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Re: [xsl] linkdiff template

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Hello Guy,

I can't see an error in your code, maybe somebody else? Did you exactly use the code below? Line 12 is the key() - no error with parantheses, commas or apostrophes?

What it does:
<xsl:key/> creates one index for each input XML. With key() you can easily access these indexed elements, here your <a/>s.
With <xsl:for-each select="document($previous)"> you do not work through the whole tree, you only switch the context to the other file, because you want to have the indexed <a/>s from there and not from your current file. With "document($previous)" you only access the root node '/', so there is only one iteration. For multiple iterations you can write "document($previous)//a", but as you said it, it's completely useless when using keys.

Regards,

Joerg



Guy McArthur wrote:

I thought I understood, but I get an "unknown error in XPath" on line 12.
Here is the xsl I'm using: (also tried putting wrapping the key() method with a boolean() method.
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<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:param name="previous"/>

<xsl:key name="links" match="a" use="@href"/>

<xsl:output method="text"/>

 <xsl:template match="a">
  <xsl:variable name="href" select="@href"/>
  <xsl:for-each select="document($previous)">
   <xsl:if test="not(key('links', $href))">
     <xsl:value-of select="$href"/>
   </xsl:if>
  </xsl:for-each>
 </xsl:template>

</xsl:stylesheet>
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Also, it would seem by looking at it, that it would walk through all the elements of document($previous) for every matching "a" element. Shouldn't building a key eleminate the need to walk through the tree each time?

Guy

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