Im trying to get the sitemap of an external page so that i can return it within umbraco.

Is it enough to add the:  

<?xe.target ..domain.com/user.xml?user=77765&return=xml&per_page=1000?>

at the 2nd line or is there more to it? (i guess there is since it isnt working atm).

Ive been looking through w3school and google and somehow i can't find anything usefull to me regarding linking to an external xml file.

The code as it is now (there is hardly any)

<?xml version="1.0" encoding="UTF-8"?>
<?xe.target ..http://www.domain.com/user.xml?user=5555&return=xml&per_page=100?>
<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"/>
<xsl:template match="/">

  <xsl:for-each select="listings/listing">
    <xsl:value-of select="description"/>
  </xsl:for-each>
  
</xsl:template>
</xsl:stylesheet>

Thanks in advance