Re: [xsl] can't get xsl:sort to work

Subject: Re: [xsl] can't get xsl:sort to work
From: UlyLee <ulyleeka@xxxxxxxxx>
Date: Mon, 19 Sep 2005 04:37:51 -0700 (PDT)
> You don't want the body of the for-each to be
> content of the xsl:sort so
> make that
> 
> <xsl:sort select="@entry"/>
> 
> and remove the line
> 
> </xsl:sort>
> 
Thanks, i never thought of that :D it was just as
simple as removing the end tag. you guys are a great
help for newbies like me... thanks!

> Note that you have multiple uses of // which is
> likely to make this code
> very inefficient in practice. (Unless your XSLT
> system does some very
> extensive automatic rewrites.
> 
> select="//complexarticle |
> //simplearticle | //dummyarticle">
> 
> means search the whole document to arbitrary depth
> to find all the
> elements of that name. Are these elements really
> arbitrarily deeply
> nested? 

yup!

> Also <xsl:variable name="x"
> select="count(.//xref)"/> You don't need to
> find all the xrefs and count them, you can just use
> last() which is the
> same number and already calculated.

i already tried using that but unfurtunately because
as you have said... i'm doing a deep searching,
arbitrarily thats why i sometimes get two nodes at
last()

but again thanks for the help and suggestions.

UlyLee


	
		
______________________________________________________ 
Yahoo! for Good 
Donate to the Hurricane Katrina relief effort. 
http://store.yahoo.com/redcross-donate3/ 

Current Thread