Hi,



You can iterate the elements with lower case names and sort selecting 
the corresponding upper case element:



<?xml version="1.0" encoding="UTF-8"?>

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

  <xsl:template match="/">

    <result>

    <xsl:for-each select="*/*[translate(name(), 'ABCDE', 'abcde') = 
name()]">

      <xsl:sort select="../*[name() = translate(name(current()), 
'abcde', 'ABCDE')]"/>

      <no><xsl:value-of select="."/></no>

    </xsl:for-each>

    </result>

  </xsl:template>

</xsl:stylesheet>



<?xml version="1.0" encoding="UTF-8"?>
<result><no>30</no><no>40</no><no>10</no><no>50</no><no>20</no></result>

Best Regards,
George
-----------------------------------------------
George Cristian Bina
<oXygen/> XML Editor & XSLT Editor/Debugger
www.---.com


Kenny Bogoe (BogoeMD) wrote:
Hi,



I have this structure where the content of tags <A> to <E> have to determine
the order of tag-content <a> to <e>. The result needs to be as shown below.

I can manage to make a stylesheet with an <xsl:if> for each <A> to <E>
asking whether the number is 1, 2, 3, 4 or 5, but I am sure there must be a
simpler way to do this. I can not use position() since the position af the
tags can change in the structure.

<Root>
<a>10</a>
<b>20</b>
<c>30</c>
<d>40</d>
<e>50</e>
...

...
<A>3</A>
<B>5</B>
<C>1</C>
<D>2</D>
<E>4</E>
</Root>

Result I need:



<no>30</no> <!-- C = 1 has to be first with the value of c (which is 30) -->
<no>40</no>
<no>10</no>
<no>50</no>
<no>30</no>


Thanks.
Kenny Bogoe