need to transform xml into visual sitemap

Hi!

Any specific code would be appreciated in converting XML sitemap into visual HTML sitemap.
I guess XSLT would be the best way.
http://www.topsecurityinc.com/sitemap.xml
TrueBlueAsked:
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Geert BormansInformation ArchitectCommented:
This is all the XSLT you need
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xd="http://www.oxygenxml.com/ns/doc/xsl" version="1.0">
    <xsl:output method="html" indent="yes"/>
    <xsl:strip-space elements="*"/>
<xsl:template match="urlset">
    <html>
        <body>
            <xsl:apply-templates select="url">
                <xsl:sort select="priority" order="descending" data-type="text"/>
            </xsl:apply-templates>
        </body>
    </html>
</xsl:template>
    <xsl:template match="url">
        <p>
             <a href="{loc}">
                <xsl:value-of select="loc"/>
            </a>
        </p>
    </xsl:template>
</xsl:stylesheet>

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TrueBlueAuthor Commented:
Gertone,
Thank you for the XSLT.
Any idea why this is not displaying anything?

<%
      Dim NewXML, re, oXmlSrc, oXml, oNode, oXslSrc, oXsl, oXslTemplate
      
      ' If you're using an XSLTemplate and XSLProcessor object,
      '   you have to use the FreeThreaded versions of DOMDocument
      Set oXmlSrc = Server.CreateObject("MSXML2.FreeThreadedDOMDocument.4.0")
      Call oXmlSrc.setProperty("ServerHTTPRequest", True)            
      oXmlSrc.async = False
      
      Set oXslSrc = Server.CreateObject("MSXML2.FreeThreadedDOMDocument.4.0")
      Call oXslSrc.setProperty("ServerHTTPRequest", True)            
      oXslSrc.async = False
      
      Set oXslTemplate = Server.CreateObject("MSXML2.XSLTemplate.4.0")

      ' If I haven't posted the form first, then load my unaltered XML.
      ' If I'm posting info to resort or edit, then build my XML from that.
      
    ' On Error Resume Next  'debugging code if table does not appear uncomment this line

      'xmlResponse = Replace(xmlResponse, "<?xml version=""1.0""?>", "<?xml version=""1.0"" encoding=""utf-8""?>")

            'Call oXmlSrc.loadXML (xmlResponse)      
            Call oXmlSrc.load (Server.MapPath("sitemap.xml"))
          'Response.Write "<xmp>" & oXmlSrc.xml & "</xmp>"    'debugging code
          'newXML = oXmlSrc.xml
        'Set re = new regexp
        're.pattern = " xmlns.+?>"
        'newXML = re.replace(newXML,">")
        'Call oXmlSrc.loadXML(newXML)
  if oXslsrc.load (Server.MapPath("sitemap.xsl")) then
     Set oXslTemplate.stylesheet = oXslsrc
     Set oXslProcessor = oXslTemplate.createProcessor()
     oXslProcessor.input = oXmlsrc
     Response.Write oXmlsrc
    else
       Response.Write "Could Not Retrieve SiteMap.xsl stylesheet"
end if %>
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TrueBlueAuthor Commented:
Gertone,

I think I have the ASP fixed, but no data is being displayed.
Will this XSL run using the MSXML 6 processor?
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Geert BormansInformation ArchitectCommented:
THis will run fine on any XSLT processor.
msxml6 will be fine, but make sure you have msxml6 isntalled, since you point to it directly

I don't like this
    'xmlResponse = Replace(xmlResponse, "<?xml version=""1.0""?>", "<?xml version=""1.0"" encoding=""utf-8""?>")
there is no need to add UTF_8 there since it is teh default
and you seem to cast an xml object to a string

you might need some debugging since you need to check wheiter XSLT and XML are all loaded correctly

and of course, I don't know whether your XML sitemap has a namespace
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TrueBlueAuthor Commented:
Gertone,

I think I fixed most of the aforementioned items you, except for the namespace.
http://www.topsecurityinc.com/sitemap.xml
Any suggestions as to the proper namespace for this xml page?
TIA
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Geert BormansInformation ArchitectCommented:
yeah, did not spot the nemaespace in that source before

here you go
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:sm="http://www.sitemaps.org/schemas/sitemap/0.9"
   version="1.0">
    <xsl:output method="html" indent="yes"/>
    <xsl:strip-space elements="*"/>
    <xsl:template match="sm:urlset">
        <html>
            <body>
                <xsl:apply-templates select="sm:url">
                    <xsl:sort select="sm:priority" order="descending" data-type="text"/>
                </xsl:apply-templates>
            </body>
        </html>
    </xsl:template>
    <xsl:template match="sm:url">
        <p>
            <a href="{sm:loc}">
                <xsl:value-of select="sm:loc"/>
            </a>
        </p>
    </xsl:template>
</xsl:stylesheet>

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