[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message]

Re: Finding sequences of same element

Subject: Re: Finding sequences of same element
From: Mukul Gandhi <mukul_gandhi@xxxxxxxxx>
Date: Wed, 9 Feb 2005 00:38:06 -0800 (PST)
xsl same elements
Hi Simon,
  Please try this XSL -

<?xml version="1.0"?> 
<xsl:stylesheet
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">
 
 <xsl:output method="xml" indent="yes" /> 
  
 <xsl:template match="/A">  
  <A>
   <xsl:for-each select="*">
     <xsl:choose>
       <xsl:when test="not(self::B)">
         <xsl:copy-of select="." />
       </xsl:when>
       <xsl:when test="name(preceding-sibling::*[1])
!= 'B'">
         <D>
           <xsl:copy-of select="." />
           <xsl:call-template name="create-group">
             <xsl:with-param name="list"
select="following-sibling::*" />
           </xsl:call-template>
         </D>
       </xsl:when>
     </xsl:choose>
   </xsl:for-each>
  </A> 
 </xsl:template>
 
 <xsl:template name="create-group">
   <xsl:param name="list" />
      
   <xsl:if test="name($list[1]) = 'B'">
     <xsl:copy-of select="$list[1]" />
     <xsl:call-template name="create-group">
       <xsl:with-param name="list"
select="$list[position() > 1]" />
     </xsl:call-template>
   </xsl:if>   
 </xsl:template>
 
</xsl:stylesheet>

Regards,
Mukul

--- Simon Kissane <skissane@xxxxxxxxx> wrote:

> Hi
> 
> Suppose I have an input document:
> <A><B X="1"/><B X="2"/><B X="3"/><C X="4"/><B
> X="5"/><B X="6"/><B X="7"/></A>
> 
> Now, suppose I wish to group together consecutive B
> elements, giving a
> result document like this:
> <A><D><B X="1"/><B X="2"/><B X="3"/></D><C
> X="4"/><D><B X="5"/><B
> X="6"/><B X="7"/></D></A>
> 
> How can I do this? (I would prefer not to use
> recursive templates, but
> rather for-each, if that is at all possible...)
> 
> I think I can find the initial element of these
> sequences, like so:
>   B[not(preceding-sibling::*) or
> preceding-sibling::*[0][not(self::B)]]
> This, I think, should select all B for which there
> are either no
> preceeding sibling elements, or for which the
> immediately preceeding
> sibling element is not a B element. Thus, in the
> above example, it
> would pick B[@X=1] and B[@X=5].
> 
> But, given each initial sequence element, how can I
> find the remaining
> nodes in the sequence?  With the initial sequence
> element as the
> context node, I could do:
>     .|following-sibling::B
> But that will also pick up B[@X=5] and B[@X=6] when
> the context node is B[@X=1].
> 
> Is there a predicate test I could use on
> following-sibling::B to
> restrict it only to the current sequence of B
> elements?
> 
> Thanks
> 
> Simon Kissane



		
__________________________________ 
Do you Yahoo!? 
The all-new My Yahoo! - What will yours do?
http://my.yahoo.com 

Current Thread

PURCHASE STYLUS STUDIO ONLINE TODAY!

Purchasing Stylus Studio from our online shop is Easy, Secure and Value Priced!

Buy Stylus Studio Now

Download The World's Best XML IDE!

Accelerate XML development with our award-winning XML IDE - Download a free trial today!

Don't miss another message! Subscribe to this list today.
Email
First Name
Last Name
Company
Subscribe in XML format
RSS 2.0
Atom 0.3
Site Map | Privacy Policy | Terms of Use | Trademarks
Free Stylus Studio XML Training:
W3C Member
Stylus Studio® and DataDirect XQuery ™are products from DataDirect Technologies, is a registered trademark of Progress Software Corporation, in the U.S. and other countries. © 2004-2013 All Rights Reserved.