XML to XSL conversion. Need to keep track of rows based on an if condition

[Cindy Jones]

I am using XSL to convert an XML file to a csv file.
The XML file in the format below:

<file>
<record>
<name>Johnny</name>
<age>32</age>
</record>
<record>
<name>Jimmy</name>
<age>5</age>
</record>
<record>
<name>Baby Moon</name>
<age>1</age>
</record>
<record>
<name>Lisa</name>
<age>25</age>
</record>
</file>
My xsl file should be implemented such that all records where age <20 are ignored.

<xsl:stylesheet
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="2.0">
<xsl:output method="html"/>
<xsl:template match="/">
<xsl:for-each select="file/record">
<xsl:if test="age > 20">
<xsl:value-of select="position()"/>
<xsl:value-of select="name"/>
<xsl:text>,</xsl:text>
<xsl:value-of select="age"/>
<xsl:text> </xsl:text>
</xsl:if>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>

I need to keep track of the records that have been picked and keep a sequence going. Following is the expected result.

1,Johnny,32
2,Lisa,25

but I am getting
1,Johnny,32
4,Lisa,25

How can I fix this?

Thanks!
[ May 13, 2008: Message edited by: Cindy Jones ]

[Paul Clapham]

The position() function returns the element's position in the nodelist selected by xsl:for-each. That's why you're getting 1 and 4, because you consider only positions 1 and 4. What I would do is this:

<xsl:for-each select="file/record[age &gt; 20]">

and drop the xsl:if element because you're just selecting the nodes you are interested in.

[Cindy Jones]

Thanks! Let me check if it solves my real problem I had posted a simplistic version.