Sorting problem (Paging)

pat123 · February 13th, 2006, 06:42 AM

Hi all,

I want to sort my results by a date attribute, but also break the results up into pages (18 per page). I have done all this but my sort only works for one page at a time. How can I sort all of the results first. Any help will be greatly appreciated.

XSL:

Code:

<xsl:template match="/">

<html>
<link rel="stylesheet" type="text/css" href="portfolio_style.css" />
<body>
<h1>Portfolio</h1>


<xsl:call-template name="pagination" />    

<div id="portfolio">    

<xsl:for-each select="portfolio/campaign[position() &gt; $vFirstRecord][position() &lt;= $pRecordsPerPage]">

 
<xsl:sort select="substring(@date,7,4)" order="descending"/> 
 
<xsl:sort select="substring(@date,4,2)" order="descending"/> 

<xsl:sort select="substring(@date,1,2)" order="descending"/> 

<xsl:call-template name="campaign"/>

</xsl:for-each>
</div>

</body>
</html>
</xsl:template>

mhkay · February 13th, 2006, 07:21 AM

You want a two-phase transformation. Capture the results of the sort in a variable

<xsl:variable name="sorted-set">
  <xsl:for-each>
    <xsl:sort>

</xsl:variable>

and then group the result into pages:

<xsl:for-each select="xx:node-set($sorted-set)/*[position() mod $N = 1]">
  <page>
    <xsl:copy-of select=".|following-sibling::*[position() < $N]"/>
  </page>
</xsl:for-each>

The xx:node-set() function is implementation-dependent; with XSLT 2.0 it isn't needed (and you can do the grouping a bit more elegantly using for-each-group).

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

pat123 · February 13th, 2006, 07:41 AM

Thanks for the quick reponse.

I don't wish to group the results into pages if i can avoid it, this is what i would like to do:

Code:

// Get a new result sorted correctly by date
<xsl:variable name="sorted-set">

<xsl:for-each select="portfolio/campaign">

 
<xsl:sort select="substring(@date,7,4)" order="descending"/> 
 
<xsl:sort select="substring(@date,4,2)" order="descending"/> 

<xsl:sort select="substring(@date,1,2)" order="descending"/> 

<xsl:call-template name="campaign"/>

</xsl:for-each>

</xsl:variable>

// The loop through that result    (I am using php5 to parse the xml, what would i replace the xx with)
<xsl:for-each select="xx:node-set($sorted-set)/[position() &gt; $vFirstRecord][position() &lt;= $pRecordsPerPage]">

<xsl:call-template name="campaign"/>

</xsl:for-each>

Is this possible?

mhkay · February 13th, 2006, 09:01 AM

You said in your first post that you wanted to split the results into pages, and in your second post that you didn't, so I give up.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

pat123 · February 13th, 2006, 09:14 AM

Sorry about that misunderstanding.

I do want to split the results up into pages, but not like <page>...</page>, just using position() and by passing parameters to it.

The xsl is passed a $vFirstRecord (start position for results) and a $pRecordsPerPage parameter, when a user clicks on the next and back buttons the $vFirstRecord changes accordingly. For example,

$vFirstRecord = 1
$pRecordsPerPage = 18

Shows nodes 1 to 18.
The $vFirstRecord parameter will be reset to 19, and appended to the url.

(Sorry if i'm stating the obvious)

I want to sort the xml file by date, then loop through the results showing 18 at a time, but keep the sort order. At the moment it is paging correctly but not sorting correctly.

Any ideas?

mhkay · February 13th, 2006, 11:14 AM

You can do this in a single pass:

<xsl:for-each select="(unsorted nodes)">
  <xsl:sort select="(sort key)">
    <xsl:if test="position() >= $start and position() ≤ $end">
      (output the value_
    </xsl:if>
</xsl:for-each>

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

pat123 · February 13th, 2006, 11:56 AM

Thanks, but now I can only view the first page, when i click on the next button it returns no results.

Code:

            <xsl:for-each select="portfolio/campaign">
              <xsl:sort select="substring(@date,7,4)" order="descending"/> 

              <xsl:sort select="substring(@date,4,2)" order="descending"/> 

              <xsl:sort select="substring(@date,1,2)" order="descending"/> 

                <xsl:if test="position() &gt; $vFirstRecord and position() &lt;= $pRecordsPerPage">
                            <xsl:call-template name="campaign"/>
                            <xsl:call-template name="newRow" select="campaign[position() mod 6 = 0]"/>
                </xsl:if>
            </xsl:for-each>

The parameters are being passed to the xsl, I am able to print them out as far as the xsl:if statement.

pat123 · February 13th, 2006, 12:08 PM

Sorry just seen my mistake. Forgot to add the records per page to start position: Here is the working code:

[code]
            <xsl:for-each select="portfolio/campaign">

              <xsl:sort select="substring(@date,7,4)" order="descending"/>

              <xsl:sort select="substring(@date,4,2)" order="descending"/>

              <xsl:sort select="substring(@date,1,2)" order="descending"/>

                    <xsl:if test="position() > $vFirstRecord and position() <= $vFirstRecord + $pRecordsPerPage">
                            <xsl:call-template name="campaign"/>

                    </xsl:if>

            </xsl:for-each>
</code]

iamlibran · January 30th, 2007, 10:40 AM

There seems to be a bug in xsl:sort. I am using XSLT2.0. Here is the code that I am using to sort and list only 20 rows.

<xsl:for-each select="$reportRecords">
     <xsl:sort select="./descendant::*[name()=$sortcol]" order="{$sortorder}" data-type="{$sortDataType}"></xsl:sort>

    <xsl:if test="position() >= number($startRow) and position() <= number($endRow)">
    <tr>
    <xsl:apply-templates/>
    </tr>
    </xsl:if>
</xsl:for-each>

"startRow" and "endRow" are being set based on the page number. The difference will always be 20.
What it does is always picks up the first 20 records from $reportRecords and sort them. That means it picks
$reportRecords[1-20] and sorts them always. The position() is not relative to sorted result, it relates to the original set of nodelist.

mhkay · January 30th, 2007, 01:00 PM

I agree with you: whether you are using XSLT 1.0 or 2.0, position() within the for-each should reflect the position in the sorted sequence, not the unsorted sequence. Raise the issue with your XSLT vendor.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference