Hi Michael,
I'm note sure how to proceed with your suggestion. My XML/XSLT experience
is too limited. Do I have to use an input file like Andrew suggested or are
you assuming that I will have the tablular data in xml format as below?
<?xml version="1.0" encoding="ISO-8859-1"?>
<root>
<itm id="1" parentID="null" name="One" col1="bla1" col2="bla1" col3="bla1"
/>
<itm id="2" parentID="1" name="Two" col1="bla2" col2="bla2" col3="bla2" />
<itm id="3" parentID="2" name="Three" col1="bla3" col2="bla3" col3="bla3" />
<itm id="4" parentID="3" name="Four" col1="bla4" col2="bla4" col3="bla4" />
<itm id="5" parentID="1" name="Five" col1="bla5" col2="bla5 col3="bla5" />
<itm id="6" parentID="4" name="Six" col1="bla6" col2="bla6" col3="bla6" />
<itm id="7" parentID="4" name="Seven" col1="bla7" col2="bla7" col3="bla7" />
<itm id="8" parentID="7" name="Eight" col1="bla8" col2="bla8" col3="bla8" />
<itm id="9" parentID="3" name="Nine" col1="bla9" col2="bla9" col3="bla9" />
<itm id="10" parentID="9" name="Ten" col1="bla10" col2="bla10" col3="bla10"
/>
</root>
How will the root recursively find all its children? I tried to create the
template but I'm largely guessing at this point.
<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:key name="parentKey" match="itm" use="id" />
<xsl:template match="/">
<xsl:apply-templates select="key('parentKey',@id)"/>
</xsl:template>
</xsl:stylesheet>
Simon
-----Original Message-----
From: Michael Kay [mailto:mike(_at_)saxonica(_dot_)com]
Sent: February 13, 2007 9:13 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] Transforming tabular information to hierarchical
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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