Hi,

I have a function that gives me XSL for sorting

function fnGetXslStr()
    'Note the XML Need to have @Name, @ID, @RowOrder, @Description,
    dim strXsl
    strXsl = "<xsl:stylesheet xmlns:xsl=""http://www.w3.org/TR/WD-xsl"">" & _
        "<xsl:template match='/'>" & _
        "<table style=""width:auto;height:auto"" cellspacing=""1"" cellpadding=""1"">" & _
        "<xsl:for-each select='/ITEM_LIST/ITEM'>" & _
            "<xsl:sort select='@RowOrder' data-type='number'/>" & _
            "<xsl:if test=""@Deleted[.='0']"">" & _
            "<tr class=""listrow"">" & _
            "<xsl:attribute name=""id"">tr_<xsl:value-of select='@ID'/></xsl:attribute>" & _
            "<xsl:attribute name=""onclick"">javascript:fnSetSelected(""<xsl:v alue-of select='@ID'/>"")</xsl:attribute>" & _
            "<xsl:attribute name=""ondblclick"">javascript:fnEditItemWithSelec tion(""<xsl:value-of select='@ID'/>"")</xsl:attribute>" & _
            "<td valign=""top"" style=""height:auto;width:120px"">" & _
            "<xsl:value-of select='@Name'/>" & _
            "</td>" & _
            "</tr>" & _
            "</xsl:if>" & _
        "</xsl:for-each>" & _
        "</table>" & _
        "</xsl:template>" & _
    "</xsl:stylesheet>"
    fnGetXslStr = strXsl
end function

If we check the header,
<xsl:stylesheet xmlns:xsl=" http://www.w3.org/TR/WD-xsl" >
this header does not seem to work with xsl:sort.
So, I have used,
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
But this does not seem to work in IE 5.0. Actually I am not getting any error. Instead when I am transforming the XML, i get XSL as my output. But with IE 6.0 it works perfect.

How do I resolve this issue in IE 5.0