Please can someone show me how can I change the following code to only show the top 5 or whatever number of items I need from the feed ?The following code works , but it displays the whole file. Only one day spent so far on this problem and hair here is in short supply.
Thanks in advance...

Sample asp web page....
<%
myXMLfile = "http://www.24-7pressrelease.com/rss/law_and_legal.xml"

'Set getPage = Server.CreateObject("MSXML2.ServerXMLHTTP.4.0")
'Set getPage = Server.CreateObject("Microsoft.XMLHTTP")
Set getPage = Server.CreateObject("MSXML2.ServerXMLHTTP")

getPage.Open "GET", myXMLfile, false
getPage.SetRequestHeader "Content-type", "text/xml"
getPage.Send

'response.write getPage.responseXML.xml '' successfully outputs xml

'Load XSL
set xsl = Server.CreateObject("Microsoft.XMLDOM")
xsl.async = false
xsl.load(Server.MapPath("sample.xsl"))

Function RightTrimLen(theStr, leftPadding)
RightTrimLen = Right(theStr, (Len(theStr) - Len(leftPadding)))
End Function
' above function is used to get rid of encoding that is added (and causing an error)

theOutput = getPage.responseXML.transformNode(xsl)
theOutput = RightTrimLen(theOutput, "<?xml version=""1.0"" encoding=""UTF-16""?>")

response.write theOutput

Set getPage = Nothing
%>


"sample.xsl"....

<xsl:stylesheet version="1.0"
xmlnssl="http://www.w3.org/1999/XSL/Transform">

<xsl:template match="/">
<xsl:for-each select="/rss/channel/item"><h2><b>
<xsl:value-of select="title" /><br /></b></h2>
<xsl:value-of select="description" /><br /><i>
Published: <xsl:value-of select="pubDate" /><br /></i>
<a>
<xsl:attribute name="href">
<xsl:value-of select="link"/>
</xsl:attribute>
View Story
</a><br/><br/>
</xsl:for-each>
</xsl:template>

</xsl:stylesheet>