reorder nodes problem

pompluck · November 3rd, 2003, 08:29 AM

Hello there,
I have the following problem:
I have a list like
<Parent>
  <ELEMENT name="aaa" attr="bad"></ELEMENT>
  <ELEMENT name="bbb" attr="bad"></ELEMENT>
  <ELEMENT name="ccc"></ELEMENT>
  <ELEMENT name="ddd"></ELEMENT>
  <ELEMENT name="eee"></ELEMENT>
...
</Parent>
I want to reorder them into this:
<NewPar>
  <ELEMENT name="ccc"></ELEMENT>
  <ELEMENT name="ddd"></ELEMENT>
</NewPar>
<NewPar>
  <ELEMENT name="eee"></ELEMENT>
...
</NewPar>
...

This means leaving out the two bad ones, and putting every two of the following ones into a new Tag.
I am new to XSLT and I although I tried hard I have no idea how to get along without any suitable counters or something like that.

Thank you for your help.
Tony

joefawcett · November 3rd, 2003, 09:20 AM

Try this, it uses the mod operator to see whether to start a new group and adds a new document element.

Code:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output omit-xml-declaration="yes"/>  
  <xsl:template match="/">
    <data>
      <xsl:apply-templates select="Parent/ELEMENT[not(@attr = 'bad')]"/>
    </data>
  </xsl:template>
  <xsl:template match="ELEMENT">    
    <xsl:if test="position() mod 2 = 1">
      <newParent>
        <xsl:copy-of select="."/>
        <xsl:copy-of select="following-sibling::*[not(@attr = 'bad')][1]"/>
      </newParent>     
    </xsl:if>
  </xsl:template>  
</xsl:stylesheet>

Joe (MVP - xml)

pompluck · November 3rd, 2003, 10:19 AM

Wow, thank you for that quick answer.
I see there is a lot to learn for me and I'll carefully analyze your examples.

Mangling your code into my stylesheet I came across a little follow-up problem:
'Elements' are formatted. I call the formatting with
<xsl:call-template name="format_name">.
So i changed

<xsl:if test="position() mod 2 = 1">
<newParent>
   <xsl:copy-of select="."/>
   <xsl:copy-of select="following-sibling::*[not(@attr = 'bad')][1]"/>
  </newParent>
</xsl:if>

to

<xsl:if test="position() mod 2 = 1">
<newParent>
   <xsl:call-template name="format_name"> <--
   <xsl:copy-of select="following-sibling::*[not(@attr = 'bad')][1]"/>
  </newParent>
</xsl:if>

but don't know how to pass the actual sibling of 'following-sibling...' to 'call-template'. The respective first element is already formatted well.

Any further clue for me?

Thank you for your answer anyway.
tony

pompluck · November 3rd, 2003, 05:34 PM

did my homework meanwhile...:)
for documentation, it works like this (think there's a shorter way)

...
<newParent>
  <xsl:call-template name="format_name">
  <xsl:for-each select="following-sibling::*[1]"/>
    <xsl:call-template name="format_name">
  </xsl:for-each>
</newParent>
...

hope it helps someone else,
tony

joefawcett · November 5th, 2003, 09:02 AM

That's ok, as long as all your "bad" elemenets are at the beginning. You could also try using parameters, that way your "format_name" template does not have to rely on using the current node:

Code:

<xsl:template name="format_name">
  <xsl:param name="current" select="."/>

</xsl:template>

Called by:

Code:

<xsl:call-template name="format_name">
  <xsl:with-param select="."/>
</xsl:call-template>
<xsl:call-template name="format_name">
  <xsl:with-param select="following-sibling::*[not(@attr = 'bad')][1]"/>
</xsl:call-template>

If you don't pass a parameter the current node will be used.

Joe (MVP - xml)

--

Joe