[xsl] Sort and paging problem

["Fran" <franciscojose@xxxxxxxxxxx>]

Hey guys!!
 I post this problem because I'm a XSL beginner and I have looked back at
responses
on the list and taken into account previous suggestions and codes but my
code still don't
work how I want.  I am using the code below to transform the xml document.
First I'm trying to order all HOTEL elements
within attribute "price" and then I want to paginate this, but my code don't
make this exactlly. It pagings ok, but It sorts only the records that are at
page(not all the records), because I want firts order all records and then
paginate to 10 records each page.
Any suggestions would be really appreciated.

Thanks in advance
Frank Vargas

Code:

.....
<xsl:param name="RecordsPerPage" select="5" />
<xsl:param name="PageNumber" />
<xsl:variable name="start" select="((number($PageNumber) - 1) *
(number($RecordsPerPage))) + 1"/>
<xsl:variable name="end" select="($start - 1) + (number($RecordsPerPage))"/>
<xsl:variable name="count" select="(count(DISPONIBILIDAD/LHOTEL/HOTEL))"/>

<xsl:template match="/">

<xsl:for-each select="DISPONIBILITY/LHOTEL/HOTEL[position() >= $start and
position() &lt;= $end]">
	<xsl:sort select="number(@price)"/>
</xsl:for-each>

.....
 <b>Order
  <xsl:choose>
	<xsl:when test="$end >= $count and $start = 1">
		<xsl:value-of select="$start"/>-<xsl:value-of select="$count"/>
		de <xsl:value-of select="$count"/>&#160;
	</xsl:when>
      <xsl:when test="$end >= $count">
      <xsl:when .....(Code to put the records order and to put the previous
and Next button
  </xsl:choose>
  </b>
....



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