And can you say me why in the following xsl the processor doesn't step
into this template:
<xsl:template match="/" mode="main">
<xsl:apply-templates select="$sorted" mode="main"/>
</xsl:template>
The hole xsl:
<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
xmlns:msxsl="urn:schemas-microsoft-com:xslt">
<xsl:output media-type="xml"/>
<xsl:variable name="sorted">
<xsl:apply-templates mode="copy" select="data/item"/>
</xsl:variable>
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()">
<xsl:sort select="@sector"/>
<xsl:sort select="@subsector"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
<xsl:template match="/" mode="main">
<xsl:apply-templates select="$sorted" mode="main"/>
</xsl:template>
<xsl:template match="item" mode="main">
<xsl:for-each select="item">
<xsl:value-of select="."/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
Mon, 25 Jul 2005 16:18:45 +0200, Geert Josten
<Geert(_dot_)Josten(_at_)daidalos(_dot_)nl> писал(а):
You will need to use a node-set extension function in XSLT 1.0...
Cheers,
Geert
John Robb wrote:
Can you please help me?I want to save the sorted tree in a variable.
Whats wrong with the following xsl:
<xsl:variable name="sorted">
<xsl:apply-templates mode="copy" select="data/item"/>
</xsl:variable>
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()" mode="copy">
<xsl:sort select="@sector"/>
<xsl:sort select="@subsector"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
And then I want to perform smth. like this:$sorted/item/@sector.
How is it possible to realize such a thing?
---
GSM.By.COM - интернет-магазин сотовых телефонов стандарта GSM,
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