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August 20th, 2003, 04:57 AM

Dan Timis

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Join Date: Aug 2003

Posts: 2

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Help with XSLT

Hi,

I understand that XSLT has no side effects. I'm trying to do something that would be very simple in a procedural language and I just can't figure out how to do it.

I have an XML file, something like this:

<root>
<data tag="a">1</data>
<data tag="b">2</data>
<data tag="a">3</data>
<data tag="b">4</data>
</root>

I would like to sort the "data" elements by tag, but print the tag only once. Something like this:

a:
- 1
- 3
b:
- 2
- 4

I can of course do this:

a:
- 1
a:
- 3
b:
- 2
b:
- 4

I can also print just the first tag using <xsl:if test="position() = 1"> I tried using [position() - 1] to compare it with the previous one when position() != 1, but it didn't work.

How can I do this?

Thanks,

Dan Timis

August 21st, 2003, 08:50 AM

raman_27

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Location: Chennai(Madras)

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answer to help with XSLT

hi,
The solution for the above said problem
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"

xmlns:fo="http://www.w3.org/1999/XSL/Format">
<xsl:key name="dist" match="@tag" use="."/>

<xsl:template match="/">
<html>
<head>
<title>Distinct</title>
</head>

<body>
<p>

</p>

<xsl:apply-templates/>

</body>

</html>

</xsl:template>



<xsl:template match="root">

<table border="0" cellPadding="1" cellSpacing="1" width="50%" align="left">

<xsl:for-each select="data[count(@tag | key('dist', @tag)[1]) = 1]">

<tr>
<td><xsl:value-of select="@tag"/></td>
</tr>

</xsl:for-each>

</table>

<table border="0" cellPadding="1" cellSpacing="1" width="100%" align="center">
<xsl:for-each select="data">

<tr><td><xsl:value-of select="."/></td></tr>

</xsl:for-each>

</table>

</xsl:template>

</xsl:stylesheet>

August 21st, 2003, 04:17 PM

Dan Timis

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Join Date: Aug 2003

Posts: 2

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Thanks raman_27,

The key part was:

select="data[count(@tag | key('dist', @tag)[1]) = 1]"

I understand that key('dist', @tag) will have a node set with all the @tag's matching @tag of <data> and that key('dist', @tag)[1] will have just the first @tag.

I suppose that this:

@tag | key('dist', @tag)[1]

means a union of two sets and that the nodes in each set are references not values. Correct?

So for the first <data> @tag and key('dist', @tag)[1] are the same and the count is 1, while for the third <data> @tag and key('dist', @tag)[1] are different (even though their value is the same) and therefore count is 2.

Correct?

Thanks again,

Dan

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