You have to abandon the procedural mindset ;-) use modes.
try this:
<xsl:apply-template select="row[(_at_)year=2001]" mode="year_only"/>
<xsl:apply-template select="row[(_at_)year=2001]" mode="month"/>
<xsl:template match="row" mode="year_only">
<xsl:value-of select="@year"/><xsl:text>/</xsl:text>
</xsl:template>
<xsl:template match="row" mode="month">
<xsl:value-of select="@month"/><br/>
</xsl:template>
...just pseudo-code haven't tried it, but that's the hard part.
Grouping the years shouldn't be too hard.
simon
On Sunday, April 27, 2003, at 02:24 AM, Cheung Tin Po wrote:
hi expert,
sorry for the syntax error
<record>
<row year="2001" month="1"/>
<row year="2001" month="2"/>
<row year="2001" month="3"/>
<row year="2001" month="4"/>
<row year="2002" month="1"/>
<row year="2002" month="2"/>
<row year="2002" month="3"/>
<row year="2003" month="1"/>
<row year="2003" month="2"/>
</record>
From: "Cheung Tin Po" <timothy298(_at_)hotmail(_dot_)com>
Reply-To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
To: XSL-List(_at_)lists(_dot_)mulberrytech(_dot_)com
CC: timothy298(_at_)hotmail(_dot_)com
Subject: [xsl] Show YEAR once if the YEAR same as preious element by
xsl
Date: Sun, 27 Apr 2003 03:57:44 +0000
hi expert,
i get the following XML file
<record>
<row year="2001" month="1">
<row year="2001" month="2">
<row year="2001" month="3">
<row year="2001" month="4">
<row year="2002" month="1">
<row year="2002" month="2">
<row year="2002" month="3">
<row year="2003" month="1">
<row year="2003" month="2">
</record>
i want to transform it to html with the following result
2001/1
2
3
4
2002/1
2
3
2003/1
2
as XSLT cannot store the variable so i cannot think any method to
should the result.
:<
From Timothy
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