Write a stylesheet that has the same structure as a normal one: start with a
template that processes the root node, and call apply-templates when you
want to process its children. The only difference is that the children are
not physical XML children, but logical children found by using a key. Define
a key for nodes based on the ParentID property, and to find the logical
children of a node, use key('parentKey', @ID).
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: Simon Shutter [mailto:simon(_at_)schemax(_dot_)com]
Sent: 13 February 2007 16:32
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Transforming tabular information to hierarchical
If I have a tabular data set that defines parent-child
relationships, is it possible to transform this into a
hierarchical tree format using XSLT?
eg source:
ID ParentID Name col1 col2 col3
1 null One bla1 bla1 bla1
2 1 Two bla2 bla2 bla2
3 2 Three bla3 bla3 bla3
4 3 Four bla4 bla4 bla4
5 1 Five bla5 bla5 bla5
6 4 Six bla6 bla6 bla6
7 4 Seven bla7 bla7 bla7
8 7 Eight bla8 bla8 bla8
9 3 Nine bla9 bla9 bla9
10 9 Ten bla10 bla10 bla10
desired output:
<?xml version="1.0" encoding="utf-8"?>
<ul id='root' xmlns:bla="http://www.blablabla.com/bla";>
<li id='1' bla:col1='bla1' bla:col2='bla1' bla:col3='bla1'>
One
<ul>
<li bla:col1='bla2' bla:col2='bla2' bla:col3='bla2'>
Two
<ul>
<li bla:col1='bla3' bla:col2='bla3' bla:col3='bla3' />Three
<ul>
<li id='4' bla:col1='bla4' bla:col2='bla4'
bla:col3='bla4' />Four
<ul>
<li id='6' bla:col1='bla6' bla:col2='bla6'
bla:col3='bla6'>Six</li>
<li id='7' bla:col1='bla7' bla:col2='bla7'
bla:col3='bla7'>
Seven
<ul>
<li id='8' bla:col1='bla8' bla:col2='bla8'
bla:col3='bla8'>Eight</li>
</ul>
</li>
</ul>
<li id='9' bla:col1='bla9' bla:col2='bla9' bla:col3='bla9'>
Nine
<ul>
<li id='10' bla:col1='bla10' bla:col2='bla10'
bla:col3='bla10'>Ten</li>
</ul>
</li>
</ul>
</ul>
</li>
<li id='5' bla:col1='bla5' bla:col2='bla5'
bla:col3='bla5'>Five</li>
</ul>
</li>
</ul>
Thanks, Simon
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