restructuring xml

bobtail · June 18th, 2008, 02:35 AM

Hi,

I can do basic xml translations but I've become stuck when trying to restructure an xml document.

The document is in the following format

Code:

<model>
 <root node name="example" ID="4" type="A"/>
  <item>
   <name>xyz</name>
   <ID>7</ID>
   <type>A</type>
   <parentname>example</parentname>
   <parentID>4</parentID>
  </item>
  <item>
   <name>bob</name>
   <ID>5</ID>
   <type>B</type>
   <parentname>example</parentname>
   <parentID>4</parentID>
  </item>
  <item>
   <name>cat</name>   
   <ID>10</ID>
   <type>A</type>
   <parentname>xyz</parentname>
   <parentID>7</parentID>
  </item>
</model>

Basically you have a model and all these items including a root item listed. I want to restructure this so items are contained within their parents such to this

Code:

<model>
 <nonterm name="example" ID="4">
  <item>
   <name>xyz</name>
   <ID>7</ID>
   <type>A</type>
   <parentname>example</parentname>
   <parentID>4</parentID>
  </item>
  <item>
   <name>bob</name>
   <ID>5</ID>
   <type>B</type>
   <parentname>example</parentname>
   <parentID>4</parentID>
  </item>
 </nonterm>

 <nonterm name="xyz" ID="7">
  <item>
   <name>cat</name>
   <ID>10</ID>
   <type>A</type>
   <parentname>xyz</parentname>
   <parentID>7</parentID>
  </item>
 </nonterm>

</model>

Any ideas what the best way to do this would be? This is just an example of the format. Just a note that the original order and number of the items essentially has not order with respect to hierarchy.

Thanks in advance.

mhkay · June 18th, 2008, 03:19 AM

The basic approach to this is to do a recursive descent using xsl:apply-templates in exactly the same way as you would normally, except that you apply-templates to the logical children of a node instead of the physical children. In your example you can find the logical children of the context <item> element using <xsl:apply-templates select="key('parent', ID)"/> where the key is defined as <xsl:key name="parent" match="item" use="parentID"/>.

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

Volder · June 18th, 2008, 05:53 AM

maybe it is not the best approach, but it is giving the needed result:

Code:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:key name="parent" match="item|root" use="ID"/>
    <xsl:template match="model">
        <model>
            <xsl:apply-templates select="item/ID|root/@ID"/>
        </model>
    </xsl:template>
    <xsl:template match="item/ID|root/@ID">
        <xsl:if test="count(//item[parentID=current()])>0">
            <nonterm ID="{current()}">
                <xsl:attribute name="name">
                    <xsl:choose>
                        <xsl:when test="name(..)='root'">
                            <xsl:value-of select="../@name"/>
                        </xsl:when>
                        <xsl:otherwise>
                            <xsl:value-of select="../name"/>
                        </xsl:otherwise>
                    </xsl:choose>
                </xsl:attribute>
                <xsl:copy-of select="//item[parentID=current()]"/>
            </nonterm>
        </xsl:if>
    </xsl:template>
</xsl:stylesheet>

bobtail · June 18th, 2008, 06:37 AM

Thanks to the first and second poster. The first poster for the right idea and the second poster who came in with a solution as I was struggling with how to use the key function.

Thank you. Works well!

Volder · June 18th, 2008, 06:48 AM

Quote:

quote:...and the second poster who came in with a solution as I was struggling with how to use the key function.

well, actually, my example is not using key function at all.
<xsl:key ... /> is redundantly left there by me after my attempts to implement it through key() func.
so you can erase it without any harm to the solution.