I can switch error reporting off?


4. Assume that I have xmlDoc, xmlXPathContext (constructed from this
xmlDoc), and xmlNode (somewhere from this xmlDoc). I want to perform
XPath-request as if this node was the root of the document. What is
the best way to do this? Currently I replace root node, perform the
request, and place the root node back:
http://www.everfall.com/paste/id.php?wzgzmx95uza8

Is there any better solution for this?


Ok, that’s it for now, thanks for help :)

Denis.