Muenchian Grouping with in a Muenchian Group ???

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April 4th, 2008, 04:59 AM

waxdart

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Join Date: Apr 2008

Posts: 2

Time spent in forums: 1 h 7 m 30 sec
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Muenchian Grouping with in a Muenchian Group ???

I've not idea if this is possible. I have a big flat xml file which I cannot change. So I'm hoping XSLT will help.

I found a url jenitennison dot com which has got me part of the way; but now I'm faffing.

Here is the out put I'm trying to get
Its grouped by Location, then Course and then I'd like every instance of date for its respective grouping.

Location A
Course A - Date1
- Date2
- Date3
Course B - Date4
- Date5
- Date6

Location B
Course A - Date7
- Date8
- Date9
Course B - Date10
- Date11
- Date12

The xsl and xml I've gone back to is just like the example given on jenitennison dot com I spent all day yesterday trying to understand; but with no luck. Any help or link to help is more than welcome

I get this which is almost there but not great.

Location A
Course A - Date1
Course A - Date2

Course B - Date4
Course B - Date5

Location B
Course A - Date7
Course A - Date8
Course B - Date10
Course B - Date11

<xsl:key name="course-by-location" match="contact" use="location" />

<xsl:template match="records">
<xsl:for-each select="contact[count(. | key('course-by-location', location)[1]) = 1]">
<xsl:sort select="location" />
<xsl:value-of select="location" /> 

<xsl:for-each select="key('course-by-location', location)">
<xsl:sort select="course" />
<xsl:value-of select="course" /> - <xsl:value-of select="date" /> 
</xsl:for-each> 

</xsl:for-each>

</xsl:template>

<records>
<contact id="0001">
<date>Date1</date>
<course>Course A</course>
<location>Location A</location>
</contact>
<contact id="0002">
<date>Date2</date>
<course>Course A</course>
<location>Location A</location>
</contact>
<contact id="0003">
<date>Date3</date>
<course>Course D</course>
<location>Location A</location>
</contact>
...
</records>

April 4th, 2008, 09:14 AM

jkmyoung

Contributing User

Join Date: Sep 2004

Posts: 474

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Reputation Power: 52

second level grouping:

Code:

<xsl:key name="date-by-course-location" match="contact" use="concat(location, '-', course)"/>

<xsl:for-each select="key('course-by-location', location)[count(.|key('date-by-course-location', concat(location,'-', course))[1]) = 1]">
	<xsl:sort select="course"/>
	<xsl:value-of select="course"/>
	<xsl:for-each select="key('date-by-course-location', concat(location,'-', course))">
		<xsl:text>- </xsl:text>
		<xsl:value-of select="date"/>
		<br/>
	</xsl:for-each>
</xsl:for-each>

April 4th, 2008, 10:44 AM

waxdart

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Join Date: Apr 2008

Posts: 2

Time spent in forums: 1 h 7 m 30 sec
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:) its works - it works

Thank you. I wouldn't have got the concat.

For those who are just starting out and learn best from example I have posted the full thing for you to use and change

Code:

<?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="date-by-course-location" match="contact" use="concat(location, '-', course)"/>
<xsl:key name="course-by-location" match="contact" use="location" />

<xsl:template match="records">
	<xsl:for-each select="contact[count(. | key('course-by-location', location)[1]) = 1]">
		<xsl:sort select="location" />
		<b><xsl:value-of select="location" /></b><br />
		<xsl:for-each select="key('course-by-location', location)[count(.|key('date-by-course-location', concat(location,'-', course))[1]) = 1]">
			<xsl:sort select="course"/>
				<xsl:value-of select="course"/>
			<xsl:text> - </xsl:text> 
			<xsl:for-each select="key('date-by-course-location', concat(location,'-', course))">
				<xsl:value-of select="date"/>
				<xsl:text> </xsl:text> 
			</xsl:for-each>
			<br />
		</xsl:for-each>
		<br />		
	</xsl:for-each>
</xsl:template>

</xsl:stylesheet>

<tags>  xslt

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