xsl-list
[Top] [All Lists]

RE: [xsl] xalan and replace "regex"

2007-12-27 17:30:33

The reason I am hesitant is this problem. I don't want to 
rewrite everything for saxon.
I am using docbook-xsl-1.73.2. And this is the latest one. 
The document I am trying to compile is this :

<snip/>

I don't know if it get's any simpler.

There's nothing wrong with your source document - no-one ever suggested
there was! - the problem is in the stylesheet.

Basically, if you want to use the XSLT 2.0 replace() function, which is what
you were trying to do, then you will need an XSLT 2.0 processor, and you
will need the XSLT 2.0 versions of the docbook stylesheets. See
http://wiki.docbook.org/topic/DocBookXsl2Stylesheets - but I don't know what
the current state of play is. If you don't want to make that move, you won't
be able to use replace() or xsl:analyze-string. It's your choice.

Michael Kay
http://www.saxonica.com/


--~------------------------------------------------------------------
XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--