Sort xml file

victorcorey · December 3rd, 2007, 04:47 PM

Hello all,

I am new to xslt and I am trying to do something I think would be simple.

I am loading an xml file using the Microsoft.XMLDOM client-side. I need to sort this and I think I need to use XSLT. I would like to pass in a parameter (the node being sorted) and direction and then render the same xml (just sorted)

Help.
Thank you,
Victor

--
Victor Corey

mhkay · December 3rd, 2007, 07:26 PM

Yes, this seems a reasonable approach. Which part of it are you having difficulty with?

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

victorcorey · December 4th, 2007, 01:32 AM

A couple of areas :-(

1. I think my xslt is incorrect for the output xml. It renders correctly but adds the html, head, and body tags.

Code:

<?xml version="1.0" encoding="UTF-8" ?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xs="http://www.w3.org/2001/XMLSchema">
  <xsl:output omit-xml-declaration="no" method="xml" indent="yes"/>
  <xsl:param name="SortCol" />
  <xsl:param name="SortDir" />
  <xsl:param name="SortType" />
  <xsl:template match="worldheritage">
  <xsl:value-of select="$SortDir" />
    <xsl:for-each select="row">
        <xsl:sort order="{$SortDir}" select="*[name()=$SortCol]" data-type="{$SortType}" />
            <xsl:copy-of select="."/>
    </xsl:for-each>
  </xsl:template>
</xsl:stylesheet>

2. Client-side how do I pass parameters? I'm using the Microsoft.XMLDOM parser.

Code:

xmlDoc = new ActiveXObject("Microsoft.XMLDOM");
xmlDoc.async = false;
xmlDoc.load(sFile);
xmlObj = xmlDoc.documentElement;

I need to add the xslt and pass in the parameters.

Thanks,
Victor

--
Victor Corey

Martin Honnen · December 4th, 2007, 09:50 AM

If you want to use parameters then you need to use the addParameter method of the XSLTProcessor object, see the MSXML SDK http://msdn2.microsoft.com/en-us/library/ms762312.aspx

victorcorey · December 4th, 2007, 01:31 PM

OK. I think I am close. My xslt almost works. It sorts the data but only shows the row nodes not the root node.

Code:

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml" omit-xml-declaration="yes" indent="yes" />
  <xsl:param name="SortCol" />
  <xsl:param name="SortDir" />
  <xsl:param name="SortType" />
  <xsl:template match="worldheritage">
    <xsl:for-each select="row">
      <xsl:sort order="{$SortDir}" select="*[name()=$SortCol]" data-type="{$SortType}" />
      <xsl:copy-of select="."/>
    </xsl:for-each>
  </xsl:template>
</xsl:stylesheet>

Here's the xml data:

Code:

<worldheritage promo="UNESCO">
<row id="27" unique="30" trans="0" lat="37.26166667" long="-108.48555560">
    <criteria>(iii)</criteria>
    <inscribeDate>1978</inscribeDate>
    <code>us</code>
    <region>North America</region>
    <location>Colorado</location>
    <state>United States of America</state>
    <site>Mesa Verde National Park</site>
    <description>A great concentration of ancestral Pueblo Indian dwellings, built from the 6th to the 12th century, can be found on the Mesa Verde plateau in south-west Colorado at an altitude of more than 2,600 m. Some 4,400 sites have been recorded, including villages built on the Mesa top. There are also imposing cliff dwellings, built of stone and comprising more than 100 rooms.</description>
  </row>
  .
  .
  .
</worldheritage>

I just need to sort the xml by state node and return the xml.

Thanks,
Victor

--
Victor Corey

Martin Honnen · December 4th, 2007, 01:40 PM

I think you simply want to copy the root node from the XML input document e.g. you need

Code:

  <xsl:template match="worldheritage">
   <xsl:copy>
    <xsl:for-each select="row">
      <xsl:sort order="{$SortDir}" select="*[name()=$SortCol]" data-type="{$SortType}" />
      <xsl:copy-of select="."/>
    </xsl:for-each>
   </xsl:copy>
  </xsl:template>