Hi manikandan,

You can use the umbraco.library:GetXmlNodeByXPath() extension for that - NOTE: The examples on that page use the old XMLSchema...

If you can't figure out how to do it, let us know and we'll help you get it to work.

/Chriztian

Hi chriztian,

Actually i am new bee for umbraco. i don't know where to call "umbraco.library:GetXmlNodeByXPath() " this function,Please look at the below xslt code. and tell me where to apply. thanks in advance

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE xsl:stylesheet [ <!ENTITY nbsp "&#x00A0;"> ]>
<xsl:stylesheet
  version="1.0"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:msxml="urn:schemas-microsoft-com:xslt"
  xmlns:umbraco.library="urn:umbraco.library" xmlns:Exslt.ExsltCommon="urn:Exslt.ExsltCommon" xmlns:Exslt.ExsltDatesAndTimes="urn:Exslt.ExsltDatesAndTimes" xmlns:Exslt.ExsltMath="urn:Exslt.ExsltMath" xmlns:Exslt.ExsltRegularExpressions="urn:Exslt.ExsltRegularExpressions" xmlns:Exslt.ExsltStrings="urn:Exslt.ExsltStrings" xmlns:Exslt.ExsltSets="urn:Exslt.ExsltSets"
  exclude-result-prefixes="msxml umbraco.library Exslt.ExsltCommon Exslt.ExsltDatesAndTimes Exslt.ExsltMath Exslt.ExsltRegularExpressions Exslt.ExsltStrings Exslt.ExsltSets ">


<xsl:output method="xml" omit-xml-declaration="yes"/>

<xsl:param name="currentPage"/>
<!-- start writing XSLT --> 


<xsl:template match="/">
<xsl:param name="sourcePaths" select="macro/sourcePath" />
  <xsl:param name="siteimage" select="concat(concat('$currentPage','/',$sourcePaths),'/','imageurl/')" />
<xsl:variable name="mediaId" select="number($siteimage)" />
<xsl:if test="$mediaId > 0">
        <xsl:variable name="mediaNode" select="umbraco.library:GetMedia($mediaId, 0)" />
        <xsl:if test="$mediaNode/umbracoFile">
            <img src="{$mediaNode/umbracoFile}" height="{@umbracoHeight}" width="{@umbracoWidth}" alt="($mediaNode/nodeName)" />

        </xsl:if>
    </xsl:if>   
</xsl:template>
</xsl:stylesheet>

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