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Re: Combining two XBEL XML files

Subject: Re: Combining two XBEL XML files
From: "Aaron Gray" <aaronngray.lists@xxxxxxxxxxxxxx>
Date: Sat, 12 Jul 2008 19:02:45 +0100
Re:  Combining two XBEL XML files
Aaron Gray wrote:

Its just working with two documents at the same time, sorting them, recursing down folders in syncronism and getting back up is quite a complex task. looking for some guidance.


Here is an attempt in XSLT 2.0 but I have no files to test that with so that is completely untested. The primary input document should be one bookmark file, the URL of the second should be passed in as the parameter sec-file:


<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:my="http://example.com/2008/my" exclude-result-prefixes="my" version="2.0">

<xsl:output method="xml" indent="yes"/>

  <xsl:param name="sec-file" select="'bookmark2.xml'"/>
  <xsl:variable name="sec-doc" select="doc($sec-file)"/>

<xsl:function name="my:folder-merge" as="element()*">
<xsl:param name="folders" as="element()*"/>
<xsl:for-each-group select="$folders" group-by="title">
<xsl:sort select="current-grouping-key()"/>
<folder>
<title>
<xsl:value-of select="current-grouping-key()"/>
</title>
<xsl:sequence select="my:folder-merge(current-group()/folder)"/>
<xsl:sequence select="my:bookmark-merge(current-group()/bookmark)"/>
</folder>
</xsl:for-each-group>
</xsl:function>


  <xsl:function name="my:bookmark-merge" as="element()*">
    <xsl:param name="bookmarks" as="element()*"/>
    <xsl:for-each-group select="$bookmarks" group-by="@href">
      <xsl:sort select="current-group()[1]/title"/>
      <xsl:sequence select="current-group()[1]"/>
    </xsl:for-each-group>
  </xsl:function>

<xsl:template match="xbel">
<xsl:copy>
<xsl:sequence select="my:folder-merge(folder | $sec-doc/xbel/folder)"/>
<xsl:sequence select="my:bookmark-merge(bookmark | $sec-doc/xbel/bookmark)"/>
</xsl:copy>
</xsl:template>


</xsl:stylesheet>

Great I have prepared some files.


http://www.cybercomms.org/Tests/XBEL/All.xbel

       http://www.cybercomms.org/Tests/XBEL/1.xbel
       http://www.cybercomms.org/Tests/XBEL/2.xbel

By combining 1.xbel and 2.xbel you should get All.xbel providing duplicates are removed too :)

I will have a play with your code when I get some spare time to get on to this.

Thanks,

Aaron

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