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RE: Traverse XML Source

2004-08-17 21:43:04
Hi Karl,
  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:strip-space elements="*" />

<xsl:template match="node()">   
   <xsl:for-each select="@*">
     <xsl:attribute name="{name()}">
       <xsl:value-of select="." />
     </xsl:attribute>
   </xsl:for-each>
   <xsl:if test="count(child::*) > 0">
     <ul>
       <xsl:value-of select="name()" />
       <xsl:apply-templates />
     </ul>
   </xsl:if>
   <xsl:if test="count(child::*) = 0">
     <li>
       <xsl:value-of select="name()" />
       <xsl:apply-templates />
     </li>
   </xsl:if>
</xsl:template>
  
</xsl:stylesheet>

Regards,
Mukul

--- "Karl J. Stubsjoen" <karl(_at_)meetscoresonline(_dot_)com>
wrote:

That is a farely predicted result... I don't know to
what depth the child
nodes shall occur... but each parent node should be
a <ul> and each child
node with child nodes should be a <UL>... with the
possibility of there
being sibling <li>'s.
Here is a better XML source:

<ABC>
  <A>
    <AB/><CD/><EF/>
  </A>
  <B>
    <AB/><CD/><EF/>
  </B>
  <C/>
  <D/>
  <E/>
</ABC>

And the result:
<ul>ABC
  <ul>A
    <li>AB</li>
    <li>CD</li>
    <li>EF</li>
  </ul>
  <ul>B
    <li>AB</li>
    <li>CD</li>
    <li>EF</li>
  </ul>
  <li>C</li>
  <li>D</li>
  <li>E</li>
  <li>F</li>
</ul>


-----Original Message-----
From: cking [mailto:cking(_at_)telenet(_dot_)be]
Sent: Tuesday, August 17, 2004 2:24 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] Traverse XML Source


Hi Karl,

there are a few problems with your XSL source:
it contains two templates with the same match
attribute
  <xsl:template match="*">
which basically means: match any element.
Template match attributes should be unique,
otherwise the
processor does not know which of the two templates
to use.

further, your second template has no effect, because
match="@*"
means: match any attribute, and your XML source tree
doesn't
have any element with attributes.

you didn't show us the output you expect, but
something like:

<xsl:template match="/stuff/abc/def">
  <ul><xsl:apply-templates/></ul>
</xsl:template>

<xsl:template match="path | web | uid | pwd">
  <li><xsl:value-of select="."/></li>
</xsl:template>

will (hopefully) give you this output:

<ul>
  <li>\\path-to-server</li>
  <li>http://my-web-path</li>
  <li>9999</li>
  <li>monkeysee</li>
</ul>

HTH,
Anton Triest

Tuesday, August 17, 2004 10:46 PM,
karl(_at_)meetscoresonline(_dot_)com wrote:

Hello,
I would like a very simple XSLT transformation
that traverses an XML
source and creates an HTML list output.  Is this
an identity
transformation?
So far, I have only been able to get anything to
happen on the document
root, and nothing below is traversed.
Thanks for the help.  ~karl

Here is what I have so far,
XML Source:
<stuff>
<abc>
<def>
<path>\\path-to-server</path>
<web>http://my-web-path</web>
<uid>9999</uid>
<pwd>monkeysee</pwd>
</def>
</abc>
</stuff>

XSL Source:
<xsl:template match="*">
  <xsl:copy>
    <xsl:apply-templates select="@*" />
    <xsl:apply-templates />
  </xsl:copy>
</xsl:template>


<xsl:template match="@*">
  <li><xsl:copy-of select="." /></li>
</xsl:template>

<xsl:template match="*">
  <li><xsl:copy-of select="." /></li>
</xsl:template>

</xsl:stylesheet>



                
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