[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Q on copying
Can anyone please tell me how to copy XML into certain sorted order while keeping the attributes of the root element as is? ie the following is my XML <Root att1="" att2=""> <Info sort="2"/> <Info sort="5"/> <info sort="1"/> <other sort="3"/> <other sort="5"/> <other sort="2"/> </Root> I want to turn that XML into <Root att1="" att2=""> <info sort="1"/> <Info sort="2"/> <Info sort="5"/> <other sort="3"/> <other sort="4"/> <other sort="5"/> </Root> This is what I have so far.. <xsl:template name="Root"> <Root> <xsl:for-each select="info"> <xsl:sort select="sort"/> <xsl:copy-of select="."/> </xsl:for-each> <xsl:for-each select="other"> <xsl:sort select="sort"/> <xsl:copy-of select="."/> </xsl:for-each> </Root> </xsl:template> Thanx! Rosa XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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